Question: For trapezoid ABCD shown above, AB = 24, AD = 23, and BC = 16. What is the length of segment CD?

Answer: 25

1. We know that in trapezoid ABCD, ##AD## is the base. To find the length of CD, we can use the properties of the trapezoid to decompose it into a rectangle and a triangle. Consider drawing an auxiliary line perpendicular to ##base AD##. This auxiliary line can connect point C with ##base AD## to form a new point E such that ##CE## is perpendicular to AD. Thus, ##ABCE## forms a rectangle.

2. Since ##ABCE## is a rectangle, we have AB = CE. Using the known AB = 24 and the properties of the rectangle, CE = AB, we get CE = 24.

3. In the trapezoid, knowing AD = 23, the total ##base## AD can be represented as ##AE## (the ##base## of the rectangle) and ED (forming the right-angle ##base## of △CED), i.e., AD = ##AE## + ED. Since ##ABCE## is a rectangle, ##AE## = 16 (##AE## = BC), we get 23 = 16 + ED, thus ED = 7.

4. By calculating ED, we apply the Pythagorean theorem in the right triangle CED to find CD.

5. The Pythagorean theorem (expressed here as CD^2 = CE^2 + ED^2) is applied to △CED. Substituting the known values, we get CD^2 = 24^2 + 7^2.

6. Solving this equation, we get CD^2 = 576 + 49 = 625.

7. Finding the positive result for CD, we get CD = √625 = 25.

8. Through the above reasoning, the final answer is 25.

梯形; 线段; 垂线; 直角三角形; 平行线

Definition of Trapezoid

A quadrilateral is a trapezoid if and only if it has exactly one pair of parallel sides.

<note>Side AD</note> and <note>side BC</note> are parallel, while <note>side AB</note> and <note>side CD</note> are not parallel. Therefore, according to the definition of a trapezoid, the quadrilateral ABCD is a trapezoid because it has exactly one pair of parallel sides.

Definition of Rectangle

<note>Quadrilateral ABCE is a rectangle</note>, with its interior angles <note>∠ABC</note>, <note>∠BCE</note>, <note>∠CEA</note>, <note>∠EAB</note> all being right angles (90 degrees), and <note>side AB is parallel and equal in length to side CE</note>, <note>side BC is parallel and equal in length to side AE</note>.

Definition of Right Triangle

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

Pythagorean Theorem

<note>Right triangle CED</note>, where <note>CE is a leg</note>, <note>ED is also a leg</note>, and <note>CD is the hypotenuse</note>. Therefore, according to the Pythagorean Theorem, <note>CD^2 = CE^2 + ED^2</note>. Substituting the known values, we get <note>CD^2 = 24^2 + 7^2</note>, and finally, we find <note>CD = 25</note>.


Question: In the diagram above, angle A is congruent to angle BED, and angle C is congruent to angle D. If the ratio of the length of AB to the length of EB is 5:1, and the area of triangle BED = 5*a^2 + 10, what is area of triangle ABC?

Answer: 125*a^2+250

1. Given that angle A is congruent to angle BED, and angle C is congruent to angle D, it is known from the given conditions that △ABC and △BED are similar triangles.

2. △ABC and △BED are similar, and the similarity ratio is AB : EB = 5 : 1. According to the equality of corresponding side ratios in similar triangles, the ratio of BC : BD is also 5 : 1.

3. According to the area ratio of similar triangles, the ratio of the areas is the square of the corresponding side length ratio #### (5:1), which is 25:1.

4. From the given conditions, the area of △BED is 5a^2 + 10.

5. Based on the area ratio of 25:1, the area of △ABC is 25 * (5a^2 + 10).

6. Calculate the area of △ABC as 25 * (5a^2 + 10) = 125a^2 + 250.

7. Through the above reasoning, the final answer is 125a^2 + 250.

普通三角形; 等腰三角形; 线段; 位似

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 ABC and BED are similar triangles</note>. According to the definition of similar triangles: <note>∠A = ∠BED</note>, <note>∠C = ∠D</note>, <note>∠ABC = ∠EBD</note>; <note>AB/EB = BC/BD = AC/ED = 5/1</note>.

Theorem on the Area Ratio of Similar Triangles

If two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides.

<note>Triangle ABC and triangle BED are similar triangles</note>, the ratio of side <note>AB</note> to side <note>EB</note> is <note>5:1</note>, which means the similarity ratio is <note>5:1</note>. Therefore, <note>the area ratio of triangle ABC to triangle BED is equal to (5:1)^2 = 25:1</note>.

Similarity Theorem for Triangles (AA)

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

<note>Angle A is equal to angle BED</note>, and <note>angle C is equal to angle D</note>, so <note>triangle ABC is similar to triangle BED</note>.


Question: As shown in the figure above, triangle ABC and line l. What is the sum of y and z?

Answer: 140

1. Let the left end of line l be labeled as point D. Given that ∠DAB = 140°, ∠CAB = x°, ∠BCA = z°, ∠ABC = y°. On line l, according to the definition of a straight angle, a straight angle is 180°, we have: x° + 140° = 180°.

2. From the above equation x° + 140° = 180°, by subtracting 140°, we get x° = 40°.

3. In triangle ABC, based on the triangle angle sum theorem, the sum of the interior angles of a triangle should be 180°, which can be expressed as: x° + y° + z° = 180°.

4. Substituting the known x° = 40° into the triangle angle sum equation, we get 40° + y° + z° = 180°.

5. Solving the equation from the previous step, subtracting 40°, we get y° + z° = 140°.

6. Through the above reasoning, the final answer is y + z = 140.

三角形的外角; 普通三角形; 线段

Triangle Angle Sum Theorem

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

In <note>triangle ABC</note>, <note>∠BAC</note>, <note>∠ABC</note>, and <note>∠BCA</note> are <note>the three interior angles of triangle ABC</note>. According to the <note>Triangle Angle Sum Theorem</note>, <note>∠BAC + ∠ABC + ∠BCA = 180°</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.

In the problem diagram, <note>the ray extending to the right from AC rotates around endpoint A to form a straight line with the initial side</note>, forming <note>straight angle CAD</note>. According to the definition of a straight angle, <note>a straight angle measures 180 degrees</note>, that is, <note>angle CAD = 180 degrees</note>.


Question: What is the length of side AC to the nearest hundredth? Assume ABC is a right triangle.

Answer: 10.6

1. Given that triangle ABC is a right triangle, and ∠BAC = 28°, AB = 12, according to the Pythagorean theorem, the relationship between the legs and the hypotenuse in triangle ABC is uniquely determined.

2. In a right triangle, trigonometric functions can be used to solve for unknown sides. Here, we know the length of one side and the angle of one acute angle, so we can use trigonometric functions to solve.

3. According to the definition of trigonometric functions, cos(θ) = adjacent/hypotenuse, where θ is the acute angle, the adjacent side is what we are solving for, AC, and the hypotenuse is the known side AB.

4. To find the length of AC, apply the cosine function: cos(∠BAC) = AC/AB.

5. In this problem, cos(28°) = AC/12, so AC = 12 * cos(28°).

6. By looking up the value or using a calculator, we find cos(28°) ≈ 0.8829.

7. Therefore, AC ≈ 12 * 0.8829 ≈ 10.5948.

8. Rounding the result to the nearest hundredth, we get AC ≈ 10.60.

直角三角形; 正弦; 余弦

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

Cosine Function

The cosine function is a trigonometric function that can be defined via the sides of a right triangle. For an angle α, its cosine value is equal to the ratio of the length of the adjacent side to the hypotenuse. In the unit circle, the cosine value is the x-coordinate of the point on the circle that makes an angle α with the positive x-axis.

<note>In the diagram of this problem, in the right triangle ABC, side AC is the adjacent side of angle ∠BAC, and side AB is the hypotenuse</note>. Therefore, according to the definition of the cosine function, <note>the cosine value of angle ∠BAC</note> is equal to <note>the ratio of the adjacent side AC to the hypotenuse AB</note>, that is, <note>cos(∠BAC) = AC / AB</note>.


Question: In quadrilateral PQRS above, what is the value of a^2 + b^2?

Answer: 13

1. Given that in quadrilateral PQRS, PQ = a, PS = b, QR = 3, RS = 2. In quadrilateral PQRS, connecting QS forms two right triangles.

2. In △PQS, ∠SPQ is a right angle. According to the Pythagorean theorem, in a right triangle, the sum of the squares of the three sides is: a^2 + b^2 = (QS)^2.

3. In △RQS, ∠QRS is a right angle. According to the Pythagorean theorem, in a right triangle, the sum of the squares of the three sides is: 3^2 + 2^2 = (QS)^2.

4. Calculate the square of QS: 3^2 + 2^2 = 9 + 4 = 13.

5. Therefore, we conclude that a^2 + b^2 = 13.

直角三角形; 普通四边形

Definition of Right Triangle

In the figure of this problem, in triangle PQS, <note>angle SPQ is a right angle (90 degrees)</note>, therefore triangle PQS is a right triangle. Side PQ and side PS are the legs, and side QS is the hypotenuse. Similarly, in triangle QSR, <note>angle QRS is a right angle (90 degrees)</note>, therefore triangle QSR is a right triangle. Side QR and side RS are the legs, and side QS is the hypotenuse.

