Question: As shown in the figure, AB is the diameter of ⊙O, C is a point on ⊙O, chord AD bisects ∠BAC, intersecting BC at point E, AB=6, AD=5, then the length of DE is ()

Answer: 2.2

1. As shown in the figure, it is known that AB is the diameter of ⊙O, C is a point on ⊙O, AD is a chord, and AD bisects ∠BAC. Connect BD and CD.

2. Since AB is the diameter of ⊙O, according to (Corollary 2 of the Inscribed Angle Theorem) the inscribed angle subtended by the diameter is a right angle, so ∠ADB = 90°.

3. In △ADB, the Pythagorean theorem can be used to calculate the length of BD. In this problem, AB is the hypotenuse, AD is one of the legs, and BD is the other leg. Therefore, BD = √{AB² - AD²} = √{6² - 5²} = √{36 - 25} = √{11}.

4. Since chord AD bisects ∠BAC, ∠CAD = ∠DAB.

5. According to Corollary 1 of the Inscribed Angle Theorem, ∠CBD = ∠DAC. From step 4, it can be concluded that ∠BAD = ∠EBD.

6. In △ABD and △BED, the following angles can be observed to be equal: ∠BAD = ∠EBD, ∠ADB = ∠BDE.

7. Since the two triangles have two corresponding angles equal, according to the Similar Triangle Criterion (AA), △ABD ∽ △BED.

8. According to the definition of similar triangles, DE / DB = DB / AD, that is, DE / √{11} = √{11} / 5.

9. Solving the above proportion equation, DE = 11 / 5.

10. Through the above reasoning, it can be concluded that the length of DE is 11/5.

圆; 弦; 圆周角; 直角三角形; 等腰三角形

<note>In the diagram of this problem, in circle O, the inscribed angles corresponding to arc CD are ∠CAD and ∠CBD.</note> According to Corollary 1 of the Inscribed Angle Theorem, <note>the inscribed angles corresponding to the same arc CD, ∠CAD and ∠CBD, are equal, i.e., ∠CAD = ∠CBD.</note>

<note>In the figure of this problem, in circle O, the angle subtended by the diameter AB at the circumference ADB is a right angle (90 degrees)</note> and <note>the angle subtended at the circumference ACB is a right angle (90 degrees)</note>.

Pythagorean Theorem

In the right triangle △ADB in the problem, <note>∠ADB is a right angle (90 degrees)</note>, <note>sides AD and BD are the legs</note>, <note>side AB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AB² = AD² + BD²</note>, that is, <note>BD = √{AB² - AD²} = √{6² - 5²}</note> = √(36 - 25) = √11.

Similarity Theorem for Triangles (AA)

Two triangles are similar if two of their corresponding angles are congruent.

In the original text: <note>△ABD and △BED</note>, <note>∠BAD = ∠EBD</note>, and <note>∠ADB = ∠BDE</note>, so according to the <note>Similarity Theorem for Triangles (AA)</note>, <note>△ABD ∽ △BED</note>.

Definition of Similar Triangles

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

<note>Triangle ABD</note> and <note>Triangle BED</note> are similar triangles. According to the definition of similar triangles: <note>∠BAD = ∠EBD</note>, <note>∠ADB = ∠BDE</note>; <note>DE/BD = BD/AD</note>.

Similarity Theorem for Triangles (AA)

<note>In △ABD and △BED</note>, <note>∠BAD = ∠EBD</note>, and <note>∠ADB = ∠BDE</note>, so according to the Similarity Theorem for Triangles (AA), <note>△ABD ∽ △BED</note>.

Definition of Angle Bisector

An angle bisector is a line that originates from the vertex of an angle and divides the angle into two congruent angles.

<note>The vertex of angle BAC is point A</note>, from <note>point A a line AD is drawn</note>, this line divides <note>angle BAC into two equal angles</note>, that is, <note>angle BAD and angle CAD are equal</note>. Therefore, <note>line AD is the angle bisector of angle BAD</note>.

Definition of Right Triangle

In the figure of this problem, in triangle ADB, <note>angle ADB is a right angle (90 degrees)</note>, therefore <note>triangle ADB is a right triangle</note>. <note>Side AD and side BD are the legs</note>, <note>side AB is the hypotenuse</note>. In triangle EDB, <note>angle EDB is a right angle (90 degrees)</note>, therefore <note>triangle EDB is a right triangle</note>. <note>Side ED and side BD are the legs</note>, <note>side EB is the hypotenuse</note>.

Definition of Inscribed Angle

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


Question: As shown in the figure, given ⊙O is the circumcircle of △ABC, ∠AOB=110°, then the degree of ∠C is ()

Answer: 55°

1. Given ⊙O is the circumcircle of △ABC, point O is the center of the circle, ∠AOB=110°.

2. Since point A, point B, and point C are on ⊙O, arc AC and arc BC are two arcs of ⊙O.

3. In a circle, the central angle corresponding to the same arc and the ##inscribed angle## corresponding to the arc satisfy the relationship that the degree measure of the central angle is twice the degree measure of the ##inscribed angle##.

4. In this problem, the ##inscribed angle## corresponding to the arc of the central angle ∠AOB is ∠ACB.

5. Therefore, according to the theorem, the degree measure of the ##inscribed angle## ∠ACB is half of the central angle ∠AOB, specifically expressed as ∠ACB = 1/2 × ∠AOB.

6. Substituting the given condition ∠AOB=110°, we get ∠ACB = 1/2 × 110° = 55°.

7. Through the above reasoning, the final answer is 55°.

圆; 圆心角; 圆周角; 三角形的外角

Definition of Circle

A circle is the locus of all points in a plane that are at a fixed distance from a given point.

In the figure of this problem, <note>circle ⊙O</note>, <note>point O is the center of the circle</note>, <note>the radii are OA, OB, and OC</note>. All points in the figure that are at a distance equal to OA, OB, or OC from <note>point O</note> are on <note>circle ⊙O</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 <note>circle O</note>, <note>point A and point B are two points on the circle</note>, <note>the center of the circle is point O</note>. The angle ∠AOB formed by <note>line segments OA and OB</note> is called the <note>central angle</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 <note>circle O</note>, <note>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> respectively. Therefore, <note>angle ACB is an inscribed angle</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 diagram of this problem, in circle O, points A, B, and C are on the circle, the central angle corresponding to <note>arc AB</note> is ∠AOB=110°, and the inscribed angle is ∠ACB. According to the Inscribed Angle Theorem, ∠ACB is equal to half of the central angle ∠AOB corresponding to <note>arc AB</note>, that is, ∠ACB = 1/2 ∠AOB.


Question: In △ABC, AB=AC, D and E are on BC and AC respectively, AD=AE, ∠CDE=20°, then the degree of ∠BAD is ()

Answer: 40°

1. Given AB=AC, AD=AE, in triangle ABC, AB=AC, ##according to the definition of an isosceles triangle, triangle ABC is an isosceles triangle, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal##, so ∠ABC=∠ACB.

2. ∠ADC is an exterior angle of triangle ABD, ##according to the exterior angle theorem##, ∠ADC = ∠BAD + ∠B.

3. ∠AED is an exterior angle of triangle DEC, ##according to the exterior angle theorem##, ∠AED = ∠EDC + ∠C.

4. According to the given conditions, ∠CDE=20°, ##and since ∠ADC=∠ADE + ∠EDC, so ∠ADC=∠ADE + 20°##.

5. ##In triangle ADE, AD=AE, according to the definition of an isosceles triangle, triangle ADE is an isosceles triangle, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, so ∠ADE=∠AED, from step 3, ∠AED = ∠EDC + ∠C= 20°+∠C##.

6. From step 2 and step 4, ∠BAD + ∠B=∠ADE + 20°, that is, ∠BAD + ∠B=∠C + ##20° + 20°##=∠C + 40°.

7. ##From step 1, we know## ∠ABC=∠ACB, that is, ∠B=∠C.

8. Substitute ∠B=∠C into the equation from step 6, we get: ∠BAD + ∠C=∠C + 40°.

9. Simplify to get ∠BAD = 40°.

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

等腰三角形; 普通三角形; 三角形的外角; 线段

Definition of Isosceles Triangle

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

<note>In triangle ABC, side AB and side AC are equal</note>, therefore triangle ABC is an isosceles triangle, <note>In triangle ADE, side AD and side AE are equal</note>, therefore triangle ADE is an isosceles triangle.

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 ABD, <note>angle ADC is an exterior angle of the triangle</note>, <note>angle BAD and angle B are the two interior angles that are not adjacent to the exterior angle ADC</note>. According to the Exterior Angle Theorem of Triangle, the exterior angle ADC is equal to the sum of the two non-adjacent interior angles BAD and B, that is, <note>angle ADC = angle BAD + angle B</note>. Similarly, in triangle DEC, <note>angle AED is an exterior angle of the triangle</note>, <note>angle EDC and angle C are the two interior angles that are not adjacent to the exterior angle AED</note>. According to the Exterior Angle Theorem of Triangle, the exterior angle AED is equal to the sum of the two non-adjacent interior angles EDC and C, that is, <note>angle AED = angle EDC + angle C</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.

In the figure of this problem, one interior angle of triangle ABD is <note>∠BDA</note>, and the angle formed by extending the adjacent sides AD and BD of this interior angle is <note>∠ADC</note>, which is called the exterior angle of <note>interior angle ∠BDA</note>; one interior angle of triangle DEC is <note>∠CED</note>, and the angle formed by extending the adjacent sides DE and CE of this interior angle is <note>∠AED</note>, which is called the exterior angle of <note>interior angle ∠CED</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 isosceles triangle ABC, sides AB and AC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>∠ABC = ∠ACB</note>. <note>In the isosceles triangle ADE, sides AD and AE are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>∠ADE = ∠AED</note>.


