Question: As shown in the figure, lines PA and PB are two tangents to ⊙O, with A and B being the points of tangency. If ∠APB = 120° and the radius of ⊙O is 10, then the length of chord AB is ()

Answer: 10

1. Given PA and PB are two tangents to ⊙O, and A and B are the points of tangency respectively, connect OA and OB.

2. According to the property of tangents, OA is perpendicular to PA, OB is perpendicular to PB, therefore ∠OAP = 90° and ∠OBP = 90°.

3. It is given in the problem that ∠APB = 120°.

4. In quadrilateral AOBP, the sum of angles is 360°, so ∠OAP + ∠OBP + ∠APB + ∠AOB = 360°.

5. Substituting the given conditions, we get 90° + 90° + 120° + ∠AOB = 360°, therefore ∠AOB = 60°.

6. Since OA = OB = 10, because ∠AOB = 60°, according to the theorem of equilateral triangle determination (60-degree angle in an isosceles triangle), △AOB is an equilateral triangle.

7. In an equilateral triangle, all sides are equal, therefore the chord AB = OA = 10.

切线; 等腰三角形; 圆周角; 弦; 余弦

Definition of Tangent to a Circle

In the figure of this problem, <note>circle O and line PA have only one common point A, which is called the point of tangency</note>. Therefore, <note>line PA is the tangent to circle O</note>. Similarly, <note>circle O and line PB have only one common point B, which is called the point of tangency</note>. Therefore, <note>line PB is the tangent to circle O</note>.

Definition of Isosceles Triangle

In the figure of this problem, in triangle OAB, <note>side OA and side OB are equal</note>, therefore <note>triangle OAB is an isosceles triangle</note>.

An isosceles triangle with one interior angle measuring 60 degrees is an equilateral triangle.

<note>△OAB is an isosceles triangle</note>, with <note>sides OA and OB being equal</note>, and there is an internal angle of 60°, i.e., <note>∠AOB=60°</note>. According to the Equilateral Triangle Identification Theorem, if an isosceles triangle has an internal angle of 60°, then the lengths of its three sides are equal, and all three internal angles are 60°. Therefore, it can be determined that <note>△OAB is an equilateral triangle</note>.

Definition of Equilateral Triangle

An equilateral triangle is a triangle in which all three sides have the same length, and all interior angles are equal (each measuring 60°).

<note>In the figure of this problem, triangle OAB is an equilateral triangle</note>. <note>The lengths of side OA, side OB, and side AB are equal</note>, and <note>the degrees of angle OAB, angle OBA, and angle AOB are equal</note>, all being 60°. Among them, <note>OA = OB = 10</note>, and <note>∠AOB = 60°</note>, therefore <note>the length of chord AB is equal to 10</note>.

A tangent line to a circle is perpendicular to the radius that passes through the point of tangency.

<note>In circle O, points A and B are the points of tangency where lines PA and PB touch the circle</note>, <note>line segments OA and OB are the radii of the circle</note>. According to the <note>property of the tangent line to a circle</note>, <note>tangent PA is perpendicular to the radius OA at the point of tangency A</note>, <note>tangent PB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠OAP = 90°</note> and <note>∠OBP = 90°</note>.

Triangle Angle Sum Theorem

In the figure of this problem, in triangle OAP, <note>angle OAP, angle OPA, and angle PAO are the three interior angles of triangle OAP</note>, according to the Triangle Angle Sum Theorem, <note>angle OAP + angle OPA + angle PAO = 180°</note>; in triangle OBP, <note>angle OBP, angle BPO, and angle POB are the three interior angles of triangle OBP</note>, according to the Triangle Angle Sum Theorem, <note>angle OBP + angle BPO + angle POB = 180°</note>; in triangle OAB, <note>angle OAB, angle AOB, and angle BOA are the three interior angles of triangle OAB</note>, according to the Triangle Angle Sum Theorem, <note>angle OAB + angle AOB + angle BOA = 180°</note>.

In any quadrilateral, the sum of the four interior angles is 360°.

In the quadrilateral <note>OAOBP</note>, <note>angle OAP, angle OBP, angle APB, and angle AOB are the four interior angles of the quadrilateral</note>. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>angle OAP + angle OBP + angle APB + angle AOB = 360°</note>.


Question: As shown in the figure, line AB is tangent to ⊙O at point A, the radius of ⊙O is 2, and ∠OBA=30°. Find the length of AB.

Answer: 2√{3}

1. Given that line AB is tangent to ⊙O at point A, according to the property that the tangent is perpendicular to the radius, we get ∠OAB=90°, therefore, triangle OAB is a right triangle.

2. Given that ∠OBA=30°, and OA=2.

3. According to the tangent function, we can get tan∠OBA = OA / AB.

4. Substituting the given conditions, we get tan 30° = 2 / AB.

5. Since tan 30° = √3 / 3, substituting it into the equation, we get √3 / 3 = 2 / AB.

6. Solving this equation, we get AB = 2√3.

7. Through the above reasoning, the final answer is 2√3.

切线; 直角三角形; 圆; 正弦; 线段

Definition of Right Triangle

A triangle is classified as a right triangle if it has one interior angle measuring exactly 90 degrees.

<note>In the figure of this problem</note>, <note>in triangle OAB, angle OAB is a right angle (90 degrees)</note>, therefore <note>triangle OAB is a right triangle</note>. <note>Side OA and side AB are the legs</note>, <note>side OB is the hypotenuse</note>.

Definition of Tangent to a Circle

A line is a tangent to a circle if and only if it has exactly one point of intersection with the circle. This point of intersection is called the point of tangency.

<note>Circle O</note> and <note>line AB</note> have only one <note>common point A</note>, which is called the <note>point of tangency</note>. Therefore, <note>line AB</note> is the <note>tangent to circle O</note>.

Property of the Tangent Line to a Circle

In the figure of this problem, in circle O, point A is <note>the tangent point of line AB with the circle</note>, <note>segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line AB is perpendicular to the radius OA passing through the tangent point A</note>, that is, <note>∠OAB=90°</note>.

Definition of Tangent Function

In a right-angled triangle, the tangent value of an acute angle is equal to the length of the opposite side divided by the length of the adjacent side.

<note>In the figure of this problem, in the right triangle OAB, angle ∠OBA is an acute angle</note>, <note>side OA is the opposite leg of angle ∠OBA</note>, <note>side AB is the adjacent leg of angle ∠OBA</note>, so <note>the tangent value of angle ∠OBA is equal to the length of side OA divided by the length of side AB</note>, that is, <note>tan(∠OBA) = OA / AB</note>.


Question: As shown in the figure, points A, B, and C are three points on ⊙O. Line CD is tangent to ⊙O at point C. If ∠DCB=40°, then the degree of ∠CAB is ()

Answer: 40°

1. Given that line CD is tangent to ⊙O at point C, according to the property of the tangent to a circle, we get ∠OCD=90°.

2. From the given condition ∠DCB=40°, we get ∠OCB=∠OCD-∠DCB=90°-40°=50°.

3. Since CO=BO, according to the property of an isosceles triangle, we get ∠OBC=50°.

4. Through the triangle angle sum theorem, we get ∠COB=180°-∠OBC-∠OCB=180°-50°-50°=80°.

5. According to the inscribed angle theorem, we get ∠CAB=1/2∠COB.

6. Calculating, we get ∠CAB=1/2×80°=40°.

7. Through the above reasoning, we finally get the answer as 40°.

圆; 切线; 圆周角; 三角形的外角; 等腰三角形

<note>In circle O</note>, <note>point C is the point of tangency of line CD with the circle</note>, <note>line segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OC at the point of tangency C</note>, i.e., <note>∠OCD=90°</note>.

Definition of Isosceles Triangle

A triangle is an isosceles triangle if and only if it has at least two congruent (≅) sides.

<note>In triangle OCB, sides CO and BO are equal</note>, therefore triangle OCB is an isosceles triangle.

Properties of Isosceles Triangle

In an isosceles triangle, the angles opposite the equal sides are equal, and the sides opposite the equal angles are congruent.

<note>In the figure of this problem, in the isosceles triangle OBC, sides OB and OC are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the angles opposite the equal sides are equal, i.e., angle OBC = angle OCB</note>.

Triangle Angle Sum Theorem

The sum of the interior angles of any triangle is 180°.

<note>Angle COB, angle OCB, and angle OBC are the three interior angles of triangle COB</note>, according to the Triangle Angle Sum Theorem, <note>angle COB + angle OCB + angle OBC = 180°</note>.

Inscribed Angle Theorem

In a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

In the figure of this problem, in circle O, points A, B, and C are on the circle, the central angle corresponding to <note>minor arc CB</note> is ∠COB, and the inscribed angle is ∠CAB. According to the Inscribed Angle Theorem, ∠CAB is equal to half of the central angle ∠COB corresponding to <note>minor arc CB</note>, i.e., ∠CAB = 1/2 ∠COB.

Definition of Central Angle

An angle formed by two radii connecting two points on the circumference of a circle to the center of the circle is called a central angle.

Original text: <note>在本题图中，圆O中，点C和点B是圆上的两点，圆心是点O</note>。连线<note>OC和OB组成的角∠COB</note>称为圆心角。

Translation: <note>In the figure of this problem, in circle O, point C and point B are two points on the circle, and the center of the circle is point O</note>. The angle formed by the lines <note>OC and OB, ∠COB</note> is called the central angle.

Definition of Inscribed Angle

An inscribed angle is an angle whose vertex lies on the circumference of a circle and whose sides intersect the circle at two points.

In <note>circle O</note>, <note>the vertex of angle CAB (point A) is on the circumference</note>, <note>and the two sides of angle CAB intersect circle O at points C and B, respectively</note>. Therefore, <note>angle CAB is an inscribed angle</note>.


Question: As shown in the figure, ⊙O is the circumcircle of Rt△ABC, ∠ACB=90°, ∠A=25°, a tangent to ⊙O is drawn through point C and intersects the extension of AB at point D. Then the degree measure of ∠D is ()

Answer: 40°

1. Connect OC. According to the inscribed angle theorem, ∠AOB = 2∠ACB. From the given conditions, it can be determined that ∠ACB = 90°, therefore ∠AOB = 2 * 90° = 180°.

2. According to the given conditions, ∠A = 25°, through the inscribed angle theorem, ∠COD = 2∠A, therefore ∠COD = 2 * 25° = 50°.

3. Since CD is the tangent to ⊙O, according to the tangent-radius theorem, OC is perpendicular to CD, i.e., ∠OCD = 90°.

4. In △COD, according to the angle sum theorem, ∠COD + ∠OCD + ∠D = 180°. Substituting the specific known conditions, we get 50° + 90° + ∠D = 180°.

5. By simplifying, ∠D = 180° - 140° = 40°.

n. Through the above reasoning, the final answer is 40°.

圆; 圆周角; 切线; 直角三角形; 三角形的外角

<note>In the figure of this problem, in circle O, the vertex C of angle ACB is on the circumference</note>, <note>the two sides of angle ACB intersect circle O at points A and B respectively</note>. Therefore, <note>angle ACB is an inscribed angle</note>.

Inscribed Angle Theorem

In the figure of this problem, <note>in circle O, points A, B, and C are on the circle</note>, the central angle corresponding to arc ACB and arc AB is <note>∠AOB</note>, and the inscribed angle is <note>∠ACB</note>. According to the Inscribed Angle Theorem, ∠ACB is equal to half of the central angle ∠AOB corresponding to arc AB, that is, <note>∠ACB = 1/2 ∠AOB</note>. In circle O, points A, B, and C are on the circle, the central angle corresponding to <note>minor arc BC</note> is <note>∠BOC</note>, and the inscribed angle is <note>∠BAC</note>. According to the Inscribed Angle Theorem, ∠BAC is equal to half of the central angle ∠BOC corresponding to arc AB, that is, <note>∠BAC = 1/2 ∠BOC</note>.

