Question: As shown in the figure, in the rhombus ABCD, AB=5, ∠B=60°, then the diagonal AC equals ()

Answer: 5

1. Given that ABCD is a rhombus, so AB = BC = CD = DA = 5.

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##2.## In △ABC, AB = 5, BC = 5, and ∠B = 60°.

##3. According to the theorem of equilateral triangles (the 60-degree angle criterion for isosceles triangles)##, if a triangle has a 60° angle and two equal sides, then the triangle is an equilateral triangle.

##4.## Therefore, △ABC is an equilateral triangle.

##5.## Thus, AC = AB = 5.

菱形; 等边三角形; 正弦; 余弦

Definition of Rhombus

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

<note>In quadrilateral ABCD, all sides AB, BC, CD, and DA are equal</note>, therefore quadrilateral ABCD is a rhombus.

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

<note>△ABC is an isosceles triangle, with side AB equal to side BC</note>, and <note>∠B=60°</note>. According to the Equilateral Triangle Identification Theorem, in an isosceles triangle <note>if one interior angle is 60°</note>, then the triangle <note>has three equal sides</note>, <note>each interior angle is 60°</note>, therefore it can be determined that <note>△ABC is an equilateral triangle</note>.

Definition of Equilateral Triangle

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

In the figure of this problem, <note>triangle ABC is an equilateral triangle</note>. <note>Sides AB, BC, and AC are of equal length</note>, and <note>angles ABC, ACB, and BAC are equal in measure, each being 60°</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, CD is the chord of ⊙O, ∠ABD=59°, then ∠C equals ()

Answer: 31°

1. Given AB is the diameter of ⊙O, point D is on the circumference, according to (corollary 2 of the inscribed angle theorem) the inscribed angle subtended by the diameter is a right angle and the definition of inscribed angle, we obtain ∠ADB=90°.

2. From ∠ADB=90° and given ∠ABD=59°, according to the complementary nature of acute angles in a right triangle, we get ∠DAB=90°-∠ABD=31°.

3. According to corollary 1 of the inscribed angle theorem, in the same circle, inscribed angles subtended by the same arc are equal, given points C and A are on the circumference, according to the definition of inscribed angle, ∠C and ∠DAB are inscribed angles, so we can conclude ∠C = ∠DAB = 31°.

####

4. Through the above reasoning, the final answer is 31°.

圆; 圆周角; 弧; 弦; 直角三角形

Definition of Diameter

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

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

Definition of Inscribed Angle

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

In <note>circle O</note>, <note>the vertex D of ∠ADB is on the circumference</note>, <note>the two sides of ∠ADB intersect circle O at points A and B respectively</note>; <note>the vertex A of ∠DAB is on the circumference</note>, <note>the two sides of ∠DAB intersect circle O at points D and B respectively</note>; <note>the vertex C of ∠C is on the circumference</note>, <note>the two sides of ∠C intersect circle O at points B and D respectively</note>. Therefore, <note>∠ADB, ∠DAB, and ∠C are all inscribed angles</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.

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

Complementary Property of Acute Angles in Right Triangle

In a right triangle, the sum of the two acute angles, other than the right angle, is 90°.

In <note>right triangle ABD</note>, <note>angle ADB is a right angle (90 degrees)</note>, angles BAD and ABD are the two acute angles other than the right angle. According to the complementary property of acute angles in a right triangle, <note>the sum of angles DAB and ABD is 90 degrees, i.e., angle DAB + angle ABD = 90°</note>.

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

<note>∠DAB</note> and <note>∠BCD</note> are both inscribed angles subtended by the same arc <note>BD</note>. According to Corollary 1 of the Inscribed Angle Theorem, the inscribed angles <note>∠DAB</note> and <note>∠BCD</note> subtended by the same arc <note>BD</note> are equal, that is, <note>∠BCD</note> = <note>∠DAB</note> = 31°.


Question: The perimeter of parallelogram ABCD is 28cm. AC and BD intersect at point O. The perimeter of △AOB is 4cm larger than the perimeter of △OBC. What is the length of AB?