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 △PQS, <note>PQ = a</note>, <note>PS = b</note>, <note>∠SPQ is a right angle</note>, according to the Pythagorean Theorem, we can obtain: PQ^2 + PS^2 = QS^2, that is, <note>a^2 + b^2 = QS^2</note>; in △QRS, <note>QR = 3</note>, <note>RS = 2</note>, <note>∠QRS is a right angle</note>, therefore: QR^2 + RS^2 = QS^2, that is, <note>3^2 + 2^2 = QS^2</note>, that is, 9 + 4 = 13. So we conclude that <note>a^2 + b^2 = 13</note>.


Question: In the figure above, AB = 6 and BC = 8. What is the length of segment BD?

Answer: 4.8

1. Given that AB = 6 and BC = 8, and it can be seen in the figure that ∠BDA is a right angle. According to the property of the altitude on the hypotenuse in a right triangle, △BDA and △BCA are similar triangles.

2. According to the definition of similar triangles, in △BDA and △BCA, the corresponding sides are proportional, that is, BA/AC = BD/BC.

3. In △BCA, according to the Pythagorean theorem, AC can be found to be 10. Substituting the given conditions into this proportional relationship, BA/AC = BD/BC can be simplified to 6/10 = BD/8.

4. Solving the proportion yields BD = 6/10*8 = 48/10 = 4.8.

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

直角三角形; 垂线; 线段

Definition of Similar Triangles

<note>Triangle BDA and triangle BCA are similar triangles</note>. According to the definition of similar triangles: <note>∠BDA = ∠CBA</note>, <note>∠A = ∠A</note>, <note>∠DBA = ∠C</note>; <note>BA/AC = BD/BC</note>.

In a right triangle, the two triangles formed by the altitude drawn to the hypotenuse are similar to the original triangle.

In <note>right triangle BAC</note>, <note>angle ABC is a right angle (90 degrees)</note>, and an altitude BD is drawn from <note>vertex B to hypotenuse AC</note>. According to the <note>Property of the Altitude on the Hypotenuse in a Right Triangle</note>, <note>altitude BD divides right triangle ABC into two new right triangles ABD and BCD</note>, and <note>triangle ABD is similar to triangle ABC</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, from vertex B perpendicular to the opposite side AC, <note>line segment BD is the altitude from vertex B</note>. <note>Line segment AC and side BD form a right angle (90 degrees)</note>, which indicates that <note>line segment BD is the perpendicular distance from vertex B to the opposite side AC</note>.

Pythagorean Theorem

<note>In right triangle ABC</note>, <note>angle ABC is a right angle (90 degrees)</note>, <note>sides BA and BC are the legs</note>, <note>side AC is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AC² = BA² + BC²</note>.


Question:
In the figure above, four line segments meet at a point to form four angles. What is the value of x?

(A) 180
(B) 240
(C) 288
(D) 312
(E) 336

Answer: 288

1. Given that 4 line segments intersect at one point, forming four angles which are x/2, x/4, x/6, and x/3 respectively. According to the ##definition of a full angle##, the sum of the four angles is equal to 360 degrees.

2. Let the four angles be ∠A, ∠B, ∠C, and ∠D, where ∠A = x/2, ∠B = x/4, ∠C = x/6, ∠D = x/3.

3. Since ∠A + ∠B + ∠C + ∠D = 360° (definition of a full angle).

4. Substitute the given conditions x/2 + x/4 + x/6 + x/3 = 360.

5. Find the common denominator of these angles. The common denominator is 12, so convert these fractions to have the same denominator: x/2 = 6x/12, x/4 = 3x/12, x/6 = 2x/12, x/3 = 4x/12.

6. Thus, we have 6x/12 + 3x/12 + 2x/12 + 4x/12 = 360.

7. Combine the fractions: (6x + 3x + 2x + 4x)/12 = 360.

8. Obtain the equation: (15x)/12 = 360.

9. Solve the equation: 15x = 360 * 12.

10. Calculate: 15x = 4320.

11. Solve the equation: x = 4320 / 15.

12. Calculate: x = 288.

13. Through the above reasoning, the final answer is 288.

对顶角; 邻补角; 线段

Definition of an Angle

An angle is a geometric figure formed by two rays sharing a common endpoint. This common endpoint is called the vertex of the angle, and the two rays are referred to as the sides of the angle.

Original: <note>In the diagram of this problem, the four angles are x/2, x/4, x/6, and x/3, formed by four rays intersecting at a point, where the point is the common endpoint</note>. According to the definition of an angle, <note>An angle is a geometric figure formed by two rays with a common endpoint</note>, where the <note>common endpoint is called the vertex of the angle</note>, and the <note>two rays are called the sides of the angle</note>.

Definition of Full Angle

An angle formed when a ray revolves around its endpoint for one complete rotation is called a full angle.

The <note>common ray of x°/4 and x°/2</note> rotates around the common vertex of the angles to form a <note>full angle</note>. According to the <note>definition of full angle</note>, the degree of the full angle is 360 degrees, that is <note>∠A + ∠B + ∠C + ∠D=360 degrees</note>.


Question: The perimeter of the rectangle above is p, and the area of the rectangle is 36. If l and w are integers, what is one possible value of p?

Answer: 24

1. Given that the perimeter of the rectangle is p and its area is 36. From the formula for the area of a rectangle, we know that l 	imes w = 36.

2. List all integer combinations for length and width: ##(36, 1)##, ##(18, 2)##, ##(12, 3)##, ##(9, 4)##, ##(6, 6)##.

3. Using the formula for the perimeter of a rectangle, p = 2(l + w) , calculate the perimeter for each possible combination:

4. Combination ##(36, 1)##, where ##l = 36##, ##w = 1##, gives a perimeter: p = 2(36 + 1) = 74.

5. Combination ##(18, 2)##, where ##l = 18##, ##w = 2##, gives a perimeter:  p = 2(18 + 2) = 40.

6. Combination ##(12, 3)##, where ##l = 12##, ##w = 3##, gives a perimeter: p = 2(12 + 3) = 30.

7. Combination ##(9, 4)##, where ##l = 9##, ##w = 4##, gives a perimeter: p = 2(9 + 4) = 26.

8. Combination (6, 6), where l = 6, w = 6, gives a perimeter: p = 2(6 + 6) = 24 .

9. From the above reasoning, one possible value for the perimeter p is 24.

矩形; 线段; 普通四边形

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, the four sides of the rectangle are two equal long sides and two equal short sides. Here, the length of the long sides is denoted by <note>l</note>, and the length of the short sides is denoted by <note>w</note>. Each angle of the rectangle is a <note>right angle (90 degrees)</note>.

Formula for the Area of a Rectangle

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

In the figure of this problem, in the rectangle, <note>side l and side w are the length and width of the rectangle</note>, so <note>the area of the rectangle = l * w</note> = 36.

Formula for the Perimeter of a Rectangle

The perimeter of a rectangle is equal to twice the sum of the lengths of its longer side (length) and its shorter side (width), which can be expressed as \( P = 2(l + w) \).

<note>In the figure of this problem, the long side of the rectangle is l, and the short side is w</note>, according to the <note>formula for the perimeter of a rectangle</note>, the perimeter of the rectangle is equal to twice the sum of its long side length and short side length, that is <note>P = 2(l + w)</note>.


Question:
Which of the following two segments from the figure create an obtuse angle?
A. BC and AC
B. EF and DF
C. CD and BD
D. AB and BD
E. EA and AC

Answer: line CD and line BD

1. According to the information in the figure, in triangle ADF, ∠AFD = 90 degrees.

2. Since ∠BDC is an exterior angle of triangle ADF, ∠BDC = ∠AFD + ∠FAD = 90 degrees + ∠FAD.

3. Since ∠BDC is greater than 90 degrees, ∠BDC is an obtuse angle.

线段; 邻补角; 直角三角形

Exterior Angle Theorem of Triangle

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

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

<note>One of the interior angles of polygon ADF is ∠ADF</note>. The angle formed by extending the adjacent sides <note>AD and DF, ∠FDC</note>, is called the exterior angle of the interior angle ∠ADF.


Question: If triangle ABC in the figure above is an equilateral triangle and D is a right angle, find the value of x.

Answer: 24

1. Since triangle ABC is an equilateral triangle, then sides AB = BC = AC.

2. Given the right triangle CDE, we know that angle D is 90 degrees.

3. In the equilateral triangle ABC, all interior angles are 60 degrees. Therefore, angle ACB equals 60 degrees.

4. Because line segments BD and AE intersect, according to the relationship of vertical angles being equal, angle ACB = angle DCE = 60 degrees.

5. Triangle CDE is a special 30°-60°-90° triangle.

6. CD is the short leg of this special triangle CDE.

7. CE is the hypotenuse of the special triangle, so CE = 2 * CD = 24.

8. Based on the results obtained, x = 24.

Through the above reasoning, we obtain the answer: 24.