In the circle <note>⊙O</note>, <note>point C is the point of tangency of line CD with the circle</note>, and <note>segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OC at the point of tangency C</note>, that is, <note>∠OCD=90°</note>.

Triangle Angle Sum Theorem

In <note>△COD</note>, combining the <note>Triangle Angle Sum Theorem</note>, we have <note>∠COD + ∠OCD + ∠D = 180°</note>. According to the given conditions, <note>∠COD = 50°</note>, <note>∠OCD = 90°</note>, therefore <note>50° + 90° + ∠D = 180°</note>, finally solving for <note>∠D = 40°</note>.

Definition of Central Angle

In the figure of this problem, in circle O, <note>point C and point B are two points on the circle</note>, <note>the center of the circle is point O</note>. <note>The angle ∠COB formed by the lines OC and OB is called the central angle</note>.


Question: As shown in the figure, segment AB is the diameter of ⊙O, chord CD is perpendicular to AB, ∠CAB=20°, then ∠BOD equals ()

Answer: 40°

1. Given CD is perpendicular to AB, ∠A=20°, in the right triangle ACE, according to the property of complementary acute angles in a right triangle, ∠C=70°.

2. According to the inscribed angle theorem, ∠AOD=2*∠C=140°.

4. Since ∠AOD and ∠BOD are adjacent supplementary angles, ∠BOD=180°-∠AOD=40°.

圆; 弦; 垂线; 圆心角; 圆周角

Definition of Diameter

A diameter is a line segment that passes through the center of a circle and has its endpoints on the circle. It is the longest chord of the circle, with a length equal to twice the radius of the circle.

<note>Line segment AB is the diameter of circle O</note>, connecting <note>the center O and points A and B on the circumference</note>, with a length equal to 2 times the radius, that is, <note>AB = 2 * OA</note>.

Definition of Central Angle

In the figure of this problem, in circle <note>O</note>, point <note>A</note> and point <note>D</note> are two points on the circle, and the center of the circle is point <note>O</note>. The angle <note>∠AOD</note> formed by the lines <note>OA</note> and <note>OD</note> is called the central angle.

Definition of Inscribed Angle

In <note>circle O</note>, <note>the vertex A of angle ACD is on the circumference</note>, <note>the two sides of angle ACD intersect circle O at points A and D</note>. Therefore, <note>angle ACD is an inscribed angle</note>.

Inscribed Angle Theorem

In <note>circle O</note>, <note>points A, D, and C are on the circle</note>, <note>the central angle corresponding to arc AD is ∠AOD, and the inscribed angle is ∠ACD</note>. According to the Inscribed Angle Theorem, <note>∠ACD is equal to half of the central angle ∠AOD corresponding to arc AD</note>, that is, <note>∠ACD = 1/2 ∠AOD</note>.

Complementary Acute Angles in a Right Triangle

In a right triangle, the sum of the two non-right angles is 90°.

In the right triangle ACE, angle AEC is a right angle (90 degrees), angle A and angle C are the two acute angles other than the right angle, according to the complementary acute angles property of a right triangle, angle A and angle C add up to 90 degrees, that is 20° + ∠C = 90°.


Question: As shown in the figure, PA and PB are tangent to ⊙O at points A and B respectively. Point E is a point on ⊙O, and ∠AEB=60°. Then ∠P=()

Answer: 60°

1. Given that PA and PB are tangent to ⊙O at points A and B respectively, and point E is a point on ⊙O, ∠AEB=60°.

2. Connect OA and OB. According to the inscribed angle theorem, the central angle ∠AOB is twice the inscribed angle ∠AEB, so ∠AOB=2∠AEB=2×60°=120°.

3. Since the tangent is perpendicular to the radius, ∠OAP=90° and ∠OBP=90°.

4. In quadrilateral OAPB, the sum of the interior angles is 360°, with two angles ∠OAP and ∠OBP both being 90°.

5. Therefore, ∠P + ∠AOB = 180° (quadrilateral interior angle sum theorem).

6. Substituting the given conditions, we get ∠P + 120°= 180°.

7. Solving this, we find ∠P=180°-120°=60°.

##8##. Through the above reasoning, the final answer is ∠P=60°.

圆; 切线; 圆周角; 等腰三角形

Definition of Inscribed Angle

In the figure of this problem, in circle O, <note>the vertex E of angle AEB is on the circumference</note>, <note>the two sides of angle AEB intersect circle O at points A and B, respectively</note>. Therefore, angle AEB is an inscribed angle.

Definition of Central Angle

In <note>circle O</note>, point <note>A and point B are two points on the circle</note>, and the <note>center of the circle is point O</note>. The angle formed by the lines <note>OA and OB</note>, <note>∠AOB</note>, is called the central angle.

Inscribed Angle Theorem

In the figure of this problem, in circle O, points A, B, and E are on the circle, <note>the central angle corresponding to arc AB and arc BE is ∠AOB</note>, <note>the inscribed angle is ∠AEB</note>. According to the Inscribed Angle Theorem, <note>∠AEB is equal to half of the central angle ∠AOB corresponding to arc AB</note>, that is, <note>∠AEB = 1/2 ∠AOB</note>.

<note>In the diagram of this problem, in circle O, point A is the point of tangency between line PA and the circle, and segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent PA is perpendicular to the radius OA at the point of tangency A</note>, that is, <note>∠OAP=90°</note>. Similarly, <note>point B is the point of tangency between line PB and the circle, and segment OB is the radius of the circle</note>, so <note>the tangent PB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠OBP=90°</note>.

In <note>quadrilateral OAPB</note>, <note>angle OAP</note>, <note>angle OBP</note>, <note>angle AOB</note>, and <note>angle P</note> are the four interior angles of the quadrilateral. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>angle OAP + angle OBP + angle AOB + angle P = 360°</note>.


Question: Line AB is tangent to ⊙O at point A, as shown in the figure. If ∠OBA=60° and AB=1, then the radius of ⊙O is ()

Answer: √{3}

1. Given that line AB is tangent to circle O at point A, according to the property of the tangent to a circle, we get that OA is perpendicular to AB, so ∠OAB=90°. According to the definition of a right triangle, we get that △OBA is a right triangle.

2. Since ∠OBA=60° and OA is perpendicular to AB, according to the triangle angle sum theorem, in △OBA, ∠BOA=180°-∠OBA-∠OAB=180°-60°-90°=30°.

3. According to the property of a 30°-60°-90° triangle, side AB is the side opposite the 30° angle BOA and is half of the hypotenuse OB. Given that AB=1, we have OB=2*AB=2.

4. Since OA is perpendicular to AB and OA is the perpendicular distance from O to AB, in the right triangle OAB, according to the Pythagorean theorem, we get OA=√(OB^2-AB^2)=√(2^2-1^2)=√3.

5. In summary, the radius of circle O is √3.

切线; 直角三角形; 圆

Definition of Tangent to a Circle

<note>Circle O</note> and <note>line AB</note> have exactly one <note>common point A</note>, which is called the <note>point of tangency</note>. Therefore, <note>line AB</note> is the <note>tangent</note> to <note>circle O</note>.

Definition of Right Triangle

In the figure of this problem, in triangle OAB, <note>angle OAB is a right angle (90 degrees)</note>, therefore triangle OAB is a right triangle. <note>Side OA and side AB are the legs</note>, <note>side OB is the hypotenuse</note>.

Original: <note>In circle O, point A is the point of tangency between line AB and the circle</note>, <note>segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line AB is perpendicular to the radius OA at the point of tangency A</note>, that is, <note>∠OAB=90°</note>.

Properties of 30°-60°-90° Triangle

In a 30°-60°-90° triangle, the side opposite the 30-degree angle (the shorter leg) is half the length of the hypotenuse. The side opposite the 60-degree angle (the longer leg) is √3 times the length of the shorter leg.

In the figure of this problem, in <note>30°-60°-90° triangle OAB</note>, <note>angle BOA is 30 degrees</note>, <note>angle OBA is 60 degrees</note>, <note>angle OAB is 90 degrees</note>. <note>Side OB is the hypotenuse</note>, <note>side AB is the side opposite the 30-degree angle</note>, <note>side OA is the side opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle, <note>side AB is equal to half of side OB</note>, <note>side OA is equal to √3 times side AB</note>. That is: <note>AB=1/2*OB</note>, <note>OA= AB*√3</note>.

Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the two legs.

<note>In the right triangle OAB, angle OAB is a right angle (90 degrees)</note>, <note>sides OA and AB are the legs, and side OB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OB^2 = OA^2 + AB^2</note>.

Triangle Angle Sum Theorem

<note>In triangle OAB, angle OAB, angle OBA, and angle BOA are the three interior angles of triangle OAB</note>, according to the Triangle Angle Sum Theorem, <note>angle OAB + angle OBA + angle BOA = 180°</note>.


Question: As shown in the figure, PA and PB are tangent to ⊙O at points A and B respectively. ∠P = 70°, then ∠C is ()

Answer: 55°

1. Given that line PA and PB are tangent to circle O at points A and B respectively, according to the perpendicular tangent theorem, the tangent to circle O is perpendicular to the radius passing through the tangent point. Connecting OA and OB, we get OA perpendicular to PA and OB perpendicular to PB.

####

##2.## In quadrilateral PAOB, ##the sum of the interior angles of the quadrilateral is 360°.##

##3.## Because OA is perpendicular to PA and OB is perpendicular to PB, therefore ∠PAO=90°, ∠PBO=90°.

##4.## Since ∠P=70°, it can be concluded that ∠AOB = 180° - ∠P = 110°.

##5.## Point C is on circle O, therefore according to the inscribed angle theorem, the inscribed angle ∠ACB corresponding to the central angle ∠AOB should be half of ∠AOB.

##6.## Since ∠AOB=110°, so ∠ACB = ∠AOB / 2 = 55°.

##7.## Through the above reasoning, the final answer is ∠ACB=55°.

圆; 切线; 普通三角形; 圆周角

Definition of Tangent to a Circle

In the figure of this problem, <note>circle O and line PA have only one common point A, which is called the point of tangency</note>. Therefore, <note>line PA is the tangent to circle O</note>. Similarly, <note>circle O and line PB have only one common point B, which is called the point of tangency</note>. Therefore, <note>line PB is the tangent to circle O</note>.

In the figure of this problem, <note>quadrilateral PAOB</note>, <note>∠P, ∠A, ∠B, ∠AOB are the four interior angles of the quadrilateral</note>, according to the theorem, the sum of the four interior angles is 360°, that is, <note>∠P+∠A+∠B+∠AOB=360°</note>.

<note>In circle O, point A is the point of tangency between line PA and the circle</note>, <note>line segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PA is perpendicular to the radius OA at the point of tangency A, i.e., ∠OAP=90 degrees</note>. Similarly, <note>point B is the point of tangency between line PB and the circle</note>, <note>line segment OB is the radius of the circle</note>, <note>the tangent line PB is perpendicular to the radius OB at the point of tangency B, i.e., ∠OBP=90 degrees</note>.

Inscribed Angle Theorem

In the figure of this problem, in circle O, points A, B, and C are on the circle, <note>the central angle corresponding to arc AB and arc BA is ∠AOB</note>, <note>the inscribed angle is ∠ACB</note>. According to the Inscribed Angle Theorem, <note>∠ACB is equal to half of the central angle ∠AOB corresponding to arc AB</note>, that is, <note>∠ACB = 1/2 ∠AOB</note>.


Question: As shown in the figure, AB is the diameter of the semicircle, point O is the center of the circle, point C is a point on the extension line of AB, and CD is tangent to the semicircle at point D. If AB=6 and CD=4, then the value of sin∠C is ()

Answer: \frac{3}{5}

1. Given that AB is the diameter of the semicircle and AB=6, according to the definition of diameter, the radius OD=3.

2. Given that CD is tangent to the semicircle at point D, according to the property of the tangent line, ∠CDO=90°.

3. According to the definition of a right triangle and the Pythagorean theorem (the square of the hypotenuse of a right triangle is equal to the sum of the squares of the two legs), we have OC=√(CD^2 + OD^2) = √(4^2 + 3^2) = 5.