Question: As shown in the figure, A, B, C are three points on ⊙O, and ∠CAO=25°, ∠BCO=35°, then the degree of ∠AOB is ()

Answer: 120°

1. Extend line segment CO to intersect circle O at point D.

2. According to the given condition ∠BCO=35°, based on the ##Inscribed Angle Theorem##, we obtain ∠BOD=2##∠BCD##=70°.

3. According to the given condition ∠CAO=25°, since CO=OA, ##based on the definition of an isosceles triangle##, triangle ACO is an isosceles triangle, therefore ##by the properties of an isosceles triangle##, ∠ACO=∠CAO=25°.

4. According to the ##Inscribed Angle Theorem##, we obtain ∠AOD=2##∠ACD##=2×25°=50°.

5. ####Finally, we obtain ∠AOB=∠AOD+∠BOD=50°+70°=120°.

圆; 圆心角; 圆周角; 弧; 弦

Definition of Inscribed Angle

<note>In circle O, the vertex C of ∠ACD is on the circumference</note>, <note>the two sides of ∠ACD intersect circle O at point A and point D respectively</note>. Therefore, <note>∠ACD is an inscribed angle</note>. Similarly, <note>the vertex C of ∠BCD is on the circumference</note>, <note>the two sides of ∠BCD intersect circle O at point B and point D respectively</note>. Therefore, <note>∠BCD is also an inscribed angle</note>.

Definition of Central Angle

<note>In the diagram of this problem, in circle O, points A and D 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 OD</note> is called the central angle <note>∠AOD</note>. Similarly, <note>in circle O, points D 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>OD and OB</note> is called the central angle <note>∠BOD</note>.

Definition of Isosceles Triangle

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

Inscribed Angle Theorem

In the figure of this problem, in circle O, <note>points A, B, C, D</note> are on the circle, the central angles corresponding to arc AD and arc DB are ∠AOD and ∠BOD, and the inscribed angles are <note>∠ACD and ∠BCD</note> respectively. According to the Inscribed Angle Theorem, ∠BOD is equal to twice the inscribed angle <note>∠BCD</note> corresponding to arc DB, that is, ∠BOD = 2<note>∠BCD</note> = 2 × 35° = 70°. Similarly, ∠AOD is equal to twice the inscribed angle <note>∠ACD</note> corresponding to arc AD, that is, ∠AOD = 2<note>∠ACD</note> = 2 × 25° = 50°.

Properties of Isosceles Triangle

<note>In the isosceles triangle OAC, side OC and side OA are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>angle OAC = angle OCA</note>.


Question: As shown in the figure, AB and CD are chords of ⊙O, and AB∥CD. If ∠BAD=36°, then ∠AOC equals ()

Answer: 72°

1. Given AB∥CD, ##according to the parallel postulate of parallel lines, if two parallel lines are intersected by a third line, then the alternate interior angles are equal.## It follows that ∠ADC=∠BAD=36°.

2. ##According to the inscribed angle theorem, in a circle, the inscribed angle is equal to half of the central angle that subtends the same arc.##

3. ##In this problem, the central angle ∠AOC and the inscribed angle ∠ADC subtend the same arc AC, so it follows that ∠AOC=2×∠ADC.##

4. Substitute the given value of ∠ADC into the formula from the previous step, i.e., ∠AOC=2×36°.

5. After calculation, it follows that ∠AOC=72°.

平行线; 圆; 弦; 圆心角; 圆周角

Definition of Alternate Interior Angles

When a straight line (referred to as a transversal) intersects two parallel lines, the angles that are located between the two parallel lines but on opposite sides of the transversal are called alternate interior angles.

<note>In the figure of this problem, two parallel lines AB and CD</note> are intersected by a <note>line AD</note>, among which <note>angle ABD and angle ADC are located between the two parallel lines and on opposite sides of the intersecting line AD</note>. Therefore, <note>angle ABD and angle ADC</note> are alternate interior angles. Alternate interior angles are equal, that is, <note>angle ABD is equal to angle ADC</note>.

Definition of Inscribed Angle

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

Definition of Central Angle

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

Parallel Postulate 2 of Parallel Lines

If two parallel lines are cut by a transversal, then the corresponding angles are equal, the alternate interior angles are equal, and the consecutive interior angles on the same side of the transversal are supplementary.

In the problem diagram, two parallel lines AB and CD are intersected by a third line AD, forming the following geometric relationships:
1. Alternate interior angles: <note>角BAD和角ADC相等</note>, i.e., <note>∠BAD = ∠ADC = 36°。</note>

Inscribed Angle Theorem

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


Question: As shown in the figure, the circle ⊙A with a diameter of 10 passes through points C(0, 5) and O(0, 0). B is a point on the major arc of ⊙A to the right of the y-axis. Then the measure of ∠OBC is ()

Answer: 30°

1. Suppose ⊙A intersects the x-axis at another point D, connect CD.

2. ∵ ∠COD=90°, according to ##(corollary 2 of the inscribed angle theorem) the inscribed angle subtended by the diameter is a right angle##, CD is the diameter of ⊙A.

3. Given the diameter is 10, we get CD=10.

4. ∵ The coordinates of point C are (0,5), and the coordinates of point O are (0,0), so the length of segment OC is 5.

5. Using the definition of sine, we get sin∠ODC=OC/CD.

6. Substituting the known conditions, we get sin∠ODC=5/10=1/2.

7. ∵ sin∠ODC=1/2, ####we get ∠ODC=30°.

8. ∵ ∠OBC is the inscribed angle corresponding to the same known arc COD, according to ##corollary 1 of the inscribed angle theorem, ∠OBC is also 30°##.

9. Through the above reasoning, we finally get the measure of angle OBC as 30°.

圆; 圆周角; 弧; 直角三角形

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>line segment CO is the diameter</note> because <note>CO passes through the center A</note>, both ends <note>C and O are on the circle</note>, and the problem states that <note>the diameter is 10</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 A, <note>point A is the center of the circle</note>, <note>point C is any point on the circle</note>, <note>line segment AC is the line segment from the center to any point on the circle</note>, therefore <note>line segment AC is the radius of the circle</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 <note>right triangle OCD</note>, <note>angle ODC is an acute angle</note>, <note>side OC is the opposite side of angle ODC</note>, and <note>side CD is the hypotenuse</note>. According to the definition of the sine function, the sine value of angle ODC is equal to <note>the ratio of the opposite side OC to the hypotenuse CD</note>, that is, <note>sin(∠ODC) = OC / CD</note>.

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.

In <note>circle A</note>, the <note>angle subtended by diameter CD at the circumference, ∠COD, is a right angle (90 degrees)</note>.

In a circle, any two inscribed angles that subtend the same arc are equal.

In the diagram of this problem, <note>in circle O</note>, <note>the inscribed angles ∠OBC and ∠ODC corresponding to arc OC are equal</note>. According to Corollary 1 of the Inscribed Angle Theorem, <note>the inscribed angles ∠OBC and ∠ODC corresponding to the same arc OC are equal</note>, that is, <note>∠OBC = ∠ODC = 30°</note>.

Definition of Inscribed Angle

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


Question: As shown in the figure, AB is the diameter of ⊙O, point C is on ⊙O, if ∠A=40°, then the degree of ∠B is ()

Answer: 50°

1. Given AB is the diameter of ⊙O, point C is on ⊙O, according to the theorem of the angle subtended by a diameter, ∠ACB = 90°.

2. Given ∠CAB = 40°, according to the triangle angle sum theorem, which states that the sum of the three interior angles of a triangle is equal to 180°, we get ∠CBA = 180° - ∠ACB - ∠CAB.

3. Substitute the given angles, we get ∠CBA = 180° - 90° - 40°.

5. Through the above reasoning, the final answer is 50°.

圆; 圆周角; 直角三角形

Triangle Angle Sum Theorem

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

<note>Angle CAB</note>, <note>Angle CBA</note>, and <note>Angle ACB</note> are <note>the three interior angles of triangle ACB</note>. According to the <note>Triangle Angle Sum Theorem</note>, <note>Angle CAB + Angle CBA + Angle ACB = 180°</note>.

In the figure of this problem, in circle O, the <note>diameter AB</note> subtends a <note>right angle ∠ACB (90 degrees)</note> at the circumference.


Question: As shown in the figure, points A, B, and C are three points on ⊙O, ∠BAC=40°, then the degree of ∠BOC is ()

Answer: 80°

1. Given that points A, B, and C are three points on circle O, and ∠BAC=40°.

2. According to the inscribed angle theorem, ##in a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.##

3. In this problem, the central angle ∠BOC subtends the arc ##BC##, and the inscribed angle ∠BAC also subtends the same arc BC.

4. Therefore, according to the inscribed angle theorem, ∠BOC=2×∠BAC.

5. Substituting the given condition, we get ∠BOC=2×40°=80°.

6. Through the above reasoning, the final answer is 80°.

圆; 圆周角; 圆心角; 弧

Definition of Central Angle

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

Definition of Inscribed Angle

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

Definition of Arc

An arc is a curve on a circle that is defined by two distinct points on the circle.

In the figure of this problem, there are two points <note>B and C</note> on <note>circle O</note>, and <note>arc BC</note> is a curve connecting these two points. According to the definition of an arc, <note>arc BC</note> is a curve between <note>two points B and C on the circle</note>.

Inscribed Angle Theorem

<note>In circle O</note>, points A, B, and C are on the circle, <note>the central angle corresponding to arc BC is ∠BOC</note>, <note>the inscribed angle is ∠BAC</note>. According to the Inscribed Angle Theorem, ∠BOC is equal to twice the inscribed angle ∠BAC corresponding to arc BC, i.e., <note>∠BOC = 2 × ∠BAC</note>.


Question: As shown in the figure, if AB is the diameter of ⊙O, CD is the chord of ⊙O, ∠ABD=55°, then the degree of ∠BCD is ()

Answer: 35°

1. Given AB is the diameter of ⊙O, connect AD.

2. According to the Corollary 2 of the Inscribed Angle Theorem, the inscribed angle subtended by the diameter is a right angle, so ∠ADB equals 90°.

3. Given ∠ABD equals 55°.

4. Since ∠ADB is a right angle, using the Triangle Sum Theorem, which states that the sum of the three interior angles of a triangle is 180°, we get: ∠A + ∠ABD equals 90°.