等边三角形; 直角三角形; 线段

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 ABC is an equilateral triangle</note>. The lengths of <note>side AB</note>, <note>side BC</note>, and <note>side AC</note> are equal, and the degrees of <note>angle BAC</note>, <note>angle ABC</note>, and <note>angle ACB</note> are equal, all being <note>60°</note>.

Definition of Right Triangle

<note>Angle CDE is a right angle (90 degrees)</note>, therefore triangle CDE is a right triangle. <note>Side CD and side DE are the legs</note>, <note>side CE is the hypotenuse</note>.

Triangle Angle Sum Theorem

In the figure of this problem, in triangle ABC, <note>angle BAC</note>, <note>angle ABC</note>, and <note>angle BCA</note> are the three interior angles of triangle ABC. According to the Triangle Angle Sum Theorem, <note>angle BAC + angle ABC + angle BCA = 180°</note>. Similarly, in triangle CDE, <note>angle DCE</note>, <note>angle CDE</note>, and <note>angle DEC</note> are the three interior angles of triangle CDE. According to the Triangle Angle Sum Theorem, <note>angle DCE + angle CDE + angle DEC = 180°</note>.

Properties of 30°-60°-90° Triangle

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

<note>In the figure of this problem, in the 30°-60°-90° triangle CDE</note>, <note>angle DEC is 30 degrees</note>, <note>angle DCE is 60 degrees</note>, <note>angle CDE is 90 degrees</note>. <note>Side CE is the hypotenuse</note>, <note>side CD is the side opposite the 30-degree angle</note>, <note>side DE is the side opposite the 60-degree angle</note>. According to the properties of the 30°-60°-90° triangle, <note>side CD is equal to half of side CE</note>, <note>side DE is equal to √3 times side CD</note>. That is: <note>CD = 1/2 * CE</note>, <note>DE = CD * √3</note>.

Definition of Vertical Angles

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

<note>Two intersecting lines BD and AE intersect at point C</note>, forming two angles: <note>angle ACB, angle DCE</note>. According to the definition of vertical angles, <note>angle ACB and angle DCE are vertical angles</note>. Since <note>vertical angles are equal in measure</note>, therefore <note>angle ACB = angle DCE</note>.


Question: As shown in the figure, given that △ABC is inscribed in ⊙O and ∠BAC=50°, then the degree of ∠BOC is ()

Answer: 100°

1. Given that triangle ABC is inscribed in circle O, and ∠BAC=50°.

2. According to the inscribed angle theorem, ##an inscribed angle is equal to half of the central angle that subtends the same arc.## Therefore, the central angle corresponding to ∠BAC is ∠BOC.

3. Based on the inscribed angle theorem, we have ∠BOC equal to 2 times ∠BAC.

4. Calculating, we get ∠BOC=2×50°=100°.

5. Through the above reasoning, the final answer is 100°.

圆; 圆心角; 圆周角; 普通三角形

Definition of Inscribed Angle

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

In the figure of this problem, in circle O, <note>the vertex A of angle BAC is on the circumference</note>, <note>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

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

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

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>the central angle corresponding to arc BC is ∠BOC</note>, <note>the inscribed angle is ∠BAC</note>. According to the Inscribed Angle Theorem, <note>∠BAC is equal to half of the central angle ∠BOC corresponding to arc BC</note>, that is, <note>∠BAC = 1/2 ∠BOC</note>.


Question: In the triangle above, what is the average (arithmetic mean) of x, y, and z?

Answer: 60

1. Let the three vertices of the triangle be A, B, and C respectively. The known condition is that the three interior angles of the triangle are x°, y°, and z°. According to the triangle angle sum theorem, we get x° + y° + z° = 180°.

2. From step 1, x° + y° + z° = 180°, using the average formula (x° + y° + z°) / 3 = average, we can obtain the average as 180° / 3.

3. Through the above reasoning, the final answer is 60°.

普通三角形; 三角形的外角

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.

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

Triangle Angle Sum Theorem

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

In <note>triangle ABC</note>, <note>angle x°</note>, <note>angle y°</note>, and <note>angle z°</note> are the three interior angles of the triangle. According to the Triangle Angle Sum Theorem, <note>angle x° + angle y° + angle z° = 180°</note>.


Question: In the figure above, AC = 7 and AB = BC. What is the smallest possible integer value of AB?

Answer: 4

1. Given that the length of line AC is 7, and triangle ABC is an isosceles triangle, i.e., AB = BC. According to the properties of an isosceles triangle, the angles opposite the equal sides are also equal, ##thus ∠BAC = ∠BCA##.

####

##2##. Apply the triangle inequality theorem: the sum of any two sides of a triangle is greater than the third side, and derive the conclusions as follows.

##3##. First, according to the triangle inequality, we get: AB + BC > AC. Since AB = BC, we have 2AB > 7, so AB > 3.5.

##4##. Secondly, according to the triangle inequality, we get: AB + AC > BC. Since BC = AB, we have ##AC > 0##, and since AC = 7, ##AC = 7 > 0 always holds##.

##5##. Finally, according to the triangle inequality, we get: BC + AC > AB. Since BC = AB, we have ##AC > 0##, and since AC = 7, ##AC = 7 > 0 always holds##.

##6##. In conclusion, the smallest integer value of AB should be the smallest integer that satisfies AB > 3.5, which is 4.

##7##. Therefore, the smallest possible integer value of AB that meets all conditions is 4.

线段; 等腰三角形; 普通三角形

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 AB and side BC are equal</note>, therefore <note>triangle ABC is an isosceles triangle</note>.

Properties of Isosceles Triangle

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

<note>In the figure of this problem, in the isosceles triangle ABC, side AB and side BC are equal.</note> Therefore, according to the properties of the isosceles triangle, the angles opposite the equal sides are equal, that is, <note>∠BAC = ∠BCA</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 AB, side AC, and side BC form a triangle</note>. According to the theorem of triangle inequality, <note>side AB + side BC > side AC</note>, <note>side AB + side AC > side BC</note>, <note>side BC + side AC > side AB</note>.


Question: In the figure above, the circle has center P and radius r. Lines AB and AC are tangent to the circle. If M is the midpoint of segment AC and the measure of ∠PMC equals the measure of ∠MPC, what is the length, in terms of r, of segment PA?

Answer: r*\sqrt{5}

1. Given that segments AB and AC are tangents to the circle, according to the properties of tangents and circles, we can deduce that the lines connecting the center P to the points of tangency PB and PC are perpendicular to the tangents respectively.####

2. According to the condition that M is the midpoint of segment AC, we can deduce that AM = MC.

3. The problem states that the measure of ∠PMC equals the measure of ∠MPC.##Since PC is perpendicular to tangent AC, ∠PCN is 90°, thus triangle PMC is an isosceles right triangle, and PC = MC##

####

##4. Since AM = MC = PC = r, triangle APC is a right triangle, AC = 2MC = 2r, PC = r, therefore by the Pythagorean theorem, PA^2 equals AC^2 + PC^2. Substituting AC and PC, we get PA^2 equals = 5r^2, ##.

##5. ##Therefore PA = r * √5, the final answer is r * √5.

圆; 切线; 等腰三角形; 中点; 线段

Definition of Tangent to a Circle

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

<note>The center of the circle is P</note>, the lines <note>AB and AC</note> each intersect the circle at only one <note>common point B and C</note>, these common points are called <note>tangent points</note>. Therefore, the lines <note>AB and AC</note> are tangents to the circle.

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 AC is point M</note>. According to the definition of the midpoint of a line segment, <note>point M divides line segment AC into two equal parts</note>, that is, <note>the lengths of line segment AM and line segment MC are equal</note>. That is, <note>AM = MC</note>.

Definition of Isosceles Triangle

In the figure of this problem, in triangle PMC, <note>side PC</note> and <note>side MC</note> are equal, so triangle PMC is an isosceles triangle.

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 <note>right triangle PAC</note>, <note>angle ∠PCA is a right angle (90 degrees)</note>, <note>sides PC and AC are the legs</note>, <note>side PA is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>PA^2 = AC^2 + PC^2</note>.


Question: As shown in the figure, AB, AC, BD are tangents to ⊙O at points P, C, D respectively. If AB=5 and AC=3, then the length of BD is ()

Answer: 2

1. Given that AC and AP are tangents to ⊙O, according to the ##tangent length theorem##, AC=AP.

2. Similarly, given that BP and BD are tangents to ⊙O, according to the ##tangent length theorem##, BP=BD.

3. Based on the given conditions, AB=5, AC=3, and AC=AP, therefore AP=3.

4. From AB - AP = PB, we can calculate PB=5-3=2.

5. Since BP=BD, therefore BD=PB=2.

Definition of Tangent to a Circle

Circle <note>O</note> and lines <note>AB</note>, <note>AC</note>, and <note>BD</note> each have exactly one common point <note>P</note>, <note>C</note>, and <note>D</note> respectively. These common points are called <note>tangent points</note>. Therefore, lines <note>AB</note>, <note>AC</note>, and <note>BD</note> are tangents to circle <note>O</note>.