4. In the right triangle CDO, according to the definition of the sine function, sin(∠C) = opposite side OD / hypotenuse OC = 3 / 5.

5. Through the above reasoning, we finally obtain the value of sin∠C as 3/5.

切线; 直角三角形; 正弦; 圆; 圆心角

Definition of Diameter

<note>Segment AB is the diameter of the semicircle</note>, connecting <note>the center O and points A and B on the circumference</note>, with a length of <note>twice the radius, i.e., AB = 2 * radius</note>.

Definition of Radius

A radius is a line segment joining the center of a circle to any point on the circumference of the circle.

In the figure of this problem, in circle O, <note>point O is the center of the circle</note>, <note>point D is any point on the circle</note>, <note>line segment OD is the line segment from the center of the circle to any point on the circle</note>, therefore <note>line segment OD is the radius of the circle</note>. According to <note>AB is the diameter</note> and AB=6, <note>the radius OD is equal to half of the diameter</note>, that is, <note>OD=3</note>.

Definition of Right Triangle

In the figure of this problem, in triangle CDO, <note>angle ∠CDO is a right angle (90 degrees)</note>, therefore <note>triangle CDO is a right triangle</note>. <note>Side CD and side OD are the legs</note>, <note>side CO is the hypotenuse</note>.

Definition of Sine Function

In a right-angled triangle, the sine of an acute angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse.

In the figure of this problem, in the right triangle CDO, <note>angle ∠DCO is an acute angle</note>, <note>side OD is the opposite side of angle ∠DCO</note>, <note>side OC is the hypotenuse</note>. According to the definition of the sine function, <note>the sine of angle ∠DCO is equal to the ratio of the opposite side OD to the hypotenuse OC</note>, that is, <note>sin(∠DCO) = OD / OC = 3 / 5</note>.

Pythagorean Theorem

In the figure of this problem, <note>triangle CDO is a right triangle</note>, <note>∠CDO = 90°</note>. <note>Side OD and CD are the legs</note>, <note>side OC is the hypotenuse</note>, according to the Pythagorean Theorem, <note>we have OC^2 = OD^2 + CD^2</note>, that is <note>OC = √(3^2 + 4^2) = 5</note>.

A tangent line to a circle is perpendicular to the radius that passes through the point of tangency.

<note>In circle O, point D is the point of tangency of line CD with the circle</note>, <note>segment OD is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OD at the point of tangency D, that is, ∠CDO = 90 degrees</note>.


Question: As shown in the figure, in △ABC, ∠BAC=70°, ∠ABC=45°, point O is the center of the circumcircle of △ABC, then ∠AOB equals ()

Answer: 130°

1. Given ∠BAC=70°, ∠ABC=45°, according to the triangle angle sum theorem, we get ∠ACB = 180° - ∠BAC - ∠ABC = 180° - 70° - 45° = 65°.

2. Since point O is the center of the circumcircle of triangle ABC, ∠AOB is the central angle. According to the central angle theorem, the central angle is equal to twice the inscribed angle that subtends the same arc, i.e., ∠AOB = 2∠ACB.

3. From the conclusion obtained in the previous step, we can get ∠AOB = 2 * 65° = 130°.

4. Through the above reasoning, the final answer is ∠AOB = 130°.

圆; 圆心角; 普通三角形; 圆周角; 弦

Triangle Angle Sum Theorem

In <note>triangle ABC</note>, <note>angle BAC, angle ABC, and angle ACB are the three interior angles of triangle ABC</note>. According to the Triangle Angle Sum Theorem, <note>angle BAC + angle ABC + angle ACB = 180°</note>. Given that <note>∠BAC = 70°</note> and <note>∠ABC = 45°</note>, it can be deduced that <note>∠ACB = 180° - 70° - 45° = 65°</note>.

Inscribed Angle Theorem

<note>Point O is the center of the circumcircle of △ABC</note>, <note>∠AOB is the central angle</note>. According to the Inscribed Angle Theorem, <note>the central angle ∠AOB is equal to twice the inscribed angle ∠ACB corresponding to the arc AC</note>, i.e., ∠AOB = 2∠ACB. Given <note>∠ACB=65°</note>, therefore <note>∠AOB = 2 * 65° = 130°</note>.


Question: In the figure above, PQRS is a rectangle. The area of triangle RST is 7, and PT=2/5*PS. What is the area of PQRS?

Answer: \frac{70}{3}

1. Given that PQRS is a rectangle and the area of triangle RST is 7. According to the properties of a rectangle, PQ = RS and QR = PS.

2. Let PS = b, then PT = 2/5 * PS = 2b/5. Since ST is the segment from point T on PS to point S, then ST = PS - PT = b - 2b/5 = 3b/5.

3. Using the triangle area formula, the area of triangle RST is S = 1/2 * RS * ST * sin∠RST. Since PQRS is a rectangle, RS = PQ and ∠RST is a right angle.

4. Since RS = PQ = a, according to the form S = 1/2 * RS * ST, we get 7 = 1/2 * a * (3b/5), solving: ab = 70/3.

5. The area of rectangle PQRS is PQ * PS = a * b, according to the solved expression for ab, we get the area of PQRS as (70/3).

6. Finally, the calculated area of rectangle PQRS is 70/3.

矩形; 线段; 普通三角形

Definition of Rectangle

A quadrilateral is called a rectangle if and only if each of its interior angles is a right angle (90°) and its opposite sides are both parallel (∥) and congruent (≅).

In the figure of this problem, quadrilateral PQRS is a rectangle, with its interior angles <note>∠PQR, ∠QRS, ∠RSP, ∠SPQ all being right angles (90 degrees)</note>, and <note>sides PQ and RS are parallel and equal in length</note>, <note>sides QR and PS are parallel and equal in length</note>.

Definition of Line Segment

A line segment is a portion of a straight line that includes two endpoints and all the points between them.

<note>A line segment PQ is a part of a straight line, including endpoint P and endpoint Q and all points between them.</note> A line segment QR is a part of a straight line, including endpoint Q and endpoint R and all points between them. <note>A line segment RS is a part of a straight line, including endpoint R and endpoint S and all points between them.</note> A line segment PS is a part of a straight line, including endpoint P and endpoint S and all points between them.

Definition of Triangle

A geometric figure is called a triangle if it is formed by three non-collinear points and the line segments connecting them.

<note>Triangle RST</note> is a geometric figure composed of <note>three non-collinear points R, S, T</note> and their connecting line segments <note>RS, ST, TR</note>. <note>Points R, S, T</note> are respectively the <note>three vertices</note> of the triangle, and <note>line segments RS, ST, TR</note> are respectively the <note>three sides</note> of the triangle.

Formula for the Area of a Rectangle

The area of a rectangle is equal to its length multiplied by its width.

In the <note>rectangle PQRS</note>, <note>side PQ and side PS are the length and width of the rectangle</note>, so <note>the area of the rectangle = PQ * PS</note>.

Area Formula of a Triangle

The area of any triangle is equal to its base multiplied by its height, divided by 2.

In the figure of this problem, in triangle RST, <note>side RS is the base</note>, <note>segment ST is the height</note>. According to the area formula of a triangle, the area of triangle RST is equal to <note>base RS multiplied by height ST and then divided by 2</note>, that is, <note>Area = (RS * ST) / 2</note>.


Question: As shown in the figure, segment AB is the diameter of ⊙O, points C and D are points on ⊙O. A tangent to ⊙O is drawn through point C and intersects the extension of AB at point E. If ∠E = 50°, then ∠CDB equals ()

Answer: 20°

1. Given that segment AB is the diameter of ⊙O, connect OC. Since the tangent to ⊙O through point C intersects the extension of AB at point E, according to the ##property of the tangent to the circle##, ∠OCE=90°.

2. Given ∠E=50°, then based on ∠OCE=90° and the ##sum of the interior angles of a triangle is 180°##, we get ∠COE=180°-∠OCE-∠E=180°-90°-50°=40°.

3. According to the inscribed angle theorem, the inscribed angle is half of its corresponding central angle, so ∠CDB=##1/2##∠COE.

4. Therefore, we get ∠CDB=##1/2##×40°=20°.

5. Through the above reasoning, the final answer is ∠CDB=20°.

圆; 圆周角; 切线; 直角三角形; 三角形的外角

Definition of Tangent to a Circle

A line is a tangent to a circle if and only if it has exactly one point of intersection with the circle. This point of intersection is called the point of tangency.

<note>Circle O</note> and <note>line CE</note> have exactly one common point <note>C</note>, and this common point is called the <note>point of tangency</note>. Therefore, <note>line CE</note> is the <note>tangent to circle O</note>.

Definition of Central Angle

In the diagram of this problem, <note>in circle O</note>, <note>point B and point C are two points on the circle</note>, <note>the center of the circle is point O</note>. <note>The angle ∠BOC formed by the lines OC and OB is called the central angle</note>.

Definition of Inscribed Angle

In the figure of this problem, in circle O, <note>the vertex D of angle CDB is on the circumference</note>, <note>the two sides of angle CDB intersect circle O at points C and B respectively</note>. Therefore, <note>angle CDB is an inscribed angle</note>.

<note>In circle O, point C is the tangent point of line CE and the circle</note>, <note>segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CE is perpendicular to the radius OC passing through the tangent point C</note>, that is, <note>∠OCE=90°</note>.

Triangle Angle Sum Theorem

The sum of the interior angles of any triangle is 180°.

<note>In triangle OCE, angle OCE, angle E, and angle COE are the three interior angles of triangle OCE</note>, according to the Triangle Angle Sum Theorem, <note>angle OCE + angle E + angle COE = 180°</note>.

Inscribed Angle Theorem

In a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

<note>In the diagram of this problem, in circle O, points C and D are on the circle, and the central angles corresponding to arc BC and arc BCA are ##∠COB##</note>, <note>the inscribed angles are ∠CDB## and ∠BAC##</note>. According to the Inscribed Angle Theorem, <note>∠CDB is equal to half of the central angle ##∠COB## corresponding to the arc CD it intercepts, i.e., ∠CDB = 1/2 ##∠COB##</note>.


Question: As shown in the figure, PA and PB are tangents to ⊙O, points A and B are points of tangency, AC is the diameter of ⊙O, given ∠P=50°, then the measure of ∠ACB is ()

Answer: 65°

1. Given PA and PB are tangents to ⊙O, points A and B are points of tangency, connect OA, OB, and OC.

2. By the property of the tangent to a circle, OA is perpendicular to PA, OB is perpendicular to PB, thus ∠OAP=∠OBP=90°.

3. In quadrilateral OAPB, the sum of the interior angles of the quadrilateral is 360°, then ∠AOB=360°-∠OAP-∠OBP-∠P=360°-90°-90°-50°=130°.

4. Because OB=OC, triangle OCB is an isosceles triangle, ∠OCB=∠OBC.

5. By the exterior angle theorem of the triangle, in triangle OCB, ∠AOB=∠OCB+∠OBC, thus ∠130°=2×∠OCB.

6. Solving this equation, we get ∠OCB=∠OBC=65°.

圆; 切线; 圆周角; 等腰三角形; 三角形的外角

Definition of Diameter

<note>AC is the diameter</note>, connecting <note>the center O and points A and C on the circumference</note>, with a length that is 2 times the radius, i.e., <note>AC = 2 * OA</note>.

Definition of Isosceles Triangle

A triangle is an isosceles triangle if and only if it has at least two congruent (≅) sides.

In the diagram of this problem, in triangle OCB, <note>side OB and side OC are equal</note>, therefore <note>triangle OCB is an isosceles triangle</note>.

Properties of Isosceles Triangle

In an isosceles triangle, the angles opposite the equal sides are equal, and the sides opposite the equal angles are congruent.

In the isosceles triangle <note>OBC</note>, <note>sides OB and OC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle OBC = angle OCB</note>.