5. Substituting the given condition ∠ABD equals 55°, we get ∠A equals 90° - ∠ABD equals 90° - 55° equals 35°.

6. In ⊙O, according to Corollary 1 of the Inscribed Angle Theorem: ∠BAD and ∠BCD subtend the same arc BD, so ∠BCD equals ∠BAD.

7. Finally, we get ∠BCD equals ∠BAD equals 35°.

圆; 直角三角形; 圆周角; 弦; 线段

Definition of Inscribed Angle

The vertex A of <note>angle ∠DAB</note> is on the circumference, and the two sides of angle ∠DAB intersect circle O at <note>points D and B</note>. Therefore, angle ∠DAB is an inscribed angle. Similarly, the vertex C of <note>angle ∠BCD</note> is on the circumference, and the two sides of angle ∠BCD intersect circle O at <note>points B and D</note>. Therefore, angle ∠BCD is an inscribed angle.

Triangle Angle Sum Theorem

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

In the circle O, the diameter AB subtends the angle ∠ADB at the circumference, which is a right angle (90 degrees).

The inscribed angle <note>∠BCD</note> corresponding to <note>arc AD</note> is equal to the inscribed angle <note>∠BAD</note> corresponding to <note>arc AD</note>. According to <note>Corollary 1 of the Inscribed Angle Theorem</note>, the inscribed angles <note>∠BCD</note> and <note>∠BAD</note> corresponding to <note>the same arc AD</note> are equal, that is, <note>∠BCD = ∠BAD = 35°</note>.


Question: As shown in the figure, △ABC is inscribed in ⊙O, ∠A=15°, connect OB, then ∠OBC equals ()

Answer: 75°

1. Given △ABC inscribed in ⊙O, ∠A=15°.

2. Connect OB and OC.

3. According to the inscribed angle theorem, ##the inscribed angle corresponding to the same chord is half of the central angle##, so ∠BOC=2∠A.

4. Given ∠A=15°, we get ∠BOC=2×15°=30°.

5. Since OB=OC, △BOC is an isosceles triangle.

6. The property of isosceles triangles tells us that the two base angles are equal, therefore ∠OBC=∠OCB.

7. The triangle angle sum theorem states that ∠BOC + 2∠OBC = 180°.

8. Substituting the known conditions, 30° + 2∠OBC = 180°.

9. Solving the equation, we get 2∠OBC = 150°, therefore ∠OBC = 75°.

10. In conclusion, ∠OBC=75°.

圆周角; 圆; 线段; 弧

Definition of Isosceles Triangle

<note>Side OB and side OC are equal</note>, therefore <note>triangle BOC is an isosceles triangle</note>.

Inscribed Angle Theorem

<note>△ABC is inscribed in ⊙O</note>, ∠A=15°, and connect <note>BO and OC</note>, <note>∠BOC is the central angle, ∠A is the corresponding inscribed angle</note>, so according to the Inscribed Angle Theorem, <note>∠BOC = 2∠A</note>.

Properties of Isosceles Triangle

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

Triangle Angle Sum Theorem

Original text: <note>In triangle BOC</note>, <note>angle BOC, angle OBC, and angle OCB are the three interior angles of triangle BOC</note>. According to the Triangle Angle Sum Theorem, <note>angle BOC + angle OBC + angle OCB = 180°</note>.

Definition of Central Angle

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

Definition of Inscribed Angle

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


Question: As shown in the figure, BE and CF are the heights of △ABC. If AB=6, BC=5, EF=3, then the length of AE is ()

Answer: \frac{18}{5}

1. Given that BE and CF are the altitudes of △ABC, ##according to the definition of altitude, angle AEB and angle AFC are both 90° right angles##.

2. Since angle AEB and angle AFC are right angles, and angle A is a shared angle, according to the theorem of similarity of triangles (AA), ##if two corresponding angles of two triangles are equal, then the two triangles are similar, so △AEB is similar to △AFC##.

3. Since △AEB∽△AFC, according to the definition of similar triangles, ##it can be deduced that AE/AF = AB/AC##.

4. Similarly, since angle A is a shared angle, ##AE/AF = AB/AC, according to the theorem of similarity of triangles (SAS), it can be deduced that triangle AEF is similar to triangle ABC. According to the definition of similar triangles, it can be deduced that: EF/BC = AE/AB##.

5. Substituting the given conditions, AB = 6, BC = 5, EF = 3, calculate to get 3/5 = AE/6.

6. Solve AE by cross-multiplication to get AE = (18/5).

7. Through the above reasoning, the final answer is that the length of AE is 18/5.

垂线; 直角三角形; 线段; 普通三角形

Definition of Altitude

An altitude is a perpendicular line segment extending from a vertex of a triangle to the opposite side (or to the extension of the opposite side).

In the figure of this problem, <note>the line segment BE perpendicular to the opposite side AC from point B</note> is the altitude from vertex B. <note>The line segment CF perpendicular to the opposite side AB from vertex C</note> is the altitude from vertex C. <note>The line segment BE forms a right angle (90 degrees) with side AC</note>, <note>the line segment CF forms a right angle (90 degrees) with side AB</note>, which indicates <note>the line segments BE and CF are respectively the perpendicular distances from vertices B and C to the opposite sides AC and AB</note>.

Similarity Theorem for Triangles (AA)

Two triangles are similar if two of their corresponding angles are congruent.

In <note>triangle AEB and triangle AFC</note>, if <note>angle AEB is equal to angle AFC</note>, and <note>angle BAE is equal to angle CAF</note>, then <note>triangle AEB is similar to triangle AFC</note>.

Definition of Similar Triangles

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

<note>Triangle AEB and triangle AFC are similar triangles</note>. According to the definition of similar triangles: <note>angle AEB = angle AFC</note>, <note>angle BAE = angle CAF</note>, <note>angle ABE = angle ACF</note>; <note>AE/AF = AB/AC = BE/CF</note>. <note>Triangle AEF and triangle ABC are similar triangles</note>. According to the definition of similar triangles: <note>angle AEF = angle ABC</note>, <note>angle FAE = angle CAB</note>, <note>angle AFE = angle ACB</note>; <note>AE/AB = AF/AC = EF/BC</note>.

SAS Criterion for Similar Triangles

If two triangles have two pairs of corresponding sides in proportion and the included angle between those sides is equal, then the two triangles are similar.

<note>Side AE corresponds to side AB</note>, <note>side AF corresponds to side AC</note>, and <note>angle EAF = angle BAC</note>, so according to the Side-Angle-Side similarity criterion (SAS criterion), <note>triangle AEF is similar to triangle ABC</note>.


Question: As shown in the figure, △ABC is an inscribed triangle of ⊙O, BD is the diameter, if ∠DBC=18°, then the degree of ∠A is ()

Answer: 72°

1. As shown in the figure, connect OC.

2. ∵##OB=OC## are radii of ⊙O, ∴##∠OBC=∠OCB=18°##.

####

##3##. ∠BOC=180°-∠OBC-∠OCB=180°-18°-18°=144°.

##4##. According to the inscribed angle theorem, △ABC is an inscribed triangle of ⊙O, and the inscribed angle ∠A is equal to half of the central angle ∠BOC subtended by arc BC.

##5##. ∴∠A=1/2 ∠BOC=1/2 × 144°=72°.

##6##. Through the above reasoning, the final answer is 72°.

圆; 圆周角; 直角三角形

Definition of Radius

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

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 BOC, <note>side OB and side OC are equal</note>, therefore <note>triangle BOC is an isosceles triangle</note>.

Inscribed Angle Theorem

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

In circle O, points A, B, and C are on the circle, the central angle corresponding to arc BC is ∠BOC, and the inscribed angle is ∠BAC. According to the Inscribed Angle Theorem, ∠BAC is equal to half of the central angle ∠BOC corresponding to arc BC, i.e., ∠BAC = 1/2 ∠BOC.

Properties of Isosceles Triangle

In the figure of this problem, <note>in the isosceles triangle BOC, sides OB and OC are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the angles opposite the equal sides are equal</note>, that is, <note>∠OBC = ∠OCB = 18°</note>.

Triangle Angle Sum Theorem

In the <note>△BOC</note>, according to the <note>Triangle Angle Sum Theorem</note>, the sum of the interior angles of the triangle is equal to 180°. Therefore, <note>∠BOC=180°-∠OBC-∠OCB</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.

<note>In the diagram of this problem, in circle O, points C 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>OC</note> and <note>OB</note>, <note>∠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.

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


Question: As shown in the figure, in the ⊙O with a radius of 5, if the length of chord AB is 8, then the distance from the center to the chord OC is equal to ()

Answer: 3

1. Given that the radius OA of circle O is equal to 5, and the length of chord AB is 8. To find the distance from the center to the chord OC, we first connect OA.

2. ##Draw the perpendicular line OC from the center O to the chord AB##, let C be the foot of the perpendicular. ##According to the perpendicular bisector theorem##, OC is perpendicular to and bisects chord AB, thus AC is half of AB, i.e., AC is 4.

3. In Rt△OAC, given OA=5 and AC=4. According to the Pythagorean theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the two legs. We can find the length of OC: OC^2 = OA^2 - AC^2.

4. Substitute the known values into the Pythagorean theorem formula, we get OC^2 = 5^2 - 4^2.

5. Calculate to get OC^2 = 25 - 16, i.e., OC^2 = 9.

6. Taking the positive value of OC, we get OC = √9 = 3.

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

圆; 弦; 垂线; 直角三角形

Definition of Radius

<note>In circle O, point O is the center of the circle</note>, <note>point A is any point on the circle</note>, <note>the line segment OA is the line segment from the center of the circle to any point on the circle, thus the line segment OA is the radius of the circle</note>, and <note>OA=5</note>.

Definition of Chord

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

In the figure of this problem, in circle O, <note>points A and B are any two points on the circle</note>, <note>line segment AB connects these two points</note>, so <note>line segment AB is a chord of circle O</note>, and AB=8.