Tangent-Segment Theorem

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

In the figure of this problem, two tangents <note>AC and AP</note> are drawn from an external point A to the circle, and their tangent lengths are equal, i.e., <note>AC=AP</note>. Two tangents <note>BP and BD</note> are drawn from an external point B to the circle, and their tangent lengths are equal, i.e., <note>BP=BD</note>.


Question: As shown in the figure, △ABC is inscribed in ⊙O, ∠AOB=80°, then the measure of ∠ACB is ()

Answer: 40°

1. Given △ABC is inscribed in ⊙O, ∠AOB=80°.

####

##2. ##According to the inscribed angle theorem: the inscribed angle subtended by the same arc is equal to half of the central angle. Applying this theorem, since ∠AOB is the central angle, and the inscribed angle subtended by arc AB is ∠ACB, therefore: ∠ACB=1/2 ∠AOB.

##3. ##Substituting the given condition ∠AOB=80°, we get: ∠ACB=1/2×80°=40°.

##4. ##Through the above reasoning, the final answer is 40°.

Definition of Central Angle

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

Definition of Inscribed Angle

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

Inscribed Angle Theorem

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


Question: As shown in the figure, there are four points A, B, C, D on ⊙O, where ∠A=80°, then the degree of ∠C is ()

Answer: 100°

1. Given that quadrilateral ABCD is a cyclic quadrilateral, where ∠BAD=80°.

2. According to the theorem of cyclic quadrilateral (theorem 3), in quadrilateral ABCD, ∠BAD + ∠BCD = 180°.

3. From the given condition ∠BAD=80°, substituting into the formula from the previous step, we get: 80° + ∠BCD = 180°.

4. Solving the equation, we get: ∠BCD = 180° - 80° = 100°.

5. Therefore, the measure of ∠BCD in cyclic quadrilateral ABCD is 100°.

圆; 圆周角; 圆内接四边形; 弧

Cyclic Quadrilateral

A quadrilateral is called a cyclic quadrilateral if and only if all four vertices lie on the circumference of a single circle. This circle is referred to as the circumcircle of the quadrilateral.

<note>All four vertices A, B, C, and D of quadrilateral ABCD lie on the same circle O</note>. This circle is called <note>the circumcircle of quadrilateral ABCD</note>. Therefore, <note>quadrilateral ABCD is a cyclic quadrilateral</note>.

Corollary 3 of the Inscribed Angle Theorem: Diagonal Supplementary Theorem for Cyclic Quadrilateral

In a cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of each pair of opposite angles is equal to 180°. Additionally, any exterior angle is equal to its interior opposite angle.

<note>Quadrilateral ABCD is a cyclic quadrilateral, with vertices A, B, C, and D all on the circle</note>. According to the Diagonal Supplementary Theorem for Cyclic Quadrilateral, <note>the sum of each pair of opposite angles of quadrilateral ABCD is equal to 180°</note>. Specifically, <note>∠BAD + ∠BCD = 180°</note>, <note>∠ABC + ∠ADC = 180°</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, C is a point on ⊙O, a tangent to ⊙O is drawn through point C and intersects the extension of AB at point E, OD ⊥ AC at point D. If ∠E = 30°, CE = 6, then the value of OD is ()

Answer: √{3}

1. Connect OC. According to the property of the tangent line, the tangent CE is perpendicular to the radius OC passing through the point of tangency, i.e., ∠ECO=90°.

2. Since ∠ECO=90°, ∠E=30°, and CE=6, according to the triangle sum theorem, we can calculate ∠COE=60°. The right triangle OCE is a 30°-60°-90° triangle, so OC=rac{6}{√{3}}=2√{3}.

3. Since OC=OA, we know OA=2√{3}.

4. According to the inscribed angle theorem, ∠EOC=2∠A, therefore ∠A=30°.

5. Since OD is perpendicular to AC, i.e., ∠ADO=90°, the right triangle ODA is a 30°-60°-90° triangle. In this right triangle, OD=rac{1}{2}OA, thus OD=√{3}.

6. Through the above reasoning, the final answer is √{3}.

圆; 切线; 直角三角形; 圆周角; 正弦

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 C is the point of tangency between line CE and the circle</note>, <note>segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CE is perpendicular to the radius OC at the point of tangency C</note>, i.e., <note>∠ECO=90°</note>.

Definition of Right Triangle

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

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

Inscribed Angle Theorem

<note>In the diagram of this problem, ∠EOC is a central angle</note>, it corresponds to the inscribed angle <note>∠ACB</note>, therefore <note>∠EOC=2∠ACB</note>. Since it is known that <note>∠EOC=60°</note>, thus <note>∠ACB=30°</note>.

Definition of Central Angle

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

Definition of Inscribed Angle

<note>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, angle A is an inscribed angle.

Properties of 30°-60°-90° Triangle

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

In the <note>30°-60°-90° triangle OCE</note>, <note>angle E is 30 degrees</note>, <note>angle EOC is 60 degrees</note>, <note>angle OCE is 90 degrees</note>. <note>Side OE is the hypotenuse</note>, <note>side OC is opposite the 30-degree angle</note>, <note>side CE is opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle, <note>side OC is half of side OE</note>, <note>side CE is √3 times side OC</note>. That is: <note>OC = 1/2 * OE</note>, <note>CE = OC * √3</note>.

In the <note>30°-60°-90° triangle OAD</note>, <note>angle A is 30 degrees</note>, <note>angle DOA is 60 degrees</note>, <note>angle ODAE is 90 degrees</note>. <note>Side OA is the hypotenuse</note>, <note>side ODC is opposite the 30-degree angle</note>, <note>side AD is opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle


Question: As shown in the figure, in △ABC, AB=AC, ∠BAC=70°, ⊙O is the circumcircle of △ABC, point D is on the minor arc {AC}, then the degree of ∠D is ()

Answer: 125°

1. Given AB=AC, ∠BAC=70°, so △ABC is an isosceles triangle, and ∠ABC=∠ACB.

2. Since the two base angles of an isosceles triangle are equal, according to the triangle angle sum theorem, the sum of the angles in △ABC is 180°, thus we get ∠ABC= ∠ACB= (180°- ∠BAC) / 2 = (180° - 70°) / 2 = 55°.

3. Since quadrilateral ABCD is a cyclic quadrilateral, according to the cyclic quadrilateral opposite angle theorem (Corollary 3 of the Inscribed Angle Theorem), the opposite angles of a cyclic quadrilateral are supplementary, i.e., ∠D and ∠ABC are supplementary.

4. We can deduce ∠D=180° - ∠ABC.

5. Given ∠BAC=55°, therefore ∠D=180° - 55° = 125°.

6. Through the above reasoning, the final answer is ∠D=125°.

等腰三角形; 圆; 圆周角; 弧; 圆内接四边形

Definition of Isosceles Triangle

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

Triangle Angle Sum Theorem

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

<note>Quadrilateral ABCD is a cyclic quadrilateral of circle O, with vertices A, B, C, and D on the circle</note>. According to the Diagonal Supplementary Theorem for Cyclic Quadrilateral, the sum of each pair of opposite angles of quadrilateral ABCD is equal to 180°. Specifically, <note>∠BAD + ∠BCD = 180°</note>; <note>∠ABC + ∠ADC = 180°</note>. Therefore, <note>∠D and ∠ABC are supplementary, i.e., ∠D = 180° - ∠ABC</note>.

Properties of Isosceles Triangle

In the figure of this problem, <note>isosceles triangle ABC</note>, <note>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>angle ABC = angle ACB</note>.


Question: As shown in the figure, ⊙O is the circumcircle of △ABD. If ∠A=135°, then the degree of ∠BDO is ()

Answer: 45°

1. Find any point E on the major arc BD, and connect BE and DE.

2. According to the definition of a cyclic quadrilateral, points A, B, E, and D all lie on circle O, so quadrilateral ABDE is a cyclic quadrilateral.

3. Given that ∠A=135°, and according to the supplementary angles theorem of cyclic quadrilaterals, we have ∠BED=180°-∠BAD=180°-135°=45°.

4. According to the inscribed angle theorem, in circle O, ∠BOD is the central angle of the minor arc BD, and ∠BED is the inscribed angle of the minor arc BD, therefore ∠BOD=2∠BED.

5. Given ∠BED=45°, then ∠BOD=2×45°=90°.

6. In △BDO, since OB=OD, △BDO is an isosceles triangle. Therefore, ∠BDO=∠BOD/2=90°/2=45°.

7. Through the above reasoning, the final answer is 45°.

圆; 圆周角; 弧; 直角三角形; 圆心角

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 A of angle BAD is on the circumference</note>, and the two sides of angle BAD intersect <note>circle O at points B and D</note> respectively. Therefore, <note>angle BAD is an inscribed angle</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 B</note> and <note>point D</note> are two points on the circle, and <note>the center of the circle is point O</note>. The angle formed by the lines <note>OB</note> and <note>OD</note>, <note>∠BOD</note>, is called the central angle.