In <note>circle O</note>, <note>points A and B are the points of tangency of lines PA and PB with the circle</note>, <note>segments OA and OB are the radii of the circle</note>. According to the property of the tangent line to a circle, <note>tangent PA is perpendicular to the radius OA at the point of tangency A</note>, <note>tangent PB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠OAP=90 degrees and ∠OBP=90 degrees</note>.

In any quadrilateral, the sum of the four interior angles is 360°.

In <note>quadrilateral OAPB</note>, <note>angle OAP</note>, <note>angle OBP</note>, <note>angle P</note>, and <note>angle AOB</note> are the four interior angles of the quadrilateral. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>angle OAP + angle OBP + angle P + angle AOB = 360°</note>.

Exterior Angle Theorem of Triangle

An exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

In the figure of this problem, in triangle OCB, <note>angle AOB is an exterior angle of the triangle</note>, <note>angle OCB and angle OBC are the two interior angles that are not adjacent to the exterior angle AOB</note>, according to the Exterior Angle Theorem of Triangle, <note>the exterior angle AOB is equal to the sum of the two non-adjacent interior angles OCB and OBC</note>, that is, <note>angle AOB = angle OCB + angle OBC</note>.

Definition of Exterior Angle of a Polygon

An exterior angle of a polygon is the angle formed between one side of the polygon and the extension of an adjacent side.

The original text: <note>One interior angle of polygon OBC is ∠BOC</note>, extending the adjacent sides of this interior angle <note>OC and OB</note> forms the <note>angle ∠AOB, which is called the exterior angle of the interior angle ∠BOC</note>.


Question: As shown in the figure, AB is the diameter of circle O, DB and DC are tangents to circle O at points B and C respectively. If ∠ACE=25°, then the degree of ∠D is ()

Answer: 50°

1. Connect BC. According to the problem statement, DB and DC are tangent to circle O at points B and C, respectively.

2. According to the property of tangents, the angle between the tangent and the radius is a right angle, so ∠DBO=90° and ∠DCO=90°.

####

##3##. The problem states that ∠ACE=25°, and since AB is the diameter, ##according to the inscribed angle theorem, the angle subtended by the diameter is 90°, thus ∠ACB=90°.##

##4. Therefore, ∠ACE+∠ACO=∠BCO+∠ACO=90°, so ∠ACE=∠BCO.##

5. According to the problem statement, ##OC and OB are radii of circle O, hence OC=OB, so triangle OBC is an isosceles triangle, and ∠OBC=∠BCO=∠ACE,## ∵∠ACE=25°, ∴∠ABC=25°.

####

##6##. According to the sum of interior angles theorem, in △ABC, ∠BAC+∠ABC+∠ACB=180°.

##7##. Substituting the known conditions, we get ∠BAC+25°+90°=180°, so ∠BAC=65°.

##8##. According to ##∠DBO=90°##, we get ∠DBC=90°-∠ABC.

##9##. Substituting the known conditions, we get ∠DBC=90°-25°=65°.

####

##10. According to the tangent-segment theorem, we get DB=DC, so ∠BDC=∠DCB.##

##11##. Since in triangle BDC, ##according to the sum of interior angles theorem##, we get ∠BDC=180°-2×∠DBC.

##12##. Substituting the known conditions, we get ∠BDC=180°-2×65°=50°.

##13##. Through the above reasoning, the final answer is ∠D=50°.

圆; 切线; 圆周角; 直角三角形; 线段

In the figure of this problem, in circle O, <note>point B and point C are the points where line DB and line DC are tangent to the circle</note>, <note>segment OB and segment OC are the radii of the circle</note>. According to the property of the tangent line to a circle, <note>tangent line DB is perpendicular to the radius OB at the point of tangency B, i.e., ∠DBO=90 degrees</note>; <note>tangent line DC is perpendicular to the radius OC at the point of tangency C, i.e., ∠DCO=90 degrees</note>.

Tangent-Segment Theorem

If two tangent segments are drawn to a circle from an external point, then the lengths of these tangent segments are equal. Moreover, the line segment connecting the center of the circle to the external point bisects the angle formed by the two tangent segments.

<note>Two tangents DB and DC are drawn from an external point D to the circle</note>, and their lengths are equal, i.e., <note>DB=DC</note>.

Definition of Inscribed Angle

An inscribed angle is an angle whose vertex lies on the circumference of a circle and whose sides intersect the circle at two points.

In the figure of this problem, in circle O, <note>the vertex C of angle ACB is on the circumference</note>, <note>the two sides of angle ACB intersect circle O at points A and B respectively</note>. Therefore, angle ACB is an inscribed angle.

Corollary to the Inscribed Angle Theorem 2: The Angle Subtended by the Diameter

The angle subtended by the diameter of a semicircle is a right angle; conversely, if an inscribed angle is 90°, then the chord subtending that angle is a diameter.

The original text: <note>The inscribed angle ACB is 90 degrees</note>, so <note>the chord AB it subtends is the diameter</note>.

Triangle Angle Sum Theorem

In the figure of this problem, in triangle ABC, <note>angle BAC</note>, <note>angle ABC</note>, and <note>angle ACB</note> are the three interior angles of triangle ABC. According to the Triangle Angle Sum Theorem, <note>angle BAC + angle ABC + angle ACB = 180°</note>. In triangle BDC, <note>angle D</note>, angle DBC, and <note>angle DCB</note> are the three interior angles of triangle BDC. According to the Triangle Angle Sum Theorem, <note>angle D</note> + angle DBC + <note>angle DCB</note> = 180°.

Properties of Isosceles Triangle

<note>In the figure of this problem, in the isosceles triangle BDC</note>, <note>side BD and side DC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>angle DBC = angle DCB</note>.

Definition of Isosceles Triangle

<note>In the diagram of this problem, in triangle BCD, side DB and side DC are equal</note>, therefore triangle BCD is an isosceles triangle.


Question: As shown in the figure, AB is the diameter of ⊙O, AC is tangent to ⊙O at A, BC intersects ⊙O at point D. If ∠C=70°, then the degree of ∠AOD is ()

Answer: 40°

1. Given that AC is the tangent to ⊙O, and AB is the diameter of ⊙O. According to the property of the tangent to a circle, AC is perpendicular to AB, i.e., ∠CAB=90°.

2. Given ∠C=70°, by the sum of the interior angles of a triangle, in △ABC, we get ∠ABC=20°.

3. According to the theorem of the inscribed angle, the inscribed angle is equal to half of the central angle corresponding to the arc it intercepts. ∠ABC is the inscribed angle of arc AD, and ∠AOD is the central angle of arc AD, so ∠AOD=2*∠ABC=40°.

4. Through the above reasoning, the final answer is ∠AOD=40°.

圆; 圆周角; 圆心角; 切线; 直角三角形

Definition of Tangent to a Circle

In the figure of this problem, circle O and line AC have only one common point A, which is called the point of tangency. Therefore, line AC is the tangent to circle O.

Triangle Angle Sum Theorem

In triangle ABC, the angles CAB, ABC, and BCA are the three interior angles of triangle ABC, according to the Triangle Angle Sum Theorem, angle CAB + angle ABC + angle BCA = 180°. Given that angle CAB = 90°, angle BCA = 70°, therefore angle ABC = 20°.

A tangent line to a circle is perpendicular to the radius that passes through the point of tangency.

In the figure of this problem, in circle O, point A is the <note>tangent point of line AC and the circle</note>, and the <note>line segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, the <note>tangent line AC is perpendicular to the radius OA passing through the tangent point A</note>, that is, <note>∠CAB=90°</note>.

Inscribed Angle Theorem

In <note>circle O</note>, <note>points A, B, and D are on the circle</note>, <note>the central angle corresponding to arc AD and arc ABD is ∠AOD</note>, and <note>the inscribed angle is ∠ABD</note>. According to the Inscribed Angle Theorem, <note>∠ABD is equal to half of the central angle ∠AOD corresponding to arc AD</note>, that is, <note>∠ABD = 1/2 ∠AOD</note>.

Definition of Central Angle

An angle formed by two radii connecting two points on the circumference of a circle to the center of the circle is called a central angle.

In the figure of this problem, <note>in circle O</note>, <note>point A and point D are two points on the circle</note>, <note>the center of the circle is point O</note>. <note>The angle ∠AOD formed by the lines OA and OD</note> is called the central angle.

Definition of Inscribed Angle

In the figure of this problem, in circle O, <note>the vertex of angle ABD (point B) is on the circumference</note>, <note>the two sides of angle ABD intersect circle O at points A and D respectively</note>. Therefore, <note>angle ABD is an inscribed angle</note>.


Question: In the figure above, three lines intersect at a point. What is the value of y?

Answer: 65

1. Since three lines intersect at a point, according to the theorem of vertical angles being equal, a pair of vertical angles formed by the intersection will also be equal.

2. In the figure, another pair of vertical angles is 50 degrees, denoted as ∠AEB and ∠CED. Therefore, the measure of ∠AEC is 130 degrees.

3. Denote the unknown angle y° and its vertical angle also as y°.

4. Since BC is a straight line and point E is on BC, according to the definition of supplementary angles, ∠AEB +∠AEC = 180 degrees, that is: ∠AEB + y° + y° = 180 degrees.

5. Combine to get: 180 degrees - 50 degrees = 2y. Therefore: 2y = 130 degrees.

6. Divide both sides of the equation by 2 to simplify the value of y, obtaining the formula: y = 65.

Definition of Vertical Angles

Vertical angles are the pair of opposite angles formed when two lines intersect. These angles are equal in measure.

In the figure of this problem, <note>three straight lines intersect at one point</note>, forming several pairs of vertical angles. According to the definition of vertical angles, <note>∠50° and its vertical angle are also 50°</note>, <note>y° and its vertical angle are also y°</note>. Since the angles of vertical angles are equal, <note>∠50°=∠50°</note>, <note>y°=y°</note>.

Definition of Linear Pair of Angles

Two angles are in a linear pair if they share a common side, and their other sides are extensions of one another in opposite directions.

In the diagram of this problem, <note>angle AEB and angle AEC share a common side AE</note>, <note>their other sides EB and EC are extensions in opposite directions</note>, so <note>angle AEB and angle AEC are adjacent supplementary angles to each other</note>.


Question: As shown in the figure, in △ABC, ∠B=20°, point O is a point on side BC, with O as the center and OB as the radius, a circle is drawn, intersecting side AB at point D. Connect CD, if CD is tangent to ⊙O, then the measure of ∠DCB is ()

Answer: 50°

1. Connect point O and point D. According to the problem statement, line CD is the tangent to the circle with center O and radius OB.

2. According to the property of the tangent to a circle, ∠ODC=90°.

3. Because OB=OD, therefore △OBD is an isosceles triangle.

4. In the isosceles triangle △OBD, ∠OBD=∠ODB.

5. From the given data in the problem, ∠B=20°, therefore ∠ODB=20°.

6. In △ODC, according to the triangle angle sum theorem, ∠DCB + ∠ODC + ∠DOC = 180°.

7. In △OBD, according to the exterior angle theorem, ∠DOC=∠B+∠ODB=40°.

8. Substituting the known angles, ∠DCB + 90° + 40° = 180°.

9. Therefore, solving gives ∠DCB = 180° - 90° - 40° = 50°.

10. Further verification shows the calculation steps and conclusion are correct, the final angle is 50°.

圆; 切线; 圆周角; 普通三角形; 等腰三角形

Definition of Tangent to a Circle

<note>Circle O and line CD have exactly one common point D</note>, this common point is called the point of tangency. Therefore, <note>line CD is the tangent to circle O</note>.

Definition of Isosceles Triangle

In the figure of this problem, in triangle OBD, <note>side OB and side OD are equal</note>, therefore <note>triangle OBD is an isosceles triangle</note>.

<note>In the figure of this problem, in circle O, point D is the point of tangency where line CD touches the circle, and segment OD is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OD at the point of tangency D, i.e., ∠ODC=90°</note>.