Chord Central Distance

The perpendicular distance from the center of a circle to a chord is referred to as the chord central distance.

In the figure of this problem, in circle O, <note>point O is the center of the circle</note>, <note>line segment AB is a chord of the circle</note>, <note>line segment OC is the perpendicular segment from the center O to the chord AB</note>. According to the definition of the chord central distance, <note>the length of line segment OC is the vertical distance from the center to the chord AB</note>, which is called the chord central distance.

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

In the figure of this problem, in circle O, <note>the perpendicular OC is perpendicular to chord AB</note>, then according to the Perpendicular Diameter Theorem, <note>OC bisects chord AB</note>, that is, <note>AC=CB=1/2AB=4</note>, and <note>OC bisects the two arcs subtended by chord AB</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.

In the figure of this problem, in the right triangle OAC, <note>∠OCA is a right angle (90 degrees)</note>, <note>sides OC and AC are the legs, and side OA is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OC^2 = OA^2 - AC^2</note>.


Question: As shown in the figure, AB is the diameter of ⊙O. If ∠BDC=40°, then the measure of ∠BOC is ()

Answer: 80°

1. Given AB is the diameter of ⊙O, and ∠BDC=40°.

2. According to the inscribed angle theorem, the measure of an inscribed angle is half the measure of its corresponding central angle. Therefore, ∠BDC is an inscribed angle, corresponding to the central angle ∠BOC.

3. From the inscribed angle theorem, given ∠BDC=40°, the central angle ∠BOC is twice that, i.e., ∠BOC=2×∠BDC.

5. Through the above reasoning, the final answer is 80°.

圆; 圆周角; 圆心角; 弧

Definition of Inscribed Angle

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

Definition of Central Angle

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

Inscribed Angle Theorem

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


Question: As shown in the figure, the radius of ⊙O is 10, AB is a chord, OC ⊥ AB, and the foot of the perpendicular is E. If CE = 4, then the length of AB is ()

Answer: 16

1. Given that the radius of circle O is 10, connect OA. Because radius OC is perpendicular to chord AB, ##according to the perpendicular bisector theorem##, AE is equal to BE and is half of AB.

2. Given OC=10 and CE=4, the length of OE is OC-CE##=10-4=##6.

3. In right triangle AOE, according to the Pythagorean theorem (the square of the hypotenuse in a right triangle is equal to the sum of the squares of the other two sides), we can deduce: the square of OA is equal to the square of OE plus the square of AE.

4. Specific calculation: the square of AE is equal to the square of OA minus the square of OE, then AE##=√(OA? - OE?)=√(10? - 6?)=##8.

5. Since AE is half of AB, AB##=2AE=2×8=##16.

6. Through the above reasoning, the final answer is 16.

圆; 弦; 垂线; 直角三角形; 线段

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>, points A, B, and C are any points on the circle, <note>line segments OA, OB, and OC are segments from the center of the circle to any point on the circle</note>, therefore <note>line segments OA, OB, and OC are the radii of the circle</note>. For example, <note>OA is the radius of ⊙O</note>, and its length is 10.

Definition of Chord

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

Definition of Right Triangle

In the figure of this problem, <note>triangle AOE</note>, <note>angle AEO is a right angle (90 degrees)</note>, therefore <note>triangle AOE</note> is a right triangle. Side OE and <note>side AE</note> are the legs, <note>side OA</note> is the hypotenuse.

Perpendicular Diameter Theorem

Original: <note>In circle O, diameter OC is perpendicular to chord AB</note>, then <note>according to the Perpendicular Diameter Theorem</note>, <note>diameter OC bisects chord AB, that is, AE=BE</note>, and <note>diameter OC bisects the two arcs subtended by chord AB, that is, arc AC=arc CB</note>.

Pythagorean Theorem

<note>△AOE is a right triangle</note>, <note>∠AEO is a right angle (90 degrees)</note>, side OE and <note>AE are the legs</note>, <note>side OA is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OC? = OE? + EC?</note>. Given OC=10, OE=6, therefore <note>EC=√(10? - 6?)</note>=√(100 - 36)=√64=8.


Question: As shown in the figure, MN is tangent to ⊙O at point A, ∠AOB=60°, then ∠BAM equals ()

Answer: 30°

1. Given MN is tangent to ⊙O at point A, according to the property of the tangent, the tangent is perpendicular to the radius of the circle, so ∠MAO=90°.

2. Given ∠AOB=60°, since OB=OA, △AOB is an isosceles triangle.

3. In an isosceles triangle, the two base angles are equal. Therefore, ∠OAB=∠OBA=(180°-∠AOB)÷2, i.e., ∠OAB=∠OBA=(180°-60°)÷2=60°.

4. Since ∠MAO=90°, and ∠OAB=60°, so ∠BAM=∠MAO-∠OAB, we can get ∠BAM=90°-60°.

5. Through the above reasoning, we conclude that ∠BAM=30°.

切线; 圆心角; 圆周角

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.

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

Definition of Isosceles Triangle

<note>Side OA and side OB are equal</note>, therefore triangle AOB is an isosceles triangle.

Property of the Tangent Line to a Circle

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

<note>In circle O, point A is the point of tangency between line MN 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 MN is perpendicular to the radius OA at the point of tangency A</note>, i.e., <note>∠MAO=90°</note>.


Question: As shown in the figure, AB is the diameter of ⊙O. If ∠BAC=35°, then ∠ADC=()

Answer: 55°

1. Given that AB is the diameter of ⊙O, since the angle subtended by a diameter is a right angle, we have ∠ACB=90°.

2. ∵∠BAC=35°, from ∠ACB=90°, we get ∠CAB + ∠ACB + ∠CBA = 180°, that is, ∠CBA = 180°-35°-90° = 55°.

####

3. ∵The angles subtended by the same arc are equal, ∴∠ADC=∠ABC. Therefore, ∠ADC=55°.

圆; 圆周角; 直角三角形; 线段; 圆内接四边形

Definition of Inscribed Angle

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

In circle O, the diameter AB subtends the angle ∠ACB at the circumference, which is a right angle (90 degrees).

Triangle Angle Sum Theorem

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

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

In a circle, any two inscribed angles that subtend the same arc are equal.

<note>The inscribed angles subtended by arc AC, ∠ADC and ∠ABC, are equal</note>, therefore <note>∠ADC=∠ABC</note>.


Question: As shown in the figure, in ⊙O, ∠ABC=40°, then ∠AOC=() degrees.

Answer: 80

2. According to the inscribed angle theorem, an inscribed angle is equal to half of the central angle that subtends the same arc. That is, in a circle, the inscribed angle is equal to half of its corresponding central angle.

3. In this problem, ∠ABC is the inscribed angle, and the central angle that subtends the same arc is ∠AOC.

4. By the inscribed angle theorem, ∠AOC=2∠ABC.

5. Substituting the given condition, ∠AOC=2×40°=80°.

6. Through the above reasoning, the final answer is ∠AOC=80°.

圆; 圆周角; 圆心角; 弧; 弦

Definition of Central Angle

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

Definition of Inscribed Angle

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

Inscribed Angle Theorem

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


Question: As shown in the figure, in △ABC, AB=10, AC=8, BC=6, a moving circle passing through point C and tangent to side AB intersects CA and CB at points P and Q respectively. The minimum length of segment PQ is ()

Answer: 4.8

1.##Since it is known that AB=10, AC=8, BC=6, according to##the converse of the Pythagorean theorem##, it can be concluded that ∠ACB=90°.##

2.##According to (Corollary 2 of the Inscribed Angle Theorem), the inscribed angle subtended by the diameter is a right angle. Since it is known that ∠QCP is an inscribed angle and is 90°, it follows that QP is the diameter of the circle.##

3.##As shown in the figure, let the midpoint of PQ be F, and let the point of tangency of circle F with AB be D. Connect FD, CF, CD.##

4. According to the problem statement, FC + FD = PQ, and according to##the triangle inequality theorem##, it follows that FC + FD > CD.

5. ##When point F is on CD, then FC + FD = CD, and CD is the diameter of the circle. According to the property of the tangent to a circle, CD⊥AB, that is,##when point F is on the altitude CD from the hypotenuse AB of the right triangle ABC, PQ=CD has the minimum value.##

6. According to##the area formula of a triangle##, CD = BC × AC ÷ AB.##

7. According to the given data, CD = 6 × 8 ÷ 10 = 4.8.##

8. Through the above reasoning, the final answer is 4.8.

圆; 切线; 线段; 垂线; 内错角

Definition of Tangent to a Circle

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

Definition of Right Triangle

<note>In triangle ABC, angle ACB is a right angle (90 degrees)</note>, therefore triangle ABC is a right triangle. <note>Side AC and side BC are the legs</note>, <note>side AB is the hypotenuse</note>.

Theorem of Triangle Inequality

In any triangle, the sum of the lengths of any two sides is greater than the length of the third side, and the absolute difference of the lengths of any two sides is less than the length of the third side.

<note>Side FC, side FD, and side CD form a triangle</note>. According to the theorem of triangle inequality, <note>the sum of any two sides is greater than the third side</note>, that is, <note>side FC + side FD > side CD</note>, <note>side FC + side CD > side FD</note>, <note>side CD + side FD > side FC</note>.

Area Formula of a Triangle

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

<note>In the figure of this problem, in triangle ABC, side AB is the base, and segment CD is the height.</note> According to the area formula of a triangle, <note>the area of triangle ABC is equal to the base AB multiplied by the height CD, then divided by 2</note>. ####<note>In triangle ABC, side AC is the base, and segment BC is the height.</note> According to the area formula of a triangle, <note>the area of triangle ABC is equal to the base AC multiplied by the height BC, then divided by 2</note>####

Converse of the Pythagorean Theorem

If the sum of the squares of two sides of a triangle is equal to the square of the third side, then the triangle is a right triangle, and the angle opposite to the longest side is a right angle.

Given that the three sides of triangle △ABC are 10, 8, and 6 respectively, and satisfy <note>10² = 8² + 6²</note>, then according to the Converse of the Pythagorean Theorem, <note>△ABC is a right triangle</note>, with the angle opposite the longest side <note>10 being a right angle at ACB</note>.