<note>Quadrilateral ABDE is a cyclic quadrilateral, and the vertices A, B, D, E are all on the circle</note>. According to <note>Diagonal Supplementary Theorem for Cyclic Quadrilateral</note>, the <note>sum of each pair of opposite angles of the quadrilateral is equal to 180°</note>. Specifically, <note>∠A + ∠BED = 180°</note>, that is, <note>∠A = 135°</note>, then <note>∠BED = 180° - 135° = 45°</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 problem diagram, in circle O, points B, D, and E are on the circle, <note>arc BD</note> and <note>arc BED</note> correspond to the central angle <note>∠BOD</note> and the inscribed angle <note>∠BED</note>. According to the Inscribed Angle Theorem, <note>∠BED</note> is equal to half of the central angle <note>∠BOD</note> corresponding to <note>arc BD</note>, that is, <note>∠BED = 1/2 ∠BOD</note>.

Properties of Isosceles Triangle

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

Definition of Isosceles Triangle

In triangle OBD, sides OB and OD are equal, therefore triangle OBD is an isosceles triangle.

Triangle Angle Sum Theorem

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


Question: ⊙O is a circle with radius 1, the distance from point O to line L is 3. A tangent to ⊙O is drawn through any point P on line L, and the point of tangency is Q. If a square PQRS is constructed with PQ as one of its sides, then the minimum area of square PQRS is ()

Answer: 8

1. Let the radius of ⊙O be 1, and the distance from point O to line L be 3. Draw a tangent to ⊙O from any point P on line L, with the point of tangency being Q.

2. Connect OQ and OP, and draw OH perpendicular to L, with H being the foot of the perpendicular. Since the distance from point O to line L is 3, OH = 3. Since point Q is on the circle, OQ is the radius of ⊙O, so OQ = 1.

3. Since PQ is a tangent to ⊙O, according to the property of the tangent to a circle, OQ is perpendicular to PQ. In the right triangle PQO, by the Pythagorean theorem, PQ = √(OP^2 - OQ^2).

4. When OP is minimized, PQ is minimized. According to the definition of the distance from a point to a line, the minimum value of OP is the length of OH, which is OP = 3.

5. Substituting into the Pythagorean theorem, the minimum value of PQ is √(3^2 - 1^2) = √8 = 2√2.

6. Construct a square PQRS with PQ as a side, then the area of the square is PQ^2.

7. Therefore, the minimum area of square PQRS is (2√2)^2 = 8.

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

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, and point Q is any point on the circle</note>, the line segment <note>OQ is the line segment from the center of the circle to any point on the circle</note>, therefore <note>the line segment OQ is the radius of the circle</note>, and OQ = 1.

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°).

<note>In the figure of this problem, in quadrilateral PQRS, sides PQ, QR, RS, and SP are equal</note>, and <note>∠PQR</note>, <note>∠QRS</note>, <note>∠RSP</note>, and <note>∠SPQ are right angles (90 degrees)</note>, so PQRS is a square.

<note>In the figure of this problem, in circle ⊙O, point Q is the point of tangency of line PQ with the circle</note>, <note>line segment OQ is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PQ is perpendicular to the radius OQ at the point of tangency Q</note>, i.e., <note>∠OQP=90 degrees</note>.

Pythagorean Theorem

In the diagram of this problem, <note>right triangle PQO</note> has <note>∠OQP as a right angle (90 degrees)</note>, <note>sides OQ and PQ are the legs</note>, and <note>side OP is the hypotenuse</note>. Therefore, according to the Pythagorean Theorem, <note>OP^2 = OQ^2 + PQ^2</note>. According to the problem statement, <note>OQ = 1</note>, <note>OP = 3</note>, so <note>PQ = √(3^2 - 1^2) = √8</note>.

Distance from a Point to a Line

The distance from a point to a line is defined as the shortest distance from a point not on the line to the line, which is the perpendicular distance.

In the figure of this problem, <note>point O is a point outside line L</note>, draw <note>the perpendicular segment OH from point O to line L</note>. According to the definition of the distance from a point to a line, <note>the length of the perpendicular segment OH is the shortest distance from point O to line L</note>, which is d=OH=3.

Formula for the Area of a Square

The area of a square is equal to the square of its side length.

In the diagram of this problem, <note>square PQRS</note>, <note>side PQ is the side length of the square</note>, so <note>the area of the square = PQ ^2</note>.


Question: As shown in the figure, △ABC is an inscribed triangle of ⊙O, ∠C=30°, the radius of ⊙O is 5. If point P is a point on ⊙O, in △ABP, PB=AB, then the length of PA is ()

Answer: 5√{3}

1. Given ∠C=30°, point P is a point on ⊙O, connect OA, OB, OP, let the intersection of AP and OB be point D.

2. Since ∠C=30°, according to the theorem of inscribed angles and central angles, the inscribed angles subtended by the same arc are equal, and the inscribed angle is half of the central angle subtended by the same arc, we get ∠C=∠APB=30°, ∠AOB=2∠ACB=60°.

3. Since PB=AB, according to the properties of an isosceles triangle, we get triangle ABP is an isosceles triangle, ∠PAB=∠APB=30°.

4. In triangle APB, ∠PAB+∠APB+∠ABP=180°, substituting the known conditions we get ∠ABP=120°.

5. Since PB=AB, and because OA=OB=OP, triangles OAB and OPB are both isosceles triangles. According to the theorem of inscribed angles, we get: ∠BOP=∠AOB=2∠ACB=60°. Therefore, △OBP and △OBA are equilateral triangles. Hence, in triangle ABD, ∠DAB=30°, ∠ABD=60°, according to the sum of interior angles of a triangle being 180°, we get: ∠ADB=180°-∠DAB-∠ABD=180°-30°-60°=90°, so OB⊥AP.

6. Since OB=OA, in the equilateral triangle AOB, OA=AB=OB=5.

7. In the right triangle PBD, using the cosine rule we calculate PD=cos30°·PB=cos30°·OP, that is PD= $rac{√{3}}{2}×5=rac{5√{3}}{2}$.

8. Since PD is half of AP, AP=2PD, that is AP=5√{3}.

9. Through the above reasoning, the final answer is 5√{3}.

等腰三角形; 圆; 圆周角; 弦; 普通三角形

Definition of Inscribed Angle

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

Definition of Central Angle

<note>In circle O, points A and B are two points on the circle</note>, <note>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>.

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 diagram of this problem, <note>the radius OB is perpendicular to the chord AP in circle O</note>, then according to the Perpendicular Diameter Theorem, <note>the radius OB bisects the chord AP, that is AD=PD</note>, and <note>the radius OB bisects the two arcs subtended by the chord AP, that is minor arc AB=minor arc BP</note>.

Definition of Isosceles Triangle

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

<note>In triangle ABP, side AB and side PB are equal</note>, therefore triangle ABP is an isosceles triangle.

Inscribed Angle Theorem

<note>In circle O, points A, B, C, P are on the circle</note>, the central angle corresponding to <note>minor arc AB</note> is <note>∠AOC</note>, and the inscribed angle is <note>∠APB</note>. According to the Inscribed Angle Theorem, <note>∠APB</note> is equal to half of the central angle <note>∠AOC</note> corresponding to <note>minor arc AB</note>, that is, <note>∠APB = 1/2 ∠AOC</note>.

Triangle Angle Sum Theorem

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

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

Equilateral Triangle Identification Theorem (60-Degree Angle in an Isosceles Triangle)

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

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

Cosine Theorem

The cosine function is a trigonometric function defined in terms of the sides of a right triangle. For an angle α, its cosine value is equal to the ratio of the length of the adjacent side to the hypotenuse.

In the right triangle PBD in the figure of this problem, <note>side PB, side PD, and side BD correspond to the three sides of the triangle</note>, <note>angle PBD is the angle between side PB and side BD</note>. According to the Cosine Theorem, the square of side PD is equal to the sum of the squares of the other two sides PB and BD minus twice the product of these two sides and <note>the cosine of the angle PBD between them</note>, that is, <note>##PD^2= PB^2 + BD^2 - 2 * PB * BD * cos(angle PBD).##</note>


Question:
In the diagram above, line OA is congruent to line OB. What is the measure of arc CD?

a. 27.5 degrees
b. 55 degrees
c. 70 degrees
d. 110 degrees
e. 125 degrees

Answer: 70

1. Given that line segment OA is congruent to line segment OB, which is based on the conditions provided in the problem.

##2. In triangle AOB, sides OA and OB are equal, thus triangle AOB is an isosceles triangle.##

3. In an isosceles triangle, the base angles are equal, so ∠OAB is equal to ∠OBA.

4. The problem states that ∠A is 55 degrees, so ∠OBA is also 55 degrees.

5. The triangle angle sum theorem states that the sum of the interior angles of a triangle is 180 degrees. Therefore, we get: ∠AOB = 180 degrees - ∠OAB - ∠OBA.