Triangle Angle Sum Theorem

<note>Angle DCB</note>, <note>Angle ODB</note>, and <note>Angle ODC</note> are the three interior angles of triangle <note>BCD</note>. According to the Triangle Angle Sum Theorem, <note>Angle DCB</note> + <note>Angle ODB</note> + <note>Angle ODC</note> = 180°.

Properties of Isosceles Triangle

<note>In the figure of this problem, in the isosceles triangle OBD, sides OB and OD are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the angles opposite the equal sides are equal, i.e., ∠OBD = ∠ODB</note>.

Exterior Angle Theorem of Triangle

<note>Angle BOD is an exterior angle of the triangle</note>, <note>Angle OBD and Angle ODB are the two interior angles not adjacent to the exterior angle COD</note>, according to the Exterior Angle Theorem of Triangle, <note>the exterior angle COD is equal to the sum of the two non-adjacent interior angles OBD and ODB</note>, that is, <note>Angle COD = Angle OBD + Angle ODB</note>.

In the figure of this problem, an interior angle of polygon OBD is <note>∠BOD</note>. The angle <note>∠COD</note> formed by extending the adjacent sides <note>OB</note> and <note>OD</note> of this interior angle is called the exterior angle of the interior angle <note>∠BOD</note>.


Question: As shown in the figure, line BC is tangent to ⊙O at point A, AD is a chord of ⊙O, connect OD, if ∠DAC = 50°, then the degree of ∠ODA is ()

Answer: 40°

1. Given that line BC is tangent to circle O at point A, according to the property of the tangent, the tangent is perpendicular to the radius at the point of tangency, thus OA is perpendicular to BC.

2. Therefore, angle OAC is a right angle, ∠OAC = 90°.

3. Since it is known that ∠DAC = 50°, ####, it follows that ∠OAD = ∠OAC - ∠DAC = 90° - 50° = 40°.

4. Because OA = OD (equal radii), triangle OAD is an isosceles triangle. According to the property of isosceles triangles, its base angles are equal, thus ∠ODA = ∠OAD = 40°.

5. Through the above reasoning, the final answer is 40°.

圆; 切线; 弦; 圆心角; 圆周角

Definition of Tangent to a Circle

<note>Circle O and line BC have exactly one common point A</note>, this common point is called the point of tangency. Therefore, <note>line BC is the tangent to circle O</note>.

Definition of Radius

A radius is a line segment joining the center of a circle to any point on the circumference of the circle.

<note>In circle O, point O is the center of the circle</note>, <note>point A and point D are any points on the circle</note>, <note>line segment OA and line segment OD are segments from the center to any point on the circle</note>, therefore <note>line segment OA and line segment OD are the radii of the circle</note>.

Definition of Chord

A chord is defined as a line segment that connects any two points on a circle.

In <note>circle O</note>, <note>point A and point D are any two points on the circle</note>, <note>line segment AD connects these two points</note>, so <note>line segment AD is a chord of circle O</note>.

Definition of Isosceles Triangle

<note>Side OA and side OD are equal (both are radii of circle O)</note>, therefore triangle OAD is an isosceles triangle.

<note>In the figure of this problem, in circle O, point A is the tangent point of line BC and the circle, and segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line BC is perpendicular to the radius OA passing through the tangent point A</note>, that is, <note>∠OAC=90 degrees</note>.

Properties of Isosceles Triangle

In the isosceles triangle <note>等腰三角形OAD</note>, <note>sides OA and OD are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠ODA = ∠OAD</note>.


Question: As shown in the figure, PA and PB are tangents to ⊙O, with points of tangency at A and B. If ∠OAB=30°, then the degree of ∠P is ()

Answer: 60°

1. Given ∠OAB=30°, since OA and OB are both radii of circle O, that is, OA=OB, triangle OAB is an isosceles triangle. According to the properties of isosceles triangles, we know ∠OAB=∠OBA=30°. Based on the triangle angle sum theorem, we get ∠AOB=180°-2×30°=120°.####

2. Since PA and PB are tangents to circle ⊙O, according to the properties of tangents to a circle, we get ∠OAP=∠OBP=90°.####

3. In quadrilateral OAPB, the sum of the interior angles is 360°, from which we can obtain ∠P = 360° - ∠OAP - ∠OBP - ∠AOB = 360° - 90° - 90° - 120° = 60°.

切线; 等腰三角形; 圆

In the circle <note>⊙O</note>, point <note>A</note> and point <note>B</note> are the points of tangency of the lines <note>PA</note> and <note>PB</note> with the circle, and the line segments <note>OA</note> and <note>OB</note> are the radii of the circle. According to the property of the tangent line to a circle, the tangent line <note>PA</note> is perpendicular to the radius <note>OA</note> at the point of tangency <note>A</note>, and the tangent line <note>PB</note> is perpendicular to the radius <note>OB</note> at the point of tangency <note>B</note>, i.e., <note>∠OAP=90°</note> and <note>∠OBP=90°</note>.

Triangle Angle Sum Theorem

In the figure of this problem, in triangle OAB, <note>angle OAB, angle OBA, and angle AOB are the three interior angles of triangle OAB</note>, according to the Triangle Angle Sum Theorem, <note>angle OAB + angle ABO + angle AOB = 180°</note>.

Properties of Isosceles Triangle

<note>In the isosceles triangle OAB, sides OA and OB are equal</note>. Therefore, according to the properties of an isosceles triangle, <note>the angles opposite the equal sides are equal, i.e., angle OAB = angle OBA</note>.

Definition of Isosceles Triangle

<note>The sides OA and OB are equal</note>, therefore <note>triangle OAB is an isosceles triangle</note>.

<note>Quadrilateral OABP</note>, <note>angle AOB, angle OAP, angle OBP, and angle P are the four interior angles of the quadrilateral</note>. According to the <note>Sum of Interior Angles of a Quadrilateral Theorem</note>, the sum of these four interior angles is <note>360°</note>, that is, <note>angle AOB + angle OAP + angle OBP + angle P = 360°</note>.


Question: As shown in the figure, in ⊙O, AD and CD are chords. Connect OC and extend it to intersect the tangent passing through point A at point B. If ∠ADC=25°, then the degree of ∠ABO is ()

Answer: 40°

1. Given ∠ADC=25°. By extending line OC to intersect the tangent passing through point A at point B, and connecting OA.

2. According to the inscribed angle theorem, the measure of an inscribed angle is half the measure of its corresponding central angle, i.e., ∠AOC=2×∠ADC = 2×25° = 50°.

3. Based on the properties of a circle's tangent, when a line is tangent to a circle at point A, the tangent at the point of tangency is perpendicular to the radius passing through the point of tangency. Therefore, we can conclude ∠OAB=90°.

4. In ΔOAB, using the angle sum theorem, we have: ∠OAB + ∠ABO + ∠AOB = 180°. Where ∠OAB = 90° and ∠AOB = ∠AOC = 50°.

5. From the equation in step 4, we get: 90° + ∠ABO + 50° = 180°.

6. Rearranging the equation in step 5, we obtain: ∠ABO = 180° - 90° - 50° = 40°.

7. Through the above reasoning, the final answer is 40°.

圆; 圆周角; 切线; 弦; 对顶角

Definition of Inscribed Angle

In <note>circle O</note>, <note>the vertex D of angle ADC is on the circumference</note>, <note>the two sides of angle ADC intersect circle O at points A and C respectively</note>. Therefore, <note>angle ADC is an inscribed angle</note>.

Inscribed Angle Theorem

<note>In circle O, points A, D, and C are on the circle</note>, <note>the central angle corresponding to the minor arc AC is ∠AOC</note>, <note>the inscribed angle is ∠ADC</note>. According to the Inscribed Angle Theorem, <note>∠AOC is twice the inscribed angle ∠ADC corresponding to the minor arc AC</note>, i.e., <note>∠AOC = 2 × ∠ADC</note>.

In the figure of this problem, in circle ⊙O, point A is the point of tangency of line AB with the circle, <note>line segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>tangent line AB is perpendicular to the radius OA passing through the point of tangency A</note>, that is, <note>∠OAB=90°</note>.

Triangle Angle Sum Theorem

<note>In triangle OAB, angles ∠OAB, ∠AOB, and ∠ABO are the three interior angles of triangle OAB</note>, according to the <note>Triangle Angle Sum Theorem</note>, <note>angle ∠OAB + angle ∠AOB + angle ∠ABO = 180°</note>.


Question: As shown in the figure, point P is a point on the extension line of the diameter AB of ⊙O. A tangent line PC is drawn through point P, with point C being the point of tangency. If AO=OB=PB=1, then the length of PC is ()

Answer: √{3}

1. Given that point P is on the extension of the diameter AB of ⊙O, draw the tangent line PC from point P to ⊙O, with the tangent point being C. According to the properties of the tangent line, the tangent line PC is perpendicular to the radius OC passing through the tangent point C. Therefore, connect OC, and we get OC⊥PC.

2. According to the problem statement, OC=OB=1, and point P is on the extension of the diameter AB, so OP=OB+PB=1+1=2.

3. In the right triangle △OPC, OC is one of the legs, and ##PC## is the other leg. Therefore, according to the Pythagorean theorem, we have OP^2=OC^2+PC^2.

4. Substitute the given conditions, OP=2, OC=1, we get 2^2=1^2+PC^2.

5. Solve the equation to get PC^2=4-1=3, therefore PC=√3.

6. Through the above reasoning, we finally get the answer √3.

圆; 切线; 直角三角形; 直线; 线段

Definition of Tangent to a Circle

A line is a tangent to a circle if and only if it has exactly one point of intersection with the circle. This point of intersection is called the point of tangency.

<note>Circle O and line PC have only one common point C</note>, this common point is called the point of tangency. Therefore, <note>line PC is the tangent to circle O</note>.

Definition of Diameter

A diameter is a line segment that passes through the center of a circle and has its endpoints on the circle. It is the longest chord of the circle, with a length equal to twice the radius of the circle.

In the figure of this problem, the line segment <note>AB passes through the center O of the circle</note>, and <note>points A and B are both on circle O</note>, therefore <note>AB is the diameter of circle O</note>, with a length equal to twice the radius, i.e., <note>AB = 2 * AO</note>.

Definition of Radius

<note>In circle O, point O is the center of the circle</note>, <note>points A and B are any points on the circle</note>, <note>line segment AO and line segment OB are line segments from the center O to any point on the circle</note>, therefore <note>line segment AO and line segment OB are the radii of circle O</note>, and AO=OB##=OC##=1.

In <note>circle O</note>, <note>point C is the point of tangency of line PC with the circle</note>, and <note>segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PC is perpendicular to the radius OC at the point of tangency C</note>, i.e., <note>∠OCP=90 degrees</note>.

Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the two legs.

<note>△OPC is a right triangle</note>, with the legs <note>OC</note> and <note>PC</note>, and the hypotenuse is <note>OP</note>. According to the Pythagorean Theorem, we have <note>OP^2=OC^2+PC^2</note>. Substituting the known conditions, <note>OP=2</note>, <note>OC=1</note>, we get <note>2^2=1^2+PC^2</note>, solving for <note>PC=√3</note>.

Definition of Right Triangle

A triangle is classified as a right triangle if it has one interior angle measuring exactly 90 degrees.

<note>In the figure of this problem</note>, in triangle OCB, <note>angle OCP is a right angle (90 degrees)</note>, therefore <note>triangle OCP is a right triangle</note>. <note>Side OC and side CP are the legs</note>, <note>side OP is the hypotenuse.</note>


Question: As shown in the figure, the radius of ⊙O is 2, the distance from point O to line l is 3, and point P is a moving point on line l. If PB is tangent to ⊙O at point B, then the minimum value of PB is ()

Answer: √{5}

1. Given that the radius of ⊙O is 2, and the distance from point O to line l is 3. Draw OP′ perpendicular to line l at point P′, then OP′=3.

2. Connect OB, and note the property of the tangent line, the tangent line is perpendicular to the radius passing through the point of tangency, thus ∠PBO=90°.