Original: <note>In circle F, point D is the point where line AB is tangent to the circle</note>, <note>line segment FD 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 FD at the point of tangency D</note>, i.e., <note>∠ADF=90 degrees</note>.

Midpoint of a Line Segment

A midpoint of a line segment is the point that divides the line segment into two equal parts.

<note>The midpoint of line segment QP is point F</note>. According to the definition of the midpoint of a line segment, <note>point F divides line segment QP into two equal parts</note>, that is, <note>the lengths of line segment QF and line segment FP are equal</note>. That is, <note>QF = FP</note>.

Definition of Radius

In the figure of this problem, <note>circle F</note>, <note>point F is the center of the circle</note>, <note>point D is any point on the circle</note>, <note>line segment DF is a line segment from the center to any point on the circle</note>, therefore <note>line segment DF is the radius of the circle</note>.

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

Definition of Altitude

An altitude is a perpendicular line segment extending from a vertex of a triangle to the opposite side (or to the extension of the opposite side).

In the figure of this problem, the line segment from vertex C perpendicular to the opposite side AB (or its extension) <note>CD is the altitude from vertex C</note>. The line segment <note>CD forms a right angle (90 degrees) with side AB (or its extension)</note>, which indicates that the line segment <note>CD is the perpendicular distance from vertex C to the opposite side AB (or its extension)</note>.


Question: As shown in the figure, point O is the center of ⊙O, points A, B, and C are on ⊙O, AO∥BC, ∠AOB=40°, then the degree of ∠OAC is equal to ()

Answer: 20°

1. Given that point O is the center of ⊙O, points A, B, and C are on ⊙O, AO∥BC, and ∠AOB=40°.

2. According to the Inscribed Angle Theorem, the inscribed angle subtended by the same arc in a circle is equal to half of the central angle subtended by that arc. Given ∠AOB=40°, then ∠ACB=1/2×40°=20°.

3. From the given condition AO∥BC, according to the Parallel Postulate 2, alternate interior angles are equal. Therefore, ∠OAC=∠ACB.

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

圆; 圆心角; 圆周角; 平行线; 垂线

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.

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

Parallel Postulate 2 of Parallel Lines

If two parallel lines are cut by a transversal, then the corresponding angles are equal, the alternate interior angles are equal, and the consecutive interior angles on the same side of the transversal are supplementary.

In the diagram of this problem, <note>line segment AO∥BC</note> is intersected by <note>line AC</note>, forming the following geometric relationship: <note>alternate interior angles: ∠OAC and ∠ACB are equal</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, <note>arc AB</note> and <note>arc ACB</note> correspond to the central angle ∠AOB. According to the Inscribed Angle Theorem, ∠ACB is equal to half of the central angle ∠AOB corresponding to <note>arc AB</note>, that is, <note>∠ACB = 1/2 ∠AOB</note>.

Definition of Alternate Interior Angles

When a straight line (referred to as a transversal) intersects two parallel lines, the angles that are located between the two parallel lines but on opposite sides of the transversal are called alternate interior angles.

<note>Two parallel lines OA and BC</note> are intersected by <note>a line AC</note>, where <note>angle OAC and angle ACB are located between the two parallel lines and on opposite sides of the intersecting line AC</note>, therefore <note>angle OAC and angle ACB are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle OAC is equal to angle ACB</note>.


Question: In the diagram above, angle OBC is congruent to angle OCB. How many degrees does angle A measure?

Answer: 35

1. Given that angle OBC is congruent to angle OCB, i.e., ∠OBC = ∠OCB, ####.

2. In △OBC, angle OBC = 55°, angle OCB = 55°, ##by the triangle angle sum theorem##, we get angle BOC = 180° - ∠OBC - ∠OCB = 180° - 55° - 55° = 70°.

3. Since AB and AC are chords of the circle, and angle BOC is the central angle, the corresponding inscribed angle is ∠BAC. According to the inscribed angle theorem, the inscribed angle is half of the corresponding central angle, so ∠BAC = 1/2 * ∠BOC.

4. Therefore, ∠BAC = 1/2 * 70° = 35°.

5. Through the above reasoning, the final answer is ∠BAC = 35°.

等腰三角形; 圆周角; 圆

Triangle Angle Sum Theorem

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

Inscribed Angle Theorem

In the diagram of this problem, <note>∠BAC is the inscribed angle</note>, <note>∠BOC is the corresponding central angle</note>. According to the Inscribed Angle Theorem, the inscribed angle is equal to half of the central angle that subtends the same arc, that is, <note>∠BAC = 1/2 × ∠BOC</note>, and <note>∠BOC = 70°</note>, so <note>∠BAC = 1/2 × 70° = 35°</note>.

Definition of Inscribed Angle

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

Definition of Central Angle

In the diagram of this problem, in circle <note>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 OB and OC is called the central angle##</note>.


Question: As shown in the figure, △ABC is inscribed in ⊙O, ∠BAC=120°, AB=AC, BD is the diameter of ⊙O, AB=3, then the value of AD is ()

Answer: 3√{3}

1. Given ∠BAC=120°, and AB=AC, according to the properties of an isosceles triangle, we get ∠ABC=∠ACB.

2. According to the properties of an isosceles triangle, in △ABC, AB=AC, ∠ACB=30°, ∠ACB=∠ADB=30°.

3. (Corollary 2 of the Inscribed Angle Theorem) The inscribed angle subtended by the diameter is a right angle, when BD is the diameter of ⊙O, ∠BAD is a right angle, i.e., ∠BAD=90°.

4. Because ∠ACB=30°, according to Corollary 1 of the Inscribed Angle Theorem, ∠ACB=∠ADB, so ∠ADB=30°.

5. From steps 3 and 4, we know that in the right triangle △BAD, ∠BAD=90°, ∠ADB=30°.

6. In the right triangle, according to the definition of the tangent function, tan(∠ADB)=opposite side/adjacent side, i.e., tan(30°)=AB/AD.

7. Since tan30°=√3/3, and given AB=3, we can solve for AD=AB/tan30°.

8. Substituting the known values, we get AD=3/(√3/3)=3√3.

9. After the above reasoning, the final answer is 3√3.

圆; 等腰三角形; 圆周角; 线段; 垂线

Definition of Isosceles Triangle

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

<note>In triangle ABC</note>, <note>sides AB and AC are equal</note>, therefore triangle ABC is an isosceles triangle.

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 BAD, <note>angle BAD is a right angle (90 degrees)</note>, so <note>triangle BAD is a right triangle</note>. <note>Side AB and side AD are the legs</note>, <note>side BD is the hypotenuse</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.

In the right triangle <note>BAD</note>, <note>angle ∠ADB is an acute angle</note>, <note>side AB is the side opposite to angle ∠ADB</note>, <note>side AD is the side adjacent to angle ∠ADB</note>, so <note>the tangent of angle ∠ADB is equal to the length of side AB divided by the length of side AD</note>, that is, <note>tan(∠ADB) = AB / AD</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 figure of this problem, <note>in isosceles triangle ABC</note>, <note>sides AB and AC are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the angles opposite the equal sides are equal</note>, that is, <note>angle ABC = angle ACB</note>.

In <note>circle O</note>, the inscribed angles <note>∠ADB</note> and <note>∠ACB</note> corresponding to arc <note>DC</note> are equal. According to Corollary 1 of the Inscribed Angle Theorem, the inscribed angles <note>∠ADB</note> and <note>∠ACB</note> corresponding to the same arc <note>DC</note> are equal, i.e., <note>∠ADB = ∠ACB</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 the figure of this problem, in circle O, the inscribed angle BAD subtended by diameter BD is a right angle (90 degrees)</note>. (Or <note>In the figure of this problem, the inscribed angle BAD is 90 degrees, so the chord BD subtended by it is a diameter</note>.)

Definition of Inscribed Angle

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


Question: Triangle BDC, shown above, has an area of 48. If ABCD is a rectangle, what is the area of the circle?

Answer: 36*\pi

1. Given ABCD is a rectangle, from the figure we get: CD=8.

2. According to the definition of a rectangle, CD=AB=8, ∠BCD=90°.

3. Based on the definition of a right triangle: Triangle BCD is a right triangle, and BC and CD are the legs of the right triangle, so side BC is the base of the right triangle BCD, and side CD is the height of the right triangle BCD (or side CD is the base of the right triangle BCD, and side BC is the height of the right triangle BCD). Given the area of triangle BDC is 48, according to the area formula of a right triangle, 48=1/2×CD×BC, we get CD×BC=8×BC=96, i.e., BC=12. Again, according to the definition of a rectangle, we get BC=AD=12.

4. Since the side length AD of the rectangle coincides with the diameter AD of the circle. Therefore, the diameter of the circle is 12, according to the definition of diameter, the radius is 12/2 = 6.

5. According to the area formula of a circle: π multiplied by the square of the radius, i.e., π * r². Here, the radius r is 6. Therefore, the area of the circle is π * 6² = 36π.

矩形; 直角三角形; 圆; 圆内接四边形

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 ABCD is a rectangle, <note>with its interior angles ∠DAB, ∠ABC, ∠BCD, ∠CDA all being right angles (90 degrees)</note>, and <note>sides AB and CD are parallel and equal in length, sides AD and BC are parallel and equal in length</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.

<note>Line segment AD is the diameter of the circle</note>, because it passes through <note>the center O</note> and both ends <note>A and D are on the circle</note>. According to the definition, the length of the diameter is <note>twice the radius</note>, that is <note>AD = 2r</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 the circle, <note>point O is the center of the circle</note>, <note>point A and point D are any two 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 Right Triangle

In the figure of this problem, in triangle BCD, <note>angle BCD is a right angle (90 degrees)</note>, thus <note>triangle BCD is a right triangle</note>. <note>Side BC and side CD are the legs</note>, <note>side BD is the hypotenuse</note>.

Area Formula of Right Triangle

The area of a right triangle is equal to half the product of the lengths of the two legs that meet at the right angle. Specifically, the area (A) can be calculated using the formula: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \).