6. Substituting the values, we get ∠AOB = 180 degrees - 55 degrees - 55 degrees = 70 degrees.

####

##7.## According to the central angle theorem, the measure of the arc intercepted by a central angle is equal to the measure of the central angle.

##8.## Therefore, the measure of arc CD intercepted by ∠AOB is 70 degrees.

等腰三角形; 圆; 圆心角

Definition of Isosceles Triangle

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

Definition of Central Angle

The original text: <note>Point C and Point D</note> are two points on the circle, with the center of the circle being Point O. <note>The angle ∠COD formed by the lines OC and OD</note> is called the central angle.

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 isosceles triangle AOB</note>, sides <note>OA</note> and <note>OB</note> are equal. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠OAB = ∠OBA</note>.

Triangle Angle Sum Theorem

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


Question: As shown in the figure, △ABC is an inscribed triangle in ⊙O. If ∠ACB=30°, AB=6, then the radius of ⊙O is ()

Answer: 6

1. According to the definition of inscribed angle, ∠ACB is an inscribed angle. Given ∠ACB=30°, according to the inscribed angle theorem and the definition of central angle, it follows that ∠AOB=2∠ACB=60°.

2. Connect OA and OB, OA and OB are radii of ⊙O, therefore OA=OB.

3. Because ∠AOB=60°, △AOB is an equilateral triangle.

4. In the equilateral triangle △AOB, all three sides are equal, therefore OA=AB=6.

5. Hence, the radius of ⊙O is 6.

圆内接四边形; 圆周角; 正弦; 圆; 弦

Definition of Inscribed Angle

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

Definition of Central Angle

In <note>circle O</note>, <note>point A</note> and <note>point B are two points on the circle</note>, <note>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 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 AOB is an equilateral triangle</note>. <note>Sides OA, OB, and AB are of equal length</note>, and <note>angles AOB, OAB, and OBA are equal, each measuring 60°</note>.

Definition of Radius

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

Inscribed Angle Theorem

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


Question: As shown in the figure, △ABC is inscribed in ⊙O. Connect OA and OB. If ∠C=35°, then the degree of ∠OBA is ()

Answer: 55°

1. Given ∠C=35°, △ABC is inscribed in ⊙O.

2. Connect OA, OB, ∠AOB is the central angle, and ∠ACB is the inscribed angle subtended by the minor arc AB.

3. According to the inscribed angle theorem, the central angle is twice the inscribed angle, thus ∠AOB = 2×∠ACB = 2×35° = 70°.

4. In the triangle △OAB inscribed in the circle, OA=OB, so △OAB is an isosceles triangle.

5. In the isosceles triangle △OAB, the base angles are equal, thus ∠OAB = ∠OBA.

6. In △OAB, the sum of the interior angles is 180°, i.e., ∠AOB + ∠OAB + ∠OBA = 180°.

7. Substituting the known conditions, we get 70° + 2∠OBA = 180°.

8. Solving this equation, we get 2∠OBA = 110°, so ∠OBA = 55°.

9. Through the above reasoning, we finally obtain the measure of ∠OBA as 55°.

圆; 圆周角; 圆心角; 等腰三角形; 弧

Inscribed Polygon

A polygon is called an inscribed polygon if all its vertices lie on the circumference of a circle.

All vertices of <note>triangle ABC</note>, namely <note>A, B, and C</note>, lie on <note>circle O</note>, thus this triangle is an <note>inscribed triangle of circle O</note>.

Definition of Central Angle

<note>In the diagram of this problem, in circle O, point A and point 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

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

Definition of Isosceles Triangle

In the triangle OAB, sides OA and OB are equal, therefore the triangle OAB is an isosceles triangle.

Inscribed Angle Theorem

In the figure of this problem, in circle O, points A, B, and C are on the circle, the central angle corresponding to <note>minor arc AB</note> is <note>∠AOB</note>, and the inscribed angle is <note>∠ACB</note>. According to the Inscribed Angle Theorem, <note>∠AOB</note> is equal to twice the inscribed angle <note>∠ACB</note> corresponding to <note>minor arc AB</note>, that is, <note>∠AOB = 2 × ∠ACB = 2 × 35° = 70°</note>.

Properties of Isosceles Triangle

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

Triangle Angle Sum Theorem

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


Question: ∠ABC = 30°. BC = 10. Segment BD is an altitude. If triangle ABC is isosceles, find the length of segment AC.

Answer: 5.18

1. Triangle ABC is an isosceles triangle, i.e., BA = BC = 10. ∠A = ∠C.

2. Given ∠ABC = 30 degrees, ∠A = ∠C. According to the triangle angle sum theorem, it can be found that: ∠A = ∠C = 75 degrees.

3. According to the sine rule, we have: AC/sin(30 degrees) = BC/sin(75 degrees).

4. Substituting the values, we get x ≈ 5.18.

5. Through the above reasoning, the final answer is 5.18.

等腰三角形; 直角三角形; 正弦

Definition of Isosceles Triangle

<note>Side BA and side BC are equal</note>, therefore triangle ABC is an isosceles triangle.

Properties of Isosceles Triangle

In the diagram of this problem, in <note>isosceles triangle ABC</note>, <note>sides BA and BC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are <note>equal, that is, angle A = angle C</note>.

Triangle Angle Sum Theorem

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

Sine Theorem

In any triangle, the ratio of the length of each side to the sine of its opposite angle is constant and equal to the diameter of the circumcircle, i.e., \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r = D \) (where \( r \) is the radius of the circumcircle, and \( D \) is its diameter).

<note>Side AB, BC, and AC correspond to angles C, A, and ABC respectively</note>. According to the Sine Theorem, the ratio of the lengths of each side to the sine of their opposite angles is equal and equivalent to the diameter of the circumscribed circle, i.e., <note>AB/sin(angle C) = AC/sin(angle ABC) = BC/sin(angle A)</note> = 2r = D (<note>where r is the radius of the circumscribed circle, D is the diameter</note>).


Question:
In the diagram above, line AB is parallel to line CD, and line EF is perpendicular to line CD. What is the measure of angle x?
a. 40 degrees
b. 45 degrees
c. 50 degrees
d. 60 degrees
e. 80 degrees

Answer: 50*\degree

1. Given that line AB is parallel to line CD, line EF is perpendicular to line CD, ##∠GIL##=140°.

2. According to the theorem of corresponding angles of parallel lines, since line AB is parallel to line CD, ##∠GIL##=##∠JKD##.

4. According to the definition of a straight angle, we get ##∠JKM##=180°-##∠JKD##=180°-140°=40°.

5. Since line EF is perpendicular to line CD, ##∠LMK##=90°.

6. In ##ΔJMK##, since the sum of angles is 180°, ##∠JMK##+##∠JKM##+##∠X##=180°.

7. Substituting the known conditions: 90° + 40° + ##∠X## = 180°.

8. Simplifying the equation: ##∠X## = 180° - 90° - 40° = 50°.

9. Through the above reasoning, the final answer is 50 degrees.

平行线; 内错角; 垂线

Definition of Triangle

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

<note>Triangle JMK</note> is a geometric figure composed of three non-collinear points <note>J</note>, M, K and their connecting line segments <note>JM</note>, MK, <note>LK</note>. Points <note>J</note>, M, K are the three vertices of the <note>triangle</note>, and the line segments <note>JM</note>, MK, <note>LK</note> are the three sides of the <note>triangle</note>. Its internal angles include <note>∠JMK</note>, <note>∠MJK</note>, and <note>∠JKM</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.

<note>Line AB is parallel to line CD</note>, intersected by a third line EF, corresponding angles: <note>∠GIL and ∠JKD are equal</note>, i.e., <note>∠GIL=∠JKD=140°</note>.

Definition of Corresponding Angles

Two angles are referred to as corresponding angles if they are formed when two lines, a and b, are intersected by a third line, c, and both angles lie on the same side of the intersecting line c and on the same side of the lines a and b.

In the figure of this problem, <note>two parallel lines AB and CD</note> are intersected by <note>a line GH</note>, where <note>angle GIL and angle JKD are on the same side of the intersecting line GH</note>, on the same side of the two intersected lines AB and CD, therefore <note>angle GIL and angle JKD are corresponding angles</note>. <note>Corresponding angles are equal</note>, that is, <note>angle GIL is equal to angle JKD</note>.

Definition of Straight Angle

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

<note>Ray KD rotates around endpoint K to form a straight line with the initial side</note>, forming a <note>straight angle CKD</note>. According to the definition of a straight angle, the degree of a straight angle is <note>180 degrees, i.e., angle CKD = 180 degrees</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 ∠EMD formed by the intersection of line EF and line CD is 90 degrees</note>, therefore, according to the definition of perpendicular lines, <note>line EF and line CD are perpendicular to each other</note>.

Triangle Sum Theorem

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

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


Question: As shown in the figure, △ABC is inscribed in ⊙O. If ∠AOB=110°, then the measure of ∠ACB is ()

Answer: 55°

1. Given △ABC is inscribed in ⊙O, and ∠AOB=110°.

2. According to the inscribed angle theorem, the inscribed angle subtended by the same arc is half of the central angle subtended by that arc.