3. According to the properties of the right triangle OBP, apply the Pythagorean theorem, ##to get PB=√(OP^2-OB^2)=√(OP^2-2^2)##.

4. When point P moves to the position of point P′ on line l, the length of OP is minimized, which is 3, at this time PB is also minimized.

5. Substitute OP=3, and calculate the minimum value of PB as √(##3^2##-##2^2##)=√5.

6. Through the above reasoning, the final answer is √5.

圆; 切线; 垂线; 直线; 线段

Definition of Radius

<note>In circle O, point O is the center of the circle, point B is any point on the circle, and line segment OB is the line segment from the center to any point on the circle</note>, therefore <note>line segment OB is the radius of circle O</note>, with a length of 2.

In the problem diagram, in circle O, point B is the point of tangency of line PB with the circle, <note>line segment OB is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PB is perpendicular to the radius OB at the point of tangency B, i.e., ∠PBO=90°</note>.

Pythagorean Theorem

<note>△OBP is a right triangle</note>, in which <note>∠PBO=90°</note>, the hypotenuse is <note>OP</note>, and the two legs are <note>OB</note> and <note>PB</note>. According to the Pythagorean Theorem, ##<note>OP^2 = OB^2 + PB^2</note>##.


Question: As shown in the figure, PA, PB, CD are tangents to ⊙O, A, B, E are the points of tangency, CD intersects segments PA and PB at points C and D respectively. If ∠APB=40°, then the degree of ∠COD is ()

Answer: 70°

1. According to the problem, points A, C, E, D, B are taken on and outside circle O, as shown in the figure.

2. Connect OA, OB, and OE. According to the properties of the tangent, OA is perpendicular to PA, OB is perpendicular to PB, and OE is perpendicular to CD.

3. Since PA and PB are tangents, and A and B are points of tangency, according to the tangent length theorem, we have: DB = DE and AC = CE.

4. Since point O is the center of the circle, and the radius through point O at the points of tangency are equal: AO = OE = OB.

5. Based on the above known conditions, and the congruence of triangles in the figure, we conclude: △AOC is congruent to △EOC (according to the congruent triangles theorem (SSS)).

6. Under the same basic conditions, △EOD is congruent to △BOD (according to the congruent triangles theorem (SSS)).

7. Therefore, we conclude: ∠AOC = ∠EOC and ∠EOD = ∠BOD, because ∠AOB = ∠AOE + ∠BOE = 2∠EOC + 2∠EOD, and ∠COD = ∠EOC + ∠EOD, it can be further deduced that ∠COD is half of ∠AOB.

8. According to the given condition ∠APB = 40°, and in quadrilateral PAOB, according to the interior angle sum theorem of quadrilaterals, ∠AOB = 360° - ∠OAP - ∠OBP - ∠APB = 360° - 90° - 90° - 40° = 140°.

9. Therefore, we conclude: since ∠COD = 0.5 * ∠AOB, ∠COD = 0.5 * 140° = 70°.

10. Through the above reasoning, the final answer is 70°.

圆; 切线; 圆心角; 等腰三角形; 对顶角

Definition of Tangent to a Circle

The original text: <note>Circle O and lines PA, PB, and CD have only one common point A, B, and E</note>, these common points are called points of tangency. Therefore, <note>lines PA, PB, and CD are tangents to circle O</note>, and <note>OA⊥PA</note>, <note>OB⊥PB</note>, <note>OE⊥CD</note>.

Definition of Congruent Triangles

Two triangles are congruent if and only if their corresponding sides are equal and their corresponding angles are equal.

<note>Triangle AOC and triangle EOC are congruent triangles</note>, <note>The corresponding sides and angles of triangle AOC are equal to those of triangle EOC</note>, that is: <note>side AO = side EO</note>, side CO = side CO, side AC = side EC, and the corresponding angles are also equal: <note>angle OAC = angle OEC</note>, <note>angle AOC = angle EOC</note>, angle ACO = angle ECO. <note>Triangle EOD and triangle BOD are congruent triangles</note>, <note>The corresponding sides and angles of triangle EOD are equal to those of triangle BOD</note>, that is: <note>side EO = side BO</note>, side DO = side DO, side DE = side DB, and the corresponding angles are also equal: <note>angle OED = angle OBD</note>, <note>angle EOD = angle BOD</note>, angle EDO = angle BDO.

Triangle Congruence Theorem (SSS)

Two triangles are congruent if their three pairs of corresponding sides are equal in length.

In the figure of this problem, <note>triangle AOC</note> and <note>triangle EOC</note>, <note>side AO is equal to side EO</note>, side CO is equal to side CO, <note>side AC is equal to side EC</note>, therefore according to <note>Triangle Congruence Theorem (SSS)</note>, these two triangles are congruent. In <note>triangle EOD</note> and <note>triangle BOD</note>, <note>side EO is equal to side BO</note>, side DO is equal to side DO, <note>side DE is equal to side DB</note>, therefore according to <note>Triangle Congruence Theorem (SSS)</note>, these two triangles are congruent.

<note>In the figure of this problem, in circle O, point A is the point where line PA touches the circle, and segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent PA is perpendicular to the radius OA at the point of tangency A, i.e., ∠OAP=90 degrees</note>. Similarly, <note>point B is the point where line PB touches the circle, and segment OB is the radius of the circle</note>, <note>the tangent PB is perpendicular to the radius OB at the point of tangency B, i.e., ∠OBP=90 degrees</note>. <note>Point E is the point where line CD touches the circle, and segment OE is the radius of the circle</note>, <note>the tangent CD is perpendicular to the radius OE at the point of tangency E, i.e., ∠OEC=90 degrees</note>.

Tangent-Segment Theorem

From a point C outside the circle, two tangents CA and CE are drawn, and their tangent lengths are equal, i.e., <note>CA=CE</note>. The line connecting the center of the circle O and this point C bisects the angle between the two tangents CA and CE, i.e., <note>angle OCA = angle OCE</note>. From a point D outside the circle, two tangents DE and DB are drawn, and their tangent lengths are equal, i.e., <note>DE=DB</note>. The line connecting the center of the circle O and this point D bisects the angle between the two tangents DE and DB, i.e., <note>angle ODE = angle ODB</note>.

In any quadrilateral, the sum of the four interior angles is 360°.

In the diagram of this problem, quadrilateral PAOB, <note>angle AOB</note>, <note>angle OAP</note>, <note>angle OBP</note>, and <note>angle APB</note> are the four interior angles of the quadrilateral. According to the <note>Sum of Interior Angles of a Quadrilateral Theorem</note>, the sum of these four interior angles is <note>360°</note>, that is, <note>angle AOB + angle OAP + angle OBP + angle APB = 360°</note>.


Question: As shown in the figure, AC is the tangent to ⊙O at point C, BC is the diameter of ⊙O, AB intersects ⊙O at point D, and connect OD. If ∠BAC=55°, then the measure of ∠COD is ()

Answer: 70°

1. Given that AC is the tangent line to ⊙O at point C, and according to the property that the tangent is perpendicular to the radius passing through the point of tangency, we obtain ∠ACB=90°.

2. From the given conditions, we know ∠BAC=55°.

3. According to the triangle angle sum theorem, for ##triangle ABC##, we have ∠BAC + ∠ACB + ∠ABC = 180°, thus ∠ABC=180° - ∠BAC - ∠ACB.

4. Substituting the known values ∠BAC=55° and ∠ACB=90°, we get ∠ABC=180° - 55° - 90° = 35°.

5. According to the inscribed angle theorem, the central angle ##∠COD## corresponding to the arc ##DC## is twice the inscribed angle ∠B, thus ##∠COD##=2×##∠ABC##.

6. Substituting ∠ABC=35°, we get ##∠COD##=2×35°=70°.

####

##7##. Through the above reasoning, the final answer is 70°.

圆; 切线; 圆周角; 圆心角; 直角三角形

Definition of Tangent to a Circle

In the diagram of this problem, <note>circle O and line AC have exactly one common point C</note>, this common point is called the point of tangency. Therefore, <note>line AC is the tangent to circle O</note>.

Definition of Radius

<note>Point O is the center of the circle</note>, <note>point C and point B are any points on the circle</note>, <note>line segment OC and line segment OB are segments from the center to any point on the circle</note>, therefore <note>line segment OC and line segment OB are the radii of circle O</note>.

Definition of Diameter

In the figure of this problem, <note>BC is the diameter</note>, connecting <note>the center O</note> and <note>points B and C on the circumference</note>, with a length of <note>2 times the radius, i.e., BC = 2 * r</note>.

Definition of Inscribed Angle

In the diagram of this problem, in circle O, the <note>angle DBC</note>'s <note>vertex B</note> is on the circumference, and the two sides of <note>angle DBC</note> intersect circle O at <note>point D and point C</note> respectively. Therefore, <note>angle DBC</note> is an inscribed angle.

Definition of Central Angle

The points <note>D and C</note> are two points on the circle, and the center of the circle is <note>point O</note>. The angle <note>∠DOC</note> formed by the lines <note>OD</note> and <note>OC</note> is called the central angle.

In <note>circle O</note>, <note>point C is the point of tangency of line AC with the circle</note>, <note>segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>tangent line AC is perpendicular to the radius OC at the point of tangency C</note>, that is, <note>∠ACB=90°</note>.

Triangle Angle Sum Theorem

The sum of the interior angles of any triangle is 180°.

<note>In the figure of this problem, in triangle ABC, angle BAC, angle ACB, and angle ABC are the three interior angles of triangle ABC</note>, according to the Triangle Angle Sum Theorem, <note>angle BAC + angle ACB + angle ABC = 180°</note>.

Inscribed Angle Theorem

In a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

<note>In the given figure</note>, in circle O, <note>points B, C, and D are on the circle</note>, <note>the central angle corresponding to arc DC is ∠DOC</note>, and <note>the inscribed angle is ∠DBC</note>. According to the Inscribed Angle Theorem, <note>∠DBC is equal to half of the central angle ∠DOC corresponding to arc DC</note>, that is, <note>∠DBC = 1/2 ∠DOC</note>.


Question: As shown in the figure, PA and PB are tangents to ⊙O, AC is the diameter of ⊙O, ∠c=55°, then ∠APB equals ()

Answer: 70°

1. Given PA and PB are tangents to ⊙O, AC is the diameter of ⊙O. ##Connect OB, according to the property of tangents##, we get ∠OAP=∠OBP=90°.

2. According to the given ∠C=55°, and OC=OB, therefore in the isosceles triangle OCB, the base angles are equal, thus ∠OBC=∠OCB=55°.

3. By the property that the angle subtended by the diameter at the circumference is a right angle, we get ∠ABC=90°.

4. From the conclusion in step 2, we can calculate ∠AOB: in ##△OBC##, ∠AOB as an exterior angle is composed of ##∠OCB## and ##∠OBC##, where ∠AOB=∠OBC + ∠OCB=55° + 55°=110°.

5. In the quadrilateral AOBP, the sum of the interior angles is 360°. Since ∠OAP=90° and ∠OBP=90°, therefore ∠AOB + ∠APB=180°.

6. According to the conclusion in step 4, we get ∠APB=180° - ∠AOB=180° - 110°=70°.

7. Through the above reasoning, the final answer is 70°.

圆; 切线; 圆周角; 等腰三角形; 线段

Definition of Tangent to a Circle

Original text: <note>Circle O and line PA and PB have only one common point, which are point A and point B respectively</note>, these common points are called points of tangency. Therefore, <note>line PA and PB are tangents to circle O</note>.

Definition of Diameter

<note>AC is the diameter</note>, connecting <note>the center O</note> and <note>points A and C on the circumference</note>, with a length of <note>2 times the radius</note>, that is, <note>AC = 2 * OA</note>.

Definition of Isosceles Triangle

A triangle is an isosceles triangle if and only if it has at least two congruent (≅) sides.

In <note>triangle OCB</note>, <note>sides OC and OB are equal</note>, so <note>triangle OCB is an isosceles triangle</note>.