<note>In the figure of this problem, in the right triangle BCD, angle BCD is a right angle (90 degrees)</note>, <note>sides BC and CD are the legs</note>, one leg as the base, the other leg as the height, so the area of the right triangle is equal to half the product of these two legs, that is, area = 1/2 * <note>side CD</note> * <note>side BC</note>.

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 circle is 6</note>. According to the area formula of a circle, <note>the area of the circle A is equal to the circumference π multiplied by the square of the radius 6</note>, that is, <note>A = π * 6^2 = 36π</note>.


Question: Find W R.

Answer: 8

1. Given WR=x+6, RT=2x+6, VT=10, SW=8.

2. According to the definition of vertical angles, angle WRS and angle TRV are vertical angles, thus we have ∠WRS = ∠TRV.

3. Because ∠W=∠T (same mark in the figure), ∠WRS = ∠TRV, according to the similarity theorem (AA), we know triangle WSR and triangle RTV are similar triangles.

4. According to the definition of similar triangles, we get SW/VT=WR/RT, substituting the corresponding side lengths from the problem: 8/10 = (x+6)/(2x+6).

5. We can solve this proportion. First, perform cross multiplication to get: 10*(x+6) = 8*(2x+6).

6. Expand the cross multiplication: 10x + 60 = 16x + 48.

7. Rearrange and combine like terms to get: 10x - 16x = 48 - 60, resulting in: -6x = -12.

8. Solve for x, resulting in: x = 2.

9. Therefore, substituting x=2 to calculate WR, we get: WR=x+6=2+6=8.

等腰三角形; 线段; 三角形的外角

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, two intersecting lines SR and RT intersect at point R, forming four angles: ∠WRS, ∠TRV, ∠SRT, and ∠WRV. According to the definition of vertical angles, ∠WRS and ∠TRV are vertical angles, ∠SRT and ∠WRV are vertical angles. Therefore, ∠WRS = ∠TRV, ∠SRT = ∠WRV.

Definition of Similar Triangles

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

<note>Triangles WSR and RTV are similar triangles</note>. According to the definition of similar triangles: <note>SW/VT = WR/RT</note>.

Similarity Theorem for Triangles (AA)

Two triangles are similar if two of their corresponding angles are congruent.

<note>∠W=∠T (marked the same in the diagram)</note>, <note>∠WRS = ∠TRV</note>, so <note>triangle WSR and triangle RTV are similar triangles</note>.


Question: As shown in the figure, point C is on the semicircle O with AB as the diameter, ∠BAC=20°, then ∠BOC equals ()

Answer: 40°

####

##1.## According to the inscribed angle theorem, the central angle subtended by the same arc is twice the inscribed angle subtended by that arc, i.e., ∠BOC=2∠BAC.

##2.## Since ∠BAC=20°, substituting into the formula gives ∠BOC=2*20°=40°.

##3.## Therefore, we finally get ∠BOC=40°.

圆; 圆周角; 直角三角形; 圆心角

Definition of Diameter

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

Inscribed Angle Theorem

In <note>circle O</note>, <note>points A, B, and C are on the circle</note>, <note>the central angle corresponding to arc AB and arc AC is ∠BOC</note>, and <note>the inscribed angle is ∠BAC</note>. According to the <note>Inscribed Angle Theorem</note>, <note>∠BOC is equal to twice the inscribed angle ∠BAC corresponding to arc AB</note>, that is, <note>∠BOC = 2∠BAC</note>.


Question: As shown in the figure, in △ABC, ∠ACB=90°, CD⊥AB at D, CD=4, BC=5, then AC=()

Answer: \frac{20}{3}

1. Given CD is perpendicular to AB at point D, according to the definition of a right triangle, we know that triangle BDC is a right triangle. Also, according to the Pythagorean theorem, we can obtain: BC^2 = BD^2 + CD^2.

2. From the transformation formula, we get: BD^2 = BC^2 - CD^2. Substituting the given conditions: CD = 4, BC = 5, we can obtain BD = √{5^2 - 4^2} = √{25 - 16} = √9 = 3.

3. Since CD is perpendicular to AB, it can be deduced that ∠BDC = ∠ADC = 90°.

4. Since triangles BCD and ABC share ∠B, and because ∠BDC and ∠ACB are both 90°, according to the similarity theorem (AA), we can conclude △BDC ∽ △ABC. Also, according to the definition of similar triangles, we have BC/AC = BD/CD.

5. BC/AC = BD/CD. Given: BC = 5, BD = 3, CD = 4.

6. Substituting the values, we get: 5/AC = 3/4.

7. Solving the equation, we get AC = 4 * (5/3) = 20/3.

8. Through the above reasoning, the final answer is 20/3.

Definition of Right Triangle

In the diagram of this problem, in triangle BDC, <note>CD⊥BD</note>, <note>angle BDC is a right angle (90 degrees)</note>, therefore <note>triangle BDC is a right triangle</note>. Side <note>BD and side DC are the legs</note>, <note>side BC is the hypotenuse</note>.

Definition of Perpendicular Lines

Two lines are said to be perpendicular if and only if the angle formed by their intersection is 90 degrees.

<note>The angle ∠CDA formed by the intersection of line CD and line AB is 90 degrees</note>, therefore according to the definition of perpendicular lines, <note>line CD and line AB</note> are perpendicular to each other.

Definition of Foot of a Perpendicular

The intersection point of a perpendicular line with the segment it is perpendicular to is called the foot of the perpendicular.

Original text: <note>点D是直线CD与直线AB的交点</note>，且<note>直线CD垂直于直线AB</note>，因此<note>点D是直线CD在直线AB上的垂足</note>。

Translation: <note>Point D is the intersection of line CD and line AB</note>, and <note>line CD is perpendicular to line AB</note>, therefore <note>point D is the foot of the perpendicular from line CD to line AB</note>.

Definition of Similar Triangles

<note>△BDC and △ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠BDC = ∠BCA</note>, <note>∠CBD = ∠ACB</note>, <note>∠BCD = ∠BAC</note>; <note>BC/AC = BD/CD</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 BDC</note>, <note>∠BDC is a right angle (90 degrees)</note>, <note>sides BD and CD are the legs</note>, <note>side BC is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>BC^2 = BD^2 + CD^2</note>.

Similarity Theorem for Triangles (AA)

In the original text: <note>triangle BDC and triangle ABC</note>, if <note>angle BDC is equal to angle BCA</note>, and <note>angle CBD is equal to angle ABC</note>, then <note>triangle BDC is similar to triangle ABC</note>.


Question: Which of the following must be true about the square above? I. a = b II. AC = BD III. b = c a. I only b. II only c. I and II only d. II and III only e. I, II, and III

Answer: I, II, and III

1. Given: Quadrilateral ABCD is a square, and diagonals AC and BD intersect at point O. According to the properties of the diagonals of a square, the two diagonals are equal and bisect each other perpendicularly. Therefore, point O is the midpoint of AC and BD.

2. Since the diagonals of the square are equal, AC = BD. According to option II, this option is true.

3. From the definition of a square, all four interior angles are 90 degrees. Since the four sides of the square are equal and the diagonals bisect each other, according to the definition of an isosceles right triangle, because AD = CD and ∠ADC = 90°, triangle ADC is an isosceles right triangle. Based on the properties of isosceles triangles and the triangle angle sum theorem, we can derive that a = b = (180° - ∠ADC) / 2 = (180° - 90°) / 2 = 90° / 2 = 45°. This indicates that option I is true.

4. Similarly, because BC = CD and ∠BCD = 90°, triangle BCD is an isosceles right triangle. Based on the properties of isosceles triangles and the triangle angle sum theorem, we get: c = ∠BDC = 45°, therefore all a, b, c are equal, i.e., a = b = c = 45°, so option III is true.

正方形; 对称; 等腰三角形

Definition of Square

A quadrilateral is a square if and only if all four sides are congruent (≅) and all four interior angles are right angles (90°).

In the figure of this problem, in quadrilateral ABCD, <note>sides AB, BC, CD, and DA are equal</note>, and <note>angles ∠DAB, ∠ABC, ∠BCD, and ∠CDA are all right angles (90 degrees)</note>, so ABCD is a square.

Triangle Angle Sum Theorem

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

In the figure of this problem, in triangle ADC, <note>angle ADC</note>, <note>angle a</note>, and <note>angle b</note> are the three interior angles of triangle ADC. According to the Triangle Angle Sum Theorem, <note>angle ADC + angle a + angle b = 180°</note>. In triangle BCD, <note>angle BDC</note>, <note>angle c</note>, and <note>angle BCD</note> are the three interior angles of triangle BCD. According to the Triangle Angle Sum Theorem, <note>angle BDC + angle c + angle BCD = 180°</note>.

Definition of Isosceles Right Triangle

An isosceles right triangle is a triangle with two sides of equal length and one angle measuring 90 degrees.

In the figure of this problem, <note>triangle ADC is an isosceles right triangle</note>, where <note>angle ADC is a right angle (90 degrees)</note>, <note>sides AD and CD are equal legs</note>. <note>Triangle BDC is an isosceles right triangle</note>, where <note>angle DCB is a right angle (90 degrees)</note>, <note>sides BC and CD are equal legs</note>.

Properties of Diagonals in a Square

The diagonals of a square are the line segments that connect opposite vertices. The diagonals of a square are equal in length, and they bisect each other perpendicularly.

In the diagram for this problem, in square ABCD, <note>the diagonals AC and BD are segments connecting opposite corners</note>. According to the properties of diagonals in a square, <note>AC and BD are equal</note>, and <note>AC and BD bisect each other perpendicularly</note>, forming <note>four 90-degree angles</note> at their intersection point. Therefore, <note>AC = BD</note>, and at their intersection point, they are <note>perpendicular to each other</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 isosceles triangle ADC, sides AD and CD are equal</note>. Therefore, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, i.e., <note>angle DAC = angle DCA</note>. <note>In the isosceles triangle BCD, sides BC and DC are equal</note>. Therefore, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, i.e., <note>angle CBD = angle CDB</note>.