3. By the inscribed angle theorem, for point C on the circle, ∠ACB is half of the corresponding central angle ∠AOB, i.e., ∠ACB = 1/2 * ∠AOB.

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

圆; 圆心角; 圆周角; 普通三角形

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 diagram of this problem, in circle O, <note>point A and point B are two points on the circle</note>, <note>the center of the circle is point O</note>. <note>The angle ∠AOB formed by the lines OA and OB</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>The vertex C of angle ACB is on the circumference</note>, <note>The two sides of angle ACB intersect the circle O at points A and B respectively</note>. Therefore, angle ACB is an inscribed angle.

Inscribed Angle Theorem

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

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


Question: As shown in the figure, points A, B, and C are on ⊙O. A tangent to ⊙O is drawn through point A, intersecting the extension of OC at point P. ∠B=30°, OP=3, then the length of AP is ()

Answer: \frac{3}{2}√{3}

2. According to the given condition ∠B=30°, and the theorem of the inscribed angle (the inscribed angle is half of the central angle that subtends the same arc), we can deduce ∠AOC=2∠B=60°.

3. Since a tangent line is drawn through point A to circle ⊙O and intersects the extension of OC at point P, ∠OAP=90° (the tangent line is perpendicular to the radius).

4. Given OP=3, according to the definition of sine, sin∠AOC=sin60° (∠AOC=60°), we know sin60°=√3/2.

5. From the properties of right triangle OAP, we can obtain AP=OP×sin60°.

7. Through the above reasoning, the final length of AP is 3√3/2.

圆; 切线; 圆周角; 直角三角形; 线段

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 OAP</note>, angle <note>∠AOP</note> is an acute angle, <note>side AP is the opposite side of angle ∠AOP</note>, and side OP is the hypotenuse. According to the definition of the sine function, the sine value of angle <note>∠AOP</note> is equal to the ratio of <note>opposite side AP</note> to the hypotenuse <note>OP</note>, that is, <note>sin(∠AOP) = AP / OP</note>.

Inscribed Angle Theorem

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

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 of line AP with the circle</note>, <note>segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line AP is perpendicular to the radius OA at the point of tangency A</note>, i.e., <note>∠OAP=90 degrees</note>.

Definition of Right Triangle

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

<note>Angle OAP is a right angle (90 degrees)</note>, thus triangle OAP is a right triangle. Side <note>OA</note> and side <note>OP</note> are the legs, and side <note>AP</note> is the hypotenuse.


Question: As shown in the figure, AB is the diameter of ⊙O, point D is on the extension line of AB, a tangent to ⊙O is drawn through point D, and the point of tangency is C. If ∠A=25°, then ∠D=()

Answer: 40°

1. Connect OC. Since point C is the tangent point of ⊙O, OC is perpendicular to CD.

2. Because OC = OA (radii are equal), △OAC is an isosceles triangle.

3. According to the properties of triangles, ∠OCA is equal to ∠A, so ∠OCA = ∠A = 25°.

4. In △OAC, ∠DOC is an exterior angle of △OAC. According to the exterior angle theorem of triangles, ∠DOC = ∠OCA + ∠A = 25° + 25° = 50°.

5. According to the tangent theorem of circles, the angle between the tangent and the radius is 90°, so ∠OCD = 90°.

6. In △ODC, using the property that the sum of angles is 180°, ∠D = 180° - ∠OCD - ∠DOC = 180° - 90° - 50° = 40°.

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

圆; 切线; 圆周角; 直角三角形; 邻补角

Definition of Isosceles Triangle

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

<note>In triangle OAC, side OC and side OA are equal</note>, therefore <note>triangle OAC is an isosceles triangle</note>.

Perpendicular Diameter Theorem

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

<note>In circle O, point D on the extension line of diameter AB makes a tangent to circle O, and the point of tangency is C</note>. Connect OC. Since <note>point C is the point of tangency of circle O</note>, therefore <note>OC is perpendicular to CD</note>.

Triangle Angle Sum Theorem

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

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

In the diagram of this problem, in circle O, point C is the point of tangency between line CD and the circle, <note>segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line CD is perpendicular to the radius OC at the point of tangency C</note>, <note>i.e., ∠OCD=90°</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 OAC, sides OA 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>angle OAC = angle OCA</note>.

Exterior Angle Theorem of Triangle

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

<note>In triangle AOC, angle DOC is an exterior angle of the triangle</note>, <note>angle OCA and angle A are the two interior angles that are not adjacent to the exterior angle DOC</note>. According to the Exterior Angle Theorem of Triangle, <note>the exterior angle DOC is equal to the sum of the two non-adjacent interior angles OCA and angle A</note>, that is, <note>angle DOC = angle OCA + angle A</note>.


Question: As shown in the figure, in ⊙O with a radius of 2, C is a point on the extension of the diameter AB, CD is tangent to the circle at point D, and AD is connected. Given that ∠DAC=30°, find the length of segment CD.

Answer: 2√{3}

1. Given that point D is the point of tangency, CD is tangent to the circle at point D. According to the theorem of perpendicularity of the tangent line, OD is perpendicular to CD.

2. Let the center of the circle be O, and connect OD, OC, AD, and AC.

3. Since the angle ∠DAC is 30°, according to the inscribed angle theorem, ∠DOC=2∠DAC=2×30°=60°.

4. Because OD is perpendicular to CD, ∠ODC is 90°. In △ODC, angle ∠DCO=30° and angle ∠DOC=60°.

5. According to the properties of a 30°-60°-90° triangle, side OD is the radius with a length of 2, thus CD=OD * √3 = 2 * √3.

6. Through the above reasoning, the final length of segment CD is 2√3.

圆; 切线; 线段; 直角三角形; 正弦

Definition of Radius

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

<note>In circle O, point O is the center of the circle</note>, <note>points A, B, and D are any points on the circle</note>, <note>line segments OA, OB, and OD are segments from the center to any point on the circle</note>, therefore <note>line segments OA, OB, and OD are the radii of the circle</note>, and their lengths are all 2.

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

Inscribed Angle Theorem

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

Properties of 30°-60°-90° Triangle

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

In the <note>30°-60°-90° triangle ODC</note>, <note>angle DCO is 30 degrees</note>, <note>angle DOC is 60 degrees</note>, <note>angle ODC is 90 degrees</note>. <note>Side OC is the hypotenuse</note>, <note>side OD is opposite the 30-degree angle</note>, <note>side CD is opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle, <note>side OD is half of side OC</note>, <note>side CD is √3 times side OD</note>. That is: <note>OD = 1/2 * OC</note>, <note>CD = OD * √3</note>.

Definition of Central Angle

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

Definition of Inscribed Angle

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


Question: As shown in the figure, AB is the tangent to ⊙O, A is the point of tangency, and the extension of BO intersects ⊙O at point C. If ∠OAC = 35°, then the degree of ∠B is ()

Answer: 20°

1. Given that AB is the tangent to ⊙O at point A, according to the theorem that the radius at the point of tangency is perpendicular to the tangent, we get OA perpendicular to AB.

3. Since O is the center of the circle, and OA and OC are radii and equal, triangle OAC is an isosceles triangle.

4. According to the properties of an isosceles triangle, ∠OAC=∠OCA.

6. In triangle BAC, ∠BAO=90°, ∠OAC=∠OCA=35°, using the interior angle sum theorem, we get ∠B= 180° - ∠BAO - ∠OAC -∠OCA = 180° - 90° -35° - 35° = 20°.

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

圆; 圆周角; 切线; 对顶角; 普通三角形

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.

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

Definition of Radius

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

Definition of Isosceles Triangle

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

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

Triangle Angle Sum Theorem

In <note>triangle BOA</note>, angle BAO, <note>angle AOB</note>, and <note>angle ABO</note> are the three interior angles of <note>triangle BOA</note>. According to the Triangle Angle Sum Theorem, angle BAO + <note>angle AOB</note> + <note>angle ABO</note> = 180°. In <note>triangle ABC</note>, <note>angle BAC</note>, <note>angle BCA</note>, and <note>angle ABC</note> are the three interior angles of <note>triangle ABC</note>. According to the Triangle Angle Sum Theorem, <note>angle BAC + angle BCA + angle ABC = 180°</note>.

Properties of Isosceles Triangle

<note>In the figure of this problem, in the isosceles triangle OAC, sides OA and OC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠OCA = ∠OAC = 35°</note>.


Question: As shown in the figure, AC is the tangent to ⊙O at point C, BC is the diameter of ⊙O, AB intersects ⊙O at point D, connect OD. If ∠BAC=50°, then the measure of ∠COD is ()

Answer: 80°

1. Given AC is the tangent of ⊙O at point C, according to the ##property of the tangent to a circle##, we get BC⊥AC.

2. Since BC⊥AC, we obtain ∠BCA=90°.

3. According to the problem statement, ∠BAC=50°, by the triangle angle sum theorem, we get ∠ABC=40°.

####

##4.## According to the inscribed angle theorem, #### we get ∠COD=2∠ABC.