<note>Point A is the point of tangency between line PA and the circle</note>, <note>segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PA is perpendicular to the radius OA at the point of tangency A</note>, that is, <note>∠OAP = 90 degrees</note>. Similarly, <note>point B is the point of tangency between line PB and the circle</note>, <note>segment OB is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠OBP = 90 degrees</note>.

The angle subtended by the diameter of a semicircle is a right angle; conversely, if an inscribed angle is 90°, then the chord subtending that angle is a diameter.

<note>In circle O, the inscribed angle ∠ABC subtended by the diameter AC is a right angle (90 degrees)</note>.

Properties of Isosceles Triangle

In an isosceles triangle, the angles opposite the equal sides are equal, and the sides opposite the equal angles are congruent.

<note>In the diagram of this problem, in the isosceles triangle OCB</note>, the sides <note>OC and OB are equal</note>. Therefore, according to the properties of the isosceles triangle, the angles opposite the equal sides are equal, that is, <note>angle OCB = angle OBC</note>.

Exterior Angle Theorem of Triangle

An exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

In <note>triangle OBC</note>, <note>angle AOB</note> is an exterior angle of the triangle, <note>angle OBC</note> and <note>angle OCB</note> are the two non-adjacent interior angles to the exterior angle AOB. According to the <note>Exterior Angle Theorem of Triangle</note>, the exterior angle <note>AOB</note> is equal to the sum of the two non-adjacent interior angles OBC and OCB, that is, <note>angle AOB = angle OBC + angle OCB</note>.

<note>In quadrilateral AOBP</note>, <note>angle OAP, angle OBP, angle P, and angle AOB are the four interior angles of the quadrilateral</note>. According to the Quadrilateral Interior Angle Sum Theorem, the sum of these four interior angles is 360°, that is, <note>angle OAP + angle OBP + angle P + angle AOB = 360°</note>.

An exterior angle of a polygon is the angle formed between one side of the polygon and the extension of an adjacent side.

The interior angle of <note>polygon OBC</note> is <note>∠BOC</note>. The angle formed by extending the adjacent sides <note>OC</note> and <note>OB</note> of this interior angle is called the exterior angle of the interior angle <note>∠BOC</note>.


Question: As shown in the figure, the diameters of two concentric circles are 6cm and 10cm respectively. A chord AB of the larger circle is tangent to the smaller circle. Find the length of chord AB.

Answer: 8cm

1. Draw line segment OC perpendicular to line segment AB at point C, and connect OA, as shown in the figure according to the problem statement.

2. Since chord AB is tangent to the smaller circle, OC is the radius of the smaller circle, i.e., OC=3cm.

3. In right triangle OAC, according to the given conditions, OA is the radius of the larger circle, i.e., OA=5cm, and on the perpendicular OC, OC=3cm.

4. According to the Pythagorean theorem, in right triangle OAC, the length of AC can be calculated as AC=√(OA^2 - OC^2)=√(5^2 - 3^2)=√(25 - 9)=√16=4cm.

5. Since OC is perpendicular to AB and AB is bisected by OC, AC=BC.

6. Therefore, the length of chord AB is AB=2AC=2×4=8cm.

弦; 切线; 垂线; 圆

Definition of Chord

<note>Point A and point B are any two points on the large circle</note>, <note>line segment AB connects these two points</note>, so <note>line segment AB is a chord of the large circle</note>.

Definition of Tangent to a Circle

Original text: <note>The small circle and the line AB have only one common point C, which is called the point of tangency</note>. Therefore, <note>the line AB is the tangent to the small circle</note>.

Definition of Right Triangle

<note>In the figure of this problem, in triangle OAC, angle OCA is a right angle (90 degrees)</note>, therefore triangle OAC is a right triangle. <note>Side OC and side AC are the legs</note>, <note>side OA is the hypotenuse</note>.

Pythagorean Theorem

In the figure of this problem, applied to <note>right triangle OAC</note>, according to the <note>Pythagorean Theorem</note>, there is <note>OA^2 = OC^2 + AC^2</note>, where <note>OA is the radius of the large circle</note>, <note>OC is the radius of the small circle</note>, <note>AC is one of the right-angle sides</note>, therefore <note>OA=5cm</note>, <note>OC=3cm</note>, thus we can find <note>AC=√(5^2 - 3^2)=√(25 - 9)=√16=4cm</note>.

Perpendicular Diameter Theorem

If a diameter is perpendicular to a chord, then that diameter bisects the chord and also bisects the two arcs subtended by that chord.

<note>OC is perpendicular to chord AB</note>, and <note>OC passes through the center O</note>, therefore <note>OC is the perpendicular bisector of AB</note>, and <note>OC bisects AB</note>. According to the Perpendicular Diameter Theorem, <note>AC=BC</note>, the length of chord <note>AB is twice AC</note>, that is <note>AB=2×AC=2×4=8cm</note>.


Question: As shown in the figure, PA and PB are tangent to ⊙O at points A and B respectively, point C is on the major arc ?{ACB}, ∠P=80°, then the degree of ∠C is ()

Answer: 50°

1. Given that PA is a tangent to the circle, according to the property that a tangent is perpendicular to the radius at the point of tangency, we get ∠OAP=90°.

2. Similarly, since PB is a tangent to the circle, according to the property that a tangent is perpendicular to the radius at the point of tangency, we get ∠OBP=90°.

3. According to the theorem of the sum of interior angles of a quadrilateral, the sum of the interior angles of a quadrilateral is 360°, therefore ∠AOB=360°-∠OAP-∠OBP-∠P.

4. Substituting the given condition ∠P=80°, we get ∠AOB=360°-90°-90°-80°=100°.

5. According to the inscribed angle theorem, in the same circle, the measure of the central angle subtended by the same arc is twice the measure of the inscribed angle, therefore ∠ACB=∠AOB/2.

6. Substituting the given condition ∠AOB=100°, we get ∠ACB=100°/2=50°.

7. Through the above reasoning, the final answer is ∠C=50°.

圆周角; 圆内接四边形; 切线; 圆

Definition of Tangent to a Circle

In the figure of this problem, <note>circle O and line PA have only one common point A</note>, this common point is called the point of tangency. Therefore, <note>line PA is the tangent to circle O</note>. Similarly, <note>circle O and line PB have only one common point B</note>, this common point is called the point of tangency. Therefore, <note>line PB is the tangent to circle O</note>.

Definition of Inscribed Angle

An inscribed angle is an angle whose vertex lies on the circumference of a circle and whose sides intersect the circle at two points.

<note>In circle O, the vertex C of angle ACB is on the circumference</note>, and the two sides of angle ACB intersect <note>circle O at points A and B</note>. Therefore, <note>angle ACB is an inscribed angle</note>.

A tangent line to a circle is perpendicular to the radius that passes through the point of tangency.

In <note>circle O</note>, point <note>A is the point of tangency of line PA with the circle</note>, and <note>segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PA is perpendicular to the radius OA at the point of tangency A</note>, that is, <note>∠OAP=90°</note>. Similarly, point <note>B is the point of tangency of line PB with the circle</note>, and <note>segment OB is the radius of the circle</note>, <note>the tangent line PB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠OBP=90°</note>.

In the diagram of this problem, the four interior angles of <note>quadrilateral OAPB</note>, namely <note>angle OAP</note>, <note>angle OBP</note>, <note>angle P</note>, and <note>angle AOB</note>, sum up to 360° according to the <note>Sum of Interior Angles of a Quadrilateral Theorem</note>, that is, <note>∠OAP + ∠OBP + ∠P + ∠AOB = 360°</note>.

Inscribed Angle Theorem

The <note>arc ACB</note> corresponds to the <note>central angle ∠AOB</note>, and the degree measure of ∠AOB is <note>100°</note>. According to the Inscribed Angle Theorem, the <note>∠ACB</note> is equal to half of the <note>central angle ∠AOB</note> that it subtends, which means <note>∠ACB = 1/2 ∠AOB = 50°</note>.

Definition of Central Angle

An angle formed by two radii connecting two points on the circumference of a circle to the center of the circle is called a central angle.

Original text: <note>In circle O, points A and B are two points on the circle, and the center of the circle is point O</note>. The angle formed by the lines <note>OA and OB</note> is called the central angle <note>∠AOB</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, CD is tangent at point D, the extension of AB intersects CD at point C, if ∠ACD=40°, then ∠A=()

Answer: 25°

2. Given that ##OD## is the ##radius## of ⊙O, and CD is tangent at point D, therefore according to the ##property of the tangent to a circle##, we get ∠CDO=90°.

3. Given ∠ACD=40°, so according to the ##triangle angle sum theorem##, we can obtain ##∠BOD=##∠COD=##180°-##90°-∠ACD=50°.

4. ##Since point A is on the circumference, according to the definition of the inscribed angle##, ∠A is the inscribed angle, and the corresponding central angle is ##∠BOD##.

5. The inscribed angle is half of the corresponding central angle, therefore according to the inscribed angle theorem, ∠A=##1/2∠BOD##=25°.

6. Through the above reasoning, the final answer is 25°.

圆; 圆周角; 切线; 直角三角形; 邻补角

Definition of Radius

A radius is a line segment joining the center of a circle to any point on the circumference of the circle.

<note>Point O is the center of the circle</note>, <note>Point D is any point on the circle</note>, <note>Line segment OD is the line segment from the center of the circle to any point on the circle</note>, therefore <note>Line segment OD is the radius of the circle</note>.

Definition of Tangent to a Circle

A line is a tangent to a circle if and only if it has exactly one point of intersection with the circle. This point of intersection is called the point of tangency.

Original: <note>圆⊙O与直线CD有且只有一个公共点D</note>，这个公共点叫做<note>切点</note>。因此，<note>直线CD是圆⊙O的切线</note>。

Translation: <note>Circle ⊙O and line CD have exactly one common point D</note>, this common point is called the <note>point of tangency</note>. Therefore, <note>line CD is the tangent to circle ⊙O</note>.

Definition of Inscribed Angle

<note>In the figure of this problem</note>, in circle O, the vertex of <note>angle BAD</note> is <note>A on the circumference</note>, and the two sides of <note>angle BAD</note> intersect <note>circle O at point D</note> and <note>point B</note> respectively. Therefore, <note>angle BAD is an inscribed angle</note>.

Definition of Central Angle

<note>Point B</note> and <note>Point D</note> are two points on the circle, with the center of the circle being <note>Point O</note>. The angle <note>∠BOD</note> formed by the lines <note>OB</note> and <note>OD</note> is called the central angle.

<note>In the figure of this problem, in circle O, point D is the point of tangency where line CD touches the circle</note>, segment OD is the radius of the circle. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OD at the point of tangency D</note>, that is, <note>∠CDO=90°</note>.

Triangle Angle Sum Theorem

<note>In triangle ODC</note>, <note>angle ODC</note>, <note>angle OCD</note>, and <note>angle COD</note> are the three interior angles of triangle ODC. According to the <note>Triangle Angle Sum Theorem</note>, <note>angle ODC + angle OCD + angle COD = 180°</note>.

Inscribed Angle Theorem

<note>In circle O, points A, B, and D are on the circle</note>, the central angle corresponding to arc BD is <note>∠BOD</note>, and the inscribed angle is <note>∠BAD</note>. According to the Inscribed Angle Theorem, <note>∠BAD</note> is equal to half of the central angle <note>∠BOD</note> corresponding to arc BD, that is, <note>∠BAD = 1/2 ∠BOD</note>.


Question: As shown in the figure, P is a point on the extension of the diameter AB of ⊙O, PC is tangent to ⊙O at C, ∠P=50°, ∠A is ()

Answer: 20°

1. Connect OC, we know CP is the tangent to circle O, thus according to the property of the tangent to a circle, OC is perpendicular to CP, ∠OCP=90°.

2. In the right triangle OPC, according to the triangle angle sum theorem, given ∠P=50°, ∠OCP=90°, so we can obtain ∠POC=40°.