Question: In the figure above, point B is located on side AC. If 55 < x < 60, what is one possible value of y?

Answer: 123

1. The figure given in the problem indicates that in triangle ACD, point B is located on side AC.

2. Based on the positional relationship, triangles ABD and BCD share side BD.

4. Since triangles ABD and BCD are adjacent and share side BD, according to the definition of supplementary angles, angle ABD and angle CBD are supplementary, i.e., the sum of angle ABD and angle CBD is 180 degrees.

5. From the figure, it is known that ∠ABD = x°, and 55° < x° < 60°.

6. According to the formula 180° - x = y, y° = angle CBD = 180° - x°.

7. Since the range of x is 55 < x < 60, the calculation for the range of angle CBD is 120 < y < 125.

8. To meet the condition 55 < x < 60, choose a possible value for x, such as 57°, to determine the degree of angle ABD, i.e., the degree of y.

9. The corresponding value of y is calculated as 180° - 57° = 123°.

10. The value of y is 123, which satisfies the range 120 < y < 125.

11. Through the above reasoning, the final answer is that the value of y can be 123.

三角形的外角; 普通三角形

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.

<note>Angle ABD and angle CBD have a common side BD</note>, <note>their other sides AB and CB are extensions in opposite directions</note>, so angle ABD and angle CBD are adjacent supplementary angles.


Question: Find the value of x in the figure.

Answer: 68

1. According to the exterior angle sum theorem, the sum of the exterior angles of any polygon is 360°. 2. In the figure, we can see that (x-1)°, 79°, (x+10)°, 2x° are the exterior angles of the quadrilateral, so the sum of these exterior angles is equal to 360°, hence: (x-1)° + 79° + (x+10)° + 2x° = 360°. 3. Simplify the equation: x - 1 + 79 + x + 10 + 2x = 360. 4. Further simplification gives: 4x + 88 = 360. 5. By moving the constant value to the other side of the equation, we get: 4x = 360 - 88. 6. Calculate the result of the above equation to get: 4x = 272. 7. Finally, solve for x: x = 272 / 4 = 68. 8. Check substitution for confirmation: when x = 68, (x-1)° = 67°, 79°, (x+10)° = 78°, 2x° = 136°. 9. Since the sum of the interior angles (x-1)°, 79°, (x+10)° is 67° + 79° + 78° = 224°, and 2x° is 136°, 224° + 136° = 360°, which matches the verification.

三角形的外角; 普通三角形; 线段

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

In the diagram of this problem, starting from the point of (x-1)° in a clockwise direction, the polygon is designated as <note>polygon ABCD</note>. Extend the sides <note>AD, AB, BC, CD</note> towards the directions of <note>points A, B, C, D</note> respectively, and denote the corresponding extensions as <note>EA, FB, GC, HD</note>. The interior angles of polygon ABCD are <note>∠BAD, ∠ABC, ∠BCD, ∠CDA</note>. One of the interior angles of polygon ABCD is <note>∠BAD</note>, and the angle <note>∠EAB</note> formed by extending the adjacent sides <note>AD and AB</note> of this interior angle is called the exterior angle of the interior angle <note>∠BAD</note>. The second interior angle of polygon ABCD is <note>∠ABC</note>, and the angle <note>∠FBC</note> formed by extending the adjacent sides <note>AB and BC</note> of this interior angle is called the exterior angle of the interior angle <note>∠ABC</note>. The third interior angle of polygon ABCD is <note>∠BCD</note>, and the angle <note>∠GCD</note> formed by extending the adjacent sides <note>BC

Exterior Angle Sum Theorem of Polygon

For any polygon, the sum of its exterior angles is equal to 360°.

Original: <note>四边形的外角</note>分别是<note>(x-1)°</note>、<note>79°</note>、<note>(x+10)°</note>和<note>2x°</note>。根据<note>多边形外角和定理</note>，这些外角的和等于<note>360°</note>，即<note>(x-1)° + 79° + (x+10)° + 2x° = 360°</note>。

Translation: The exterior angles of the quadrilateral are <note>(x-1)°</note>, <note>79°</note>, <note>(x+10)°</note>, and <note>2x°</note>. According to the <note>Exterior Angle Sum Theorem of Polygon</note>, the sum of these exterior angles is <note>360°</note>, i.e., <note>(x-1)° + 79° + (x+10)° + 2x° = 360°</note>.


Question: The radius of circle J is 10 units, the radius of \\odot K is 8 units, and BC = 5.4 units. Find JK.

Answer: 12.6

1. The given conditions are that the radius of circle J is 10 units, the radius of circle K is 8 units, and BC=5.4 units. We need to find the length of JK.

2. ##JB=JC-BC=10-5.4=4.6，CK=KB-BC=8-5.4=2.6##.

3. ##JK=JB+BC+CK=4.6+5.4+2.6=12.6##

####

##4##. Finally, we get JK=12.6 units.

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, <note>in circle J, point J is the center</note>, points A and C are <note>any points on the circle</note>, <note>segments JA and JC are segments from the center to any point on the circle</note>, therefore <note>segments JA and JC are the radii of circle J</note>. Similarly, <note>in circle K, point K is the center</note>, points B and D are <note>any points on the circle</note>, <note>segments KB and KD are segments from the center to any point on the circle</note>, therefore <note>segments KB and KD are the radii of circle K</note>.


Question: Find m \angle 1.

Answer: 79

1. ##Let the three vertices of the triangle be A, B, and C. Given angles ∠ABC = 27° and ∠BCA = 52°, let ∠1 be ∠DAC. Find the degree measure of ∠1##.

2. According to the triangle angle sum theorem (the sum of the three interior angles of a triangle is equal to 180°), we can calculate the degree measure of ∠CAB.

3. Based on the given conditions, we get: ∠CAB = 180° - ∠ABC - ∠BCA = 180° - 27° - 52° = 101°.

4. ##According to the definition of adjacent supplementary angles, ∠CAD and ∠CAB share a common side AC, and AD and AB are extensions in opposite directions, so ∠CAD and ∠CAB are adjacent supplementary angles##.

5. ##Based on the definition of a straight angle,## we obtain: ##∠1=∠CAD = 180° - ∠CAB = 180° - 101° = 79°##.

6. Through the above reasoning, the final answer is ##∠1## = 79°.

普通三角形; 三角形的外角; 邻补角

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.

In the figure of this problem, <note>triangle ABC is a geometric figure composed of three non-collinear points A, B, C and their connecting line segments AB, BC, AC</note>. <note>Points A, B, C are the three vertices of the triangle</note>, <note>and line segments AB, BC, AC are the three sides of the triangle</note>.

Triangle Angle Sum Theorem

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

Definition of Linear Pair of Angles

<note>Angle CAB and angle CAD share a common side CA</note>, and their <note>other sides AB and AD are extensions in opposite directions</note>, so <note>angle CAB and angle CAD are adjacent supplementary angles</note>.

Definition of Straight Angle

A straight angle is formed when a ray rotates around its endpoint and the initial side and terminal side lie on the same line but in opposite directions. A straight angle measures 180 degrees.

<note>Ray BA rotates around endpoint A</note> to <note>form a straight line with the initial side</note>, forming <note>straight angle BAD</note>. According to the <note>definition of a straight angle</note>, <note>a straight angle measures 180 degrees</note>, that is, <note>angle BAD = 180 degrees</note>.


Question: As shown in the figure, from point P outside ⊙O, draw the tangents PA and PB to the circle, with points of tangency at A and B respectively. If ∠APB = 70°, then the degree measure of the minor arc AB enclosed by these two tangents is ()

Answer: 110°

1. Given PA and PB are tangents drawn from an external point P to ⊙O, ##according to the property of the tangent to a circle##, the points of tangency A and B are right angles, i.e., ∠PAO=∠PBO=90°.

2. Since ∠APB=70°, ##according to the theorem of the sum of the interior angles of a quadrilateral, the sum of the interior angles of quadrilateral APBO is 360°##.

3. ##Therefore, ∠AOB=360°-90°-90°-70° equals 110°##.

4. ##The problem asks for the measure of the minor arc AB formed by the two tangents, evidently according to the property of the central angle, the measure of this minor arc AB is the central angle ∠AOB which equals 110°##.

圆; 切线; 弧; 圆周角; 等腰三角形

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.

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

Definition of Radius

In the figure of this problem, in circle O, <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 O to any point on the circle</note>, therefore <note>line segment OA and line segment OB are the radii of circle O</note>.

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 point of tangency of line <note>PA with the circle</note>, and segment <note>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>∠PAO = 90 degrees</note>. Similarly, point B is the point of tangency of line <note>PB with the circle</note>, and segment <note>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>∠PBO = 90 degrees</note>.

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

In <note>quadrilateral APBO</note>, <note>angle A, angle P, angle B, and angle O are the four interior angles of the quadrilateral</note>, according to the Sum of Interior Angles of a Quadrilateral Theorem, <note>the sum of these four interior angles is 360°</note>, i.e., <note>angle A + angle P + angle B + angle O = 360°</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 original text: <note>In circle O, point A 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>OA and OB, ∠AOB</note> is called the <note>central angle</note>.

Properties of Central Angles

The measure of a central angle is equal to the measure of the arc that it intercepts.

The arc corresponding to <note>central angle ∠AOB</note> is <note>arc AB</note>. According to the properties of central angles, the measure of the central angle is equal to the measure of the corresponding arc, that is, <note>the measure of ∠AOB = the measure of arc AB</note>.


Question: Triangle L M N is an equilateral triangle, and M P bisects L N. Find the side length of triangle L M N.

Answer: 10

1. The given condition is that triangle LMN is an equilateral triangle. ##According to the definition of an equilateral triangle:## all sides are equal, i.e., LM equals MN equals LN.

2. ##Given LM=3x+1, MN=4x-2, LM=MN, i.e., 3x+1=4x-2##.