##5.## Substituting the previously calculated ∠ABC=40°, we get ∠COD=2×40°=80°.

##6.## Through the above reasoning, we finally get the answer as 80°.

圆; 切线; 圆周角; 直角三角形; 圆心角

Definition of Radius

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

Definition of Diameter

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

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

Definition of Central Angle

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

<note>In the figure of this problem, in circle O, point C is the point of tangency between line AC and the circle</note>, <note>line segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line AC is perpendicular to the radius OC at the point of tangency C</note>, that is, <note>∠BCA=90°</note>.

Triangle Angle Sum Theorem

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

Inscribed Angle Theorem

In <note>circle O</note>, points <note>B, D, and C are on the circle</note>, the central angle corresponding to <note>minor arc CD</note> is ∠COD, and the inscribed angle is <note>∠ABC</note>. According to the Inscribed Angle Theorem, <note>∠ABC</note> is equal to half of the central angle <note>∠COD</note> corresponding to <note>minor arc CD</note>, that is, <note>∠COD = 2∠ABC</note>.

Definition of Inscribed Angle

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


Question: As shown in the figure, AB is tangent to ⊙O at point B, AO intersects ⊙O at point C, and point D is on ⊙O. If ∠A = 40°, then the measure of ∠BDC is ()

Answer: 25°

1. Connect OB. Since AB is tangent to ⊙O at point B, according to the definition of tangent, OB is perpendicular to AB, thus ∠OBA = 90°.

2. From the given conditions, ∠A = 40°, so the measure of ∠BOA is 180° - ∠OBA - ∠A, that is, ∠BOA = 180° - 90° - 40° = 50°.

####

##3.## From the given conditions, AO intersects ⊙O at point C, point D is on ⊙O, ####, according to the inscribed angle theorem, ∠BDC = 1/2 × ∠BOA.

##4.## Calculated as ∠BDC = 1/2 × 50° = 25°.

##5.## Through the above reasoning, the final answer is 25°.

圆; 切线; 圆周角; 线段

Definition of Tangent to a Circle

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

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

Inscribed Angle Theorem

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

Definition of Central Angle

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

Definition of Inscribed Angle

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

Triangle Angle Sum Theorem

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


Question: As shown in the figure, PA and PB are tangent to ⊙O at points A and B respectively. The tangent line EF of ⊙O intersects PA and PB at points E and F respectively. The tangent point C is on arc AB. If the length of PA is 2, then the perimeter of △PEF is ()

Answer: 4

1. Given that PA and PB are tangent to circle O at points A and B respectively, and the tangent line EF of circle O intersects PA and PB at points E and F respectively, with the tangent point C on arc AB.

2. According to the ##tangent-segment theorem##: ##PA=PB,## AE = CE, FB = CF.

3. Since ##PA=2##, according to the problem, PA = PB = 2.

4. Calculate the perimeter of △PEF ##=PE+PF+EF=PE+PF+CE+CF##

5. According to AE = CE, FB = CF, we get: the perimeter of △PEF = PE + PF + EF = PE + PF + AE + FB = PA + PB.

6. It is deduced that the perimeter of △PEF = PE + EF + PF = PA + PB = 4.

切线; 圆; 弧; 等腰三角形; 线段

Tangent-Segment Theorem

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

In the figure of this problem, <note>two tangents PA and PB are drawn from an external point P to the circle</note>, their tangent lengths are equal, i.e., <note>PA=PB</note>. <note>The line connecting the center of the circle O and the point P</note> bisects the angle between the two <note>tangents PA and PB</note>. Similarly, <note>two tangents EA and EC are drawn from an external point E to the circle</note>, their tangent lengths are equal, i.e., <note>EA=EC</note>; <note>two tangents FB and FC are drawn from an external point F to the circle</note>, their tangent lengths are equal, i.e., <note>FB=FC</note>.

Perimeter Theorem of Triangle

The perimeter of a triangle is equal to the sum of the lengths of its three sides. That is, if the lengths of the three sides of the triangle are denoted as a, b, and c respectively, then the perimeter P is given by P = a + b + c.

According to the Perimeter Theorem of Triangle, the perimeter of a triangle is equal to the sum of the lengths of its three sides. The original text: <note>The three sides of triangle △AEF are AE, AF, and EF</note>, thus the perimeter <note>P=AE+AF+EF</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, BP is the tangent to ⊙O, AP intersects ⊙O at point C, point D is a point on arc {BC}, if ∠P=40°, then ∠ADC equals ()

Answer: 40°

1. Connect BC. According to the problem statement, PB is the tangent to circle O.

2. Since the line PB is a tangent, according to the theorem that the tangent is perpendicular to the radius at the point of tangency, PB is perpendicular to AB, thus ∠ABP=90°.

3. Since AB is the diameter of circle O, according to the corollary of the inscribed angle theorem, the inscribed angle subtended by the diameter is a right angle, ∠ACB=90°.

4. In △ABC, we have ∠ABC + ∠BAC = 90°.

5. In △ABP, we have: ∠P + ∠BAC = 90°.

6. Therefore, we get ∠ABC=∠P=40°.

7. Since point D is on the arc BC of circle O, according to the corollary of the inscribed angle theorem, ∠ADC = ∠ABC.

8. Therefore, ∠ADC=40°.

圆; 圆周角; 切线; 直角三角形; 对顶角

Definition of Inscribed Angle

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

In the figure of this problem, in circle ⊙O, point B is the point of tangency between line PB and the circle, <note>line segment OB is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠ABP=90°</note>.

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

<note>In the figure of this problem, in circle O, the inscribed angles ∠ABC and ∠ADC corresponding to arc AC are equal.</note> According to Corollary 1 of the Inscribed Angle Theorem, <note>the inscribed angles ∠ABC and ∠ADC corresponding to the same arc AC are equal, that is, ∠ABC = ∠ADC.</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 the diagram of this problem, in circle O, <note>the angle subtended by the diameter AB at the circumference ACB is a right angle (90 degrees)</note>.


Question: AB is the diameter of ⊙O, PA is tangent to ⊙O at point A, PO intersects ⊙O at point C; connect BC, if ∠P = 40°, then ∠B equals ()

Answer: 25°

1. Given PA is tangent to ⊙O at point A, according to the ##property of the tangent line to a circle##, we get ∠PAO=90°.

2. Given ∠P=40°, therefore, according to the triangle angle sum theorem, we get ##∠AOP=180°-90°-40°##=50°.

3. Since OC ##and OB are radii of ⊙O##, OC=OB, thus △OBC is an isosceles triangle.

4. In an isosceles triangle, the base angles are equal ##, i.e., ∠B=∠OCB. Since ∠AOP is the exterior angle of △OBC, according to the exterior angle theorem of a triangle, we get ∠AOP=∠B+∠OCB=2∠B##.

####

##5##. From ##∠AOP##=2∠B, we get 2∠B=50°, i.e., ∠B=25°.

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

圆; 切线; 直角三角形; 圆周角; 线段

<note>In circle O, point A is the point of tangency of line PA with the circle</note>, <note>line segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line PA is perpendicular to the radius OA at the point of tangency A</note>, i.e., <note>∠PAO=90 degrees</note>.

Triangle Angle Sum Theorem

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

Definition of Isosceles Triangle

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

Properties of Isosceles Triangle

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

Exterior Angle Theorem of Triangle

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

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


Question: As shown in the figure, AB is the diameter of ⊙O, PA is tangent to ⊙O at point A, OP intersects ⊙O at point C, and BC is connected. If ∠P = 20°, then the degree of ∠B is ()

Answer: 35°

1. Connect AC, according to the property of the tangent to a circle, we get AB⊥AP, i.e., ∠BAP=90°.

 2. Since ∠P=20°, therefore in △AOP, ∠AOP=70°.

 3. Because OA=OC, so △AOC is an isosceles triangle, ∠OAC=∠OCA, according to the sum of the interior angles of a triangle, we get ∠OAC=∠OCA=55°.

 4. Since AB is the diameter, according to the theorem that the angle subtended by the diameter is a right angle, we get ∠ACB=90°.

 5. In △ACB, ∠OAC+∠ACB+∠B=180°, so ∠B=180°-∠ACB-∠OAC=180°-90°-55°=35°.

 6. From the above reasoning, we finally get the answer as 35°.

圆; 切线; 圆周角; 直角三角形; 线段

Definition of Isosceles Triangle

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

In <note>circle O</note>, <note>point A is the point of tangency between line PA and the circle</note>, <note>segment OA is the radius of the circle</note>. According to the <note>Property of the Tangent Line to a Circle</note>, <note>the tangent line PA is perpendicular to the radius OA at the point of tangency A</note>, that is, <note>∠OAP = 90 degrees</note>.

Properties of Isosceles Triangle

<note>In the isosceles triangle AOC</note>, <note>side OA and side OC 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>.

<note>In circle O, the angle subtended by the diameter AB at the circumference, ∠ACB, is a right angle (90 degrees)</note>.

Triangle Angle Sum Theorem

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