3. Since OA=OC, in △AOC, OA=OC is an isosceles triangle with equal sides.

4. Therefore, we can conclude ∠OAC=∠OCA.

5. Since ∠POC is the exterior angle of △AOC, according to the exterior angle theorem, the exterior angle is equal to the sum of the two non-adjacent interior angles, thus ∠POC=2∠OCA.

6. Combining ∠POC=40°, we can find ∠OCA=20°, so ∠A=20°.

7. Through the above reasoning, the final answer is 20°.

圆; 切线; 圆周角; 对顶角; 线段

Definition of Diameter

A diameter is a line segment that passes through the center of a circle and has its endpoints on the circle. It is the longest chord of the circle, with a length equal to twice the radius of the circle.

In the figure of this problem, <note>AB is the diameter of ⊙O</note>, connecting <note>the center O</note> and <note>points A and B on the circumference</note>, with a length of twice the radius, that is, <note>AB = 2 * OA</note>.

Definition of Tangent to a Circle

In the figure of this problem, <note>circle O and line PC have only one common point C</note>, <note>this common point is called the point of tangency</note>. Therefore, <note>line PC is the tangent to circle O</note>.

Definition of Isosceles Triangle

In <note>triangle AOC</note>, <note>sides OA and OC are equal</note>, thus <note>triangle AOC is an isosceles triangle</note>.

Definition of Right Triangle

A triangle is classified as a right triangle if it has one interior angle measuring exactly 90 degrees.

In the figure of this problem, in triangle OPC, <note>angle OCP is a right angle (90 degrees)</note>, therefore <note>triangle OPC is a right triangle</note>. <note>Side OC and side CP are the legs</note>, <note>side OP is the hypotenuse</note>.

Exterior Angle Theorem of Triangle

In the figure of this problem, in triangle AOC, <note>angle POC is an exterior angle of the triangle</note>, <note>angle OAC and angle OCA are the two non-adjacent interior angles to the exterior angle POC</note>. According to the Exterior Angle Theorem of Triangle, <note>the exterior angle POC is equal to the sum of the two non-adjacent interior angles OAC and OCA</note>, that is, <note>angle POC = angle OAC + angle OCA</note>.

In <note>circle O</note>, <note>point C is the point of tangency between line CP and the circle</note>, and <note>segment OC is the radius of the circle</note>. According to the <note>property of the tangent line to a circle</note>, <note>the tangent line CP is perpendicular to the radius OC at the point of tangency C</note>, i.e., <note>∠OCP=90 degrees</note>.

Triangle Angle Sum Theorem

<note>In triangle OPC, angle OPC, angle OCP, and angle COP are the three interior angles of triangle OPC</note>, according to the Triangle Angle Sum Theorem, <note>angle OPC + angle OCP + angle COP = 180°</note>.

In the figure of this problem, an interior angle of polygon OAC is <note>∠AOC</note>, and the angle <note>∠POC</note> formed by extending the adjacent sides <note>OA</note> and <note>OC</note> of this interior angle is called the exterior angle of the interior angle <note>∠AOC</note>.


Question: As shown in the figure, PA and PB are tangents to ⊙O, A and B are points of tangency, AC is the diameter of O. If ∠BAC=25°, then ∠P is ()

Answer: 50°

1. Given PA and PB are tangents to ⊙O, and A and B are points of tangency, according to the properties of tangents, ∠OAP and ∠OBP are 90°.

2. Since OA and OB are radii of the circle, therefore OA=OB, so triangle OAB is an isosceles triangle.

3. According to the properties of isosceles triangles: ∠OAB=∠OBA.

4. Based on the given condition ∠BAC=25°, we get ∠BAO=∠OBA=25°.

5. In triangle OAB, according to the sum of the interior angles of a triangle being 180°, we get: ∠AOB=180° - 2×∠BAC = 180° - 2×25° = 130°.

6. From the above reasoning, we get: ∠OAP = ∠OBP = 90°, and ∠AOB=130°.

7. Since the sum of the interior angles of quadrilateral AOBP is 360°, we get: ∠P = 360° - ∠OAP - ∠OBP - ∠AOB = 360° - 90° - 90° - 130° = 50°.

8. After the above reasoning, the final answer is 50°.

圆; 圆周角; 切线; 等腰三角形; 垂线

Definition of Radius

In <note>circle O</note>, <note>point O is the center of the circle</note>, <note>point A and point B are any points on the circle</note>, <note>line segment OA and line segment OB are line segments from the center to any point on the circle</note>, therefore <note>line segment OA and line segment OB are the radii of the circle</note>.

Definition of Diameter

<note>AC is the diameter of circle O</note>, <note>AC passes through the center O and connects points A and C on the circumference</note>, with a length equal to 2 times the radius, that is, <note>AC = 2 * OA</note>.

Definition of Tangent to a Circle

In the diagram of this problem, <note>circle O</note> and <note>lines PA and PB</note> have only <note>one common point A and B</note>, which is called the point of tangency. Therefore, <note>lines PA and PB are tangents to circle O</note>.

In the figure of this problem, in circle O, point A and point B are the points where line PA and PB touch the circle, segment OA and OB are the radii of the circle. According to the property of the tangent line to a circle, tangent PA and PB are perpendicular to the radii OA and OB at the points of tangency A and B, that is, ∠OAP=90° and ∠OBP=90°.

Definition of Isosceles Triangle

<note>The sides OA and OB are equal</note>, therefore <note>triangle OAB</note> is an <note>isosceles triangle</note>.

Properties of Isosceles Triangle

<note>In the figure of this problem, in the isosceles triangle OAB, side OA and side OB are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>angle OAB = angle OBA</note>.

Triangle Angle Sum Theorem

In the figure of this problem, in triangle OAB, <note>angle AOB</note>, <note>angle OAB</note>, and <note>angle OBA</note> are the three interior angles of triangle OAB. According to the Triangle Angle Sum Theorem, <note>angle AOB + angle OAB + angle OBA = 180°</note>.

<note>In quadrilateral OAPB</note>, <note>angle AOB</note>, <note>angle P</note>, <note>angle OAP</note>, and <note>angle OBP</note> are the four interior angles of the quadrilateral. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>angle AOB + angle P + angle OAP + angle OBP = 360°</note>.


Question: As shown in the figure, given a∥b∥c, AB=1, BC=2, EF=4, find DE=()

Answer: 2

1. Given that lines a ∥ b ∥ c, and AB = 1, BC = 2, EF = 4.

2. According to the theorem of proportional segments in parallel lines, we have: \( \frac{AB}{BC} = \frac{DE}{EF} \).

3. Substituting the given conditions, we get: \( \frac{1}{2} = \frac{DE}{4} \).

4. Solving this proportion equation, we get DE = 2.

5. Through the above reasoning, the final answer is 2.

平行线; 线段; 同位角

Proportional Segments Theorem

If a line is parallel to one side of a triangle and intersects the other two sides or their extensions, then it divides the segments it intersects proportionally in relation to the corresponding segments of the original triangle.

In the figure of this problem, <note>a ∥ b ∥ c</note>, then according to the Proportional Segments Theorem, we have: <note>\( \frac{AB}{BC} = \frac{DE}{EF} \)</note>.


Question: As shown in the figure, a quadrilateral green area has circular fountains with a radius of 2 at each of its four corners. The area occupied by these four fountains in the green area is ()

Answer: 4π

1. Given that each fountain is circular with a radius r = 2. According to the formula for the area of a circle, the area of a single circle is π·r^2 = 4π.

2. Circular fountains are placed at each of the four corners of the quadrilateral, and each circle's shaded part is a sector with a central angle equal to the interior angle of the quadrilateral. According to the theorem of the sum of interior angles of a quadrilateral, the sum of the central angles of these four sectors is 360°.

3. According to the formula for the area of a sector, combining like terms, the area of the shaded part equals the area of one circle.

4. According to step 1, the area of a single circle is 4π.

5. Combining the above reasoning, the total area occupied by the four circular fountains in the green park is 4π.

Area Formula of a Circle

The area of a circle is given by the formula A = πr², where A represents the area, π (pi) is the mathematical constant approximately equal to 3.14159, and r is the radius of the circle.

<note>The radius of the circular fountain is 2</note>. According to the <note>area formula of a circle</note>, the area A of the circle is equal to <note>π times the square of the radius 2</note>, that is, <note>A = π·2^2 = 4π</note>.

Formula for the Area of a Sector

The area \( X \) of a sector can be calculated using the formula \( X = \frac{\theta}{360} \times \pi \times r^2 \), where \( \theta \) is the measure of the central angle in degrees, and \( r \) is the radius length.

In the figure of this problem, the shaded <note>sector A, the central angle A is θ degrees</note>, <note>the radius length is r</note>. According to the <note>formula for the area of a sector, the area of the sector A can be calculated using the formula A = (θ/360) * π * r?</note>, where <note>angle A is the central angle in degrees, r is the radius, with a length of 2</note>. Therefore, <note>the area of sector A is A = (angle A/360) * π * 2?</note>, similarly, the shaded sectors <note>B, C, D</note> can be calculated, combining like terms, we get <note>the sum of the areas of the four sectors = {(angle A + angle B + angle C + angle D)/360} * π * 2?</note> = (360/360) * π * 2? = π * 2? = 4π.

In any quadrilateral, the sum of the four interior angles is 360°.

<note>In the diagram of this problem, let the four vertices of the quadrilateral be A, B, C, D</note>, in <note>quadrilateral ABCD</note>, <note>angle A, angle B, angle C, angle D</note> are the four interior angles of the quadrilateral. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>angle A + angle B + angle C + angle D = 360°</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, point C is on the extension of AB, CD is tangent to ⊙O at point D. If ∠A = 35°, then ∠C = ()

Answer: 20°

1. Given AB is the diameter of ⊙O, ##connect OD and BD##.

2. Since point C is on the extension of AB, and CD is tangent to ⊙O at point D, according to the property of tangents, the tangent at point D is perpendicular to the radius OD, therefore ∠ODC=90°.

3. ##According to the inscribed angle theorem, ∠COD##=2∠A=2×35°=70°.

4. In the triangle ODC, given ∠ODC=90°, ##∠COD=70°##, therefore, using the triangle angle sum theorem, ##∠C##=180°-∠ODC-##∠COD##=180°-90°-70°=20°.

5. Through the above reasoning, it is concluded that ∠C=20°.

圆; 切线; 圆周角; 邻补角; 对顶角

Definition of Diameter

<note>AB is the diameter of circle O</note>, connecting <note>the center O and points A and B on the circumference</note>, with a length of <note>2 times the radius</note>, that is <note>AB = 2 * OA</note>.

Definition of Tangent to a Circle

<note>Circle ⊙O</note> and <note>line CD</note> have exactly one <note>common point D</note>, which is called the point of tangency. Therefore, <note>line CD</note> is the <note>tangent to circle ⊙O</note>.

Triangle Angle Sum Theorem

The sum of the interior angles of any triangle is 180°.

In the figure of this problem, in triangle ODC, angle ODC, <note>angle OCD</note>, and angle COD are the three interior angles of triangle ODC. According to the Triangle Angle Sum Theorem, angle ODC + <note>angle OCD</note> + angle COD = 180°.

Inscribed Angle Theorem

In a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

In <note>circle O</note>, <note>points A, B, and D are on the circle</note>, <note>the central angle corresponding to arc BD is ∠BOD</note>, <note>the inscribed angle is ∠A</note>. According to the <note>Inscribed Angle Theorem</note>, ∠A is equal to half of the <note>central angle corresponding to arc BD, ∠BOD</note>, that is <note>∠A = 1/2 ∠BOD</note>.

Definition of Central Angle

In <note>circle O</note>, points <note>D and B are two points on the circle</note>, and the center of the circle is <note>point O</note>. The angle ∠DOB formed by the lines <note>OD and OB</note> is called the <note>central angle</note>.

<note>In the figure of this problem, in circle O, point D is the point of tangency of line CD with the circle</note>, <note>segment OD is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OD at the point of tangency D</note>, that is, <note>∠ODC=90 degrees</note>.