####

##3. Solving the equation: 1 + 2 = 4x - 3x, 3 = x.##

##4. Substituting x=3 into 3x+1, we get the side length of LN: 3*3 + 1 = 9 + 1 = 10.##

##5##. Therefore, the side length of triangle LMN is 10, i.e., each side length is 10.

##6##. Through the above reasoning, the final answer is that the side length of triangle LMN is 10.

等边三角形; 线段; 垂线

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>Triangle L M N is an equilateral triangle</note>. <note>Sides L M, M N, and L N are of equal length</note>, and <note>angles ∠L M N, ∠M N L, and ∠N L M are equal in measure, each being 60°</note>.


Question: Find x.

Answer: 30

1. According to the definition of a right triangle, one angle of the triangle in the figure is a right angle, so this triangle is a right triangle, and the other two angles are x° and (2x)° respectively.

2. According to the triangle angle sum theorem, the sum of the three interior angles of a triangle is equal to 180°.

3. Let the degrees of the three interior angles of the triangle be x°, 2x°, and 90° respectively. According to the angle sum theorem, we have: x° + 2x° + 90° = 180°.

4. Simplify the equation x° + 2x° + 90° = 180°, we get 3x° + 90° = 180°.

5. From the equation 3x° + 90° = 180°, subtract 90° to get 3x° = 90°.

6. Solve the equation 3x° = 90°, we get x° = 30°.

7. Through the above reasoning, the final answer is x = 30.

直角三角形; 三角形的外角

Definition of Right Triangle

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

<note>One angle is a right angle (90 degrees)</note>, therefore the triangle is a right triangle. <note>Side x and side 2x are the legs</note>, <note>the side opposite the right angle is the hypotenuse</note>.

Definition of Triangle

In the figure of this problem, the triangle is composed of <note>three interior angles x°, 2x°, and 90°</note>. The <note>three vertices</note> of the triangle are the vertices of the three interior angles, and the <note>three sides</note> are the segments connecting these vertices.

Triangle Angle Sum Theorem

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

In the figure of this problem, the three interior angles of the triangle are <note>x°</note>, <note>2x°</note>, and <note>90°</note>. According to the Triangle Angle Sum Theorem, <note>x° + 2x° + 90° = 180°</note>.


Question: As shown in the figure, the side length of square ABCD is 3. Equilateral triangle PCD and equilateral triangle QCD are constructed on both sides of CD. What is the length of PQ?

Answer: 3√{3}

1. In square ABCD, side lengths AB=BC=CD=DA=3.

2. According to the problem statement, equilateral triangle PCD and equilateral triangle QCD are constructed on either side of CD.

##3. From the properties of equilateral triangles, in equilateral triangle DCQ, DC=DQ=CQ=3.##

##4. Connect PQ, noting that since both sides are equilateral triangles, PD=DQ=QC=CP, so quadrilateral PCQD is a rhombus, and PQ and CD are the diagonals of rhombus PCQD.##

##5. According to the definition of a rhombus, line segments PQ and CD are perpendicular bisectors of each other and intersect at point E, ∠DEP = ∠DEQ = 90°, DE = 1/2DC = 1.5##.

####

##6. From the properties of right triangles, since ∠DEQ=90°, triangle DEQ is a right triangle.##

7. Calculate the side lengths in the triangle: In right triangle DEQ, by the Pythagorean theorem: ##QE = √(DQ^2- ##DE^2##) = √(3^2 - 1.5^2) = √(9 - 2.25) = √6.75 = (3√3)/2##.

8. Since PQ = 2 * QE, we finally get PQ = 2 * (3√3)/2 = 3√3.

9. Through the above reasoning, we finally obtain the answer as 3√3.

等边三角形; 正方形; 旋转; 对称; 平移

Definition of Square

A quadrilateral is a square if and only if all four sides are congruent (≅) and all four interior angles are right angles (90°).

In quadrilateral <note>ABCD</note>, <note>sides AB, BC, CD, and DA</note> are equal, and <note>angles ∠ABC, ∠BCD, ∠CDA, and ∠DAB</note> are all <note>right angles (90 degrees)</note>, so <note>ABCD is a square</note>.

Definition of Equilateral Triangle

<note>Triangles PCD and QCD are equilateral triangles</note>. <note>The lengths of sides PC, CD, and PD are equal</note>, and <note>the measures of angles PCD, DPC, and PDC are equal, each being 60°</note>. Similarly, <note>the lengths of sides QC, CD, and QD are equal</note>, and <note>the measures of angles QCD, DQC, and QDC are equal, each being 60°</note>.

Definition of Right Triangle

<note>In triangle DEQ, angle DEQ is a right angle (90 degrees)</note>, therefore <note>triangle DEQ is a right triangle</note>. <note>Side DE and side EQ are the legs, and side DQ is the hypotenuse</note>.

Definition of Rhombus

A quadrilateral is a rhombus if and only if all four sides are of equal length, and its diagonals are perpendicular bisectors of each other.

<note>In the figure of this problem, in the quadrilateral DPCQ, all sides DP, CP, CQ, and DQ are equal</note>, thus the quadrilateral DPCQ is a rhombus. Additionally, <note>the diagonals PQ and CD of the quadrilateral DPCQ are perpendicular bisectors of each other</note>, <note>the diagonals PQ and CD intersect at point E</note>, and angle DEQ is a right angle (90 degrees), <note>DE=CE and QE=PE</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.

In the <note>right triangle DEQ</note>, <note>∠DEQ is a right angle (90 degrees)</note>, <note>sides DE and EQ are the legs</note>, <note>side DQ is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>DQ^2 = DE^2 + EQ^2</note>.


Question: Find x such that the quadrilateral is a parallelogram.

Answer: 4

1. Given condition: A quadrilateral is a parallelogram. According to the properties of a parallelogram, the opposite sides of a parallelogram are equal.

2. Based on the above property, we can set the left side and the right side equal, and write the equation: 2x + 3 = x + 7.

3. Solve the above equation: 2x + 3 = x + 7 -> 2x - x = 7 - 3 -> x = 4.

####

##4##. Through the above reasoning, the final answer is x = 4.

平行四边形; 平行线; 线段

Definition of Parallelogram

A quadrilateral is a parallelogram if and only if it has two pairs of opposite sides that are both parallel (∥) and congruent (≅).

In the figure of this problem, the quadrilateral is given as a parallelogram. According to the definition, this means <note>the left side (2x + 3) and the right side (x + 7) are equal and parallel</note>, <note>the top side (3y - 5) and the bottom side (y + 11) are also equal and parallel</note>.

Properties of Parallelogram Theorem

In a parallelogram, opposite angles are equal, opposite sides are equal, and the diagonals bisect each other.

<note>The opposite sides of a parallelogram, 2x + 3 and x + 7, are equal</note>, so set 2x + 3 = x + 7; <note>The opposite sides, 3y - 5 and y + 11, are equal</note>, so set 3y - 5 = y + 11.


Question: In the figure above, the ratio of the length of AB to the length of AC is 2 : 5. If AC = 25, what is the length of AB?

Answer: 10

1. The given condition is that the ratio of the length of segment AB to the length of segment AC is 2:5, that is, segment AB: segment AC = 2:5.

2. Given AC = 25, according to the definition of applying ratios, i.e., the equality relationship AB/AC = 2/5, we can find the length of AB.

3. Substitute the given AC = 25 into the equality relationship, obtaining AB/25 = 2/5.

4. Solve the equation AB/25 = 2/5, multiplying by 25 to get AB = 25*(2/5).

5. Calculate 25*(2/5), which equals 10.

6. Through the above reasoning, the final answer is AB = 10.

Definition of Line Segment

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

In the figure of this problem, <note>line segment AC is a part of a straight line, including endpoint A, endpoint C, and all points between them</note>. <note>Line segment AB has two endpoints, which are A and B respectively, and every point on line segment AB is located between endpoint A and endpoint B</note>. <note>Line segment BC has two endpoints, which are B and C respectively, and every point on line segment BC is located between endpoint B and endpoint C</note>.


Question: Find the scale factor from Q to Q'.

Answer: 3

1. In this problem, the center C is the center of scaling between the two figures. CQ is the distance before scaling, and CQ' is the distance after scaling.

2. According to the measurement, CQ = 5.

3. From the given information, the line CQ' = CQ + x = 15.

5. Therefore, the scale factor from Q to Q' can be represented by the ratio of the lengths of these two segments: scale factor = Q'C / QC = x / 5.

6. Solving the equation: scale factor = Q'C / QC = x / 5 = 10 / 5 = 3.

7. So, the scale factor from Q to Q' is 3.

Scale Factor

The scale factor refers to the ratio of the lengths of corresponding line segments of a figure after scaling to the lengths of the corresponding line segments of the original figure. If the scale factor is \( k \), then the length of each side of the scaled figure is \( k \) times the length of the corresponding side of the original figure.

In the figure of this problem, <note>figure Q'</note> is obtained by scaling <note>figure Q</note> by a <note>scale factor k</note>. According to the definition of the scale factor, the length of each side of <note>the scaled figure Q'</note> is equal to <note>the corresponding side length of the original figure Q</note> multiplied by <note>k</note>. That is: <note>scaled side length = k × original side length</note>.


Question: Find the length of AC in the isosceles triangle ABC.

Answer: 7

1. Given triangle ABC is an isosceles triangle, and AB = 9x - 1, AC = 4x + 1, BC = 5x - 0.5, ##according to the figure, it is known that## AC = BC.

2. Set the expressions for AC and BC equal, i.e., 4x + 1 = 5x - 0.5.

3. Solve the equation 4x + 1 = 5x - 0.5, rearrange terms:

4. 4x + 1 - 5x + 0.5 = 0

5. -x + 1.5 = 0

6. x = 1.5

7. Substitute x = 1.5 into AC = 4x + 1, calculate:

8. AC = 4(1.5) + 1 = 6 + 1 = 7.

9. Therefore, the length of AC is 7.

Definition of Isosceles Triangle

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

In the figure of this problem, in triangle ABC, <note>side AC and side BC are equal</note>, thus <note>triangle ABC is an isosceles triangle</note>.