Answer: 9cm

1. Given the parallelogram ABCD, according to the properties of parallelograms, the diagonals bisect each other, so OA=OC and OB=OD.

2. Given that the perimeter of the parallelogram ABCD is 28cm, according to the properties of parallelograms, opposite sides are equal, so AB=CD and AD=BC.

3. According to the formula for the perimeter, we get: AB + BC + CD + DA = 2(AB + BC) = 28cm, thus we get: AB + BC = 14cm.

4. Given that the perimeter of △AOB is 4cm greater than the perimeter of △OBC, through the ##triangle perimeter theorem## we get: AB + OA + OB -(BC + OB + OC) = AB - BC = 4cm.

5. Combining AB + BC = 14cm and AB - BC = 4cm, substituting and solving we get: 2AB = 18cm, thus AB = 9cm.

6. Through the above reasoning, the final answer is 9cm.

线段; 平行四边形; 普通三角形

Definition of Parallelogram

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

In the figure of this problem, <note>quadrilateral ABCD is a parallelogram</note>, <note>side AB is parallel and equal to side CD</note>, <note>side AD is parallel and equal to side BC</note>.

Properties of Parallelogram Theorem

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

In the figure of this problem, in the parallelogram ABCD, <note>sides AB and CD are equal</note>, <note>sides AD and BC are equal</note>; <note>diagonals AC and BD bisect each other</note>, that is, <note>the intersection point O divides the diagonal AC into two equal segments AO and OC</note>, <note>and divides the diagonal BD into two equal segments BO and OD</note>.

Perimeter of a Parallelogram

The perimeter of a parallelogram is equal to twice the sum of the lengths of its two adjacent sides. The formula is: \( P = 2(a + b) \), where \( a \) and \( b \) are the lengths of the two adjacent sides of the parallelogram.

In the parallelogram <note>ABCD</note>, <note>sides AB and BC are adjacent sides</note>, and <note>sides CD and DA are also adjacent sides</note>. According to the formula for the perimeter of a parallelogram, the perimeter is equal to twice the sum of its two pairs of adjacent sides, i.e., <note>Perimeter P = 2(AB + BC)</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.

<note>Given that the three sides of triangle △AOB are OA, OB, and AB</note>, according to the Perimeter Theorem of Triangle, <note>the perimeter of the triangle is equal to the sum of the lengths of its three sides</note>, that is, the perimeter P=OA+OB+AB. Similarly, <note>given that the three sides of triangle △OBC are OC, OB, and BC</note>, according to the Perimeter Theorem of Triangle, <note>the perimeter of the triangle is equal to the sum of the lengths of its three sides</note>, that is, the perimeter P=BC + OB + OC.


Question: As shown in the figure, in parallelogram ABCD, it is known that AB=6, BC=9, ∠B=30°, then the area of parallelogram ABCD is ()

Answer: 27

1. Draw AE perpendicular to BC through point A, intersecting BC at point E.

2. According to AE perpendicular to BC, we get ∠AEB=90°.

3. Given AB=6, ∠B=30°, ##in right triangle △AEB, according to the definition of the sine function##, AE=AB×sin(30°)=6×1/2=3.

4. According to the area formula of the parallelogram S=base×height, given BC=9, therefore the area of parallelogram ABCD is 9×3=27.

Definition of Parallelogram

<note>Side AB is parallel and equal to side CD</note>, <note>side AD is parallel and equal to side BC</note>.

Definition of Right Triangle

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

<note>In triangle AEB, angle ∠AEB is a right angle (90 degrees)</note>, therefore triangle AEB is a right triangle. <note>Side AE and side BE are the legs</note>, <note>side AB is the hypotenuse</note>.

Definition of Sine Function

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

In the <note>right triangle AEB</note>, <note>angle ABE is an acute angle</note>, <note>side AE is the opposite side to angle ABE</note>, and <note>side AB is the hypotenuse</note>. According to the definition of the sine function, the <note>sine of angle ABE</note> is equal to the <note>ratio of the opposite side AE to the hypotenuse AB</note>, that is, <note>sin(ABE) = AE / AB</note>.

Area Formula of a Parallelogram

The area of a parallelogram is calculated by multiplying the length of its base by the corresponding height, which can be expressed as \( A = \text{base length} \times \text{height} \).

In the figure of this problem, in parallelogram ABCD, <note>side BC is the base</note>, the corresponding height is <note>the vertical distance from the base BC to the opposite side AD, denoted as AE</note>. Therefore, according to the area formula of a parallelogram, the area of parallelogram ABCD is equal to <note>the length of the base BC multiplied by the corresponding height AE</note>, i.e., A = BC × AE.

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).

The line segment AE <note>from vertex A perpendicular to the opposite side BC (or its extension)</note> is the altitude from vertex A. The line segment AE forms a right angle (90 degrees) with side BC, indicating that <note>the line segment AE is the perpendicular distance from vertex A to the opposite side BC</note>.


Question: In the parallelogram ABCD shown in the figure, the diagonals AC and BD intersect at point O. Given AB=7, AC=10, and the perimeter of △ABO is 16, find the length of diagonal BD.

Answer: 8

1. Given AB = 7, AC = 10, and the perimeter of △ABO is 16, according to the formula for the perimeter of a triangle, we get AB + OA + OB = 16.

2. Substitute the given condition AB, we get 7 + OA + OB = 16, thus we get OA + OB = 9.

3. Since ABCD is a parallelogram, the two diagonals AC and BD bisect each other. According to the properties of diagonals, we get AC = 2 * OA and BD = 2 * OB.

4. According to AC = 10, we get 2 * OA = 10, thus we get OA = 5.

5. From OA + OB = 9 and OA = 5, we calculate OB = 9 - 5 = 4.

6. According to the properties of diagonals BD = 2 * OB, we get BD = 2 * 4 = 8.

7. Through the above reasoning, we finally get the answer as 8.

平行四边形; 线段; 平行线; 中点; 普通三角形

Definition of Parallelogram

<note>In the figure of this problem, quadrilateral ABCD is a parallelogram</note>, <note>sides AB and CD are parallel and equal</note>, <note>sides AD and BC are parallel and equal</note>.

Properties of Parallelogram Theorem

In the diagram of this problem, in the parallelogram <note>ABCD</note>, the opposite angles <note>∠A and ∠C</note> are equal, the opposite angles <note>∠B and ∠D</note> are equal; the sides <note>AB and CD</note> are equal, the sides <note>AD and BC</note> are equal; the diagonals <note>AC and BD</note> bisect each other, that is, the intersection point <note>O</note> divides the diagonal <note>AC</note> into two equal segments <note>AO and OC</note>, and divides the diagonal <note>BD</note> into two equal segments <note>BO and OD</note>.

Perimeter Theorem of Triangle

<note>The three sides of triangle ABO are AB, OA, and OB</note>. According to the Perimeter Theorem of Triangle, <note>the perimeter of the triangle is equal to the sum of the lengths of its three sides</note>, i.e., the perimeter <note>P=AB+OA+OB</note>.


Question: As shown in the figure, a rectangular ruler is broken along a straight line and displaced. Points E, D, B, F are on the same straight line. If ∠ADE = 125°, then the measure of ∠DBC is ()

Answer: 55°

1. According to the problem, line AD is parallel to line CB.

2. Since ∠ADE and ∠ADF are supplementary, ∠ADE + ∠ADF = 180°.

3. Given ∠ADE = 125°, therefore ∠ADF = 55°.

4. Because line AD is parallel to line CB, ∠ADF and ∠CBD are alternate interior angles, ∠ADF = ∠CBD = 55°.

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

平行线; 内错角; 对顶角; 线段

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 AD is parallel to line CB</note>, <note>∠ADF</note> and <note>∠DBC</note> are <note>alternate interior angles</note>, according to the <note>Alternate Interior Angles Theorem</note>, it can be concluded that <note>∠ADF=∠DBC</note>, which is 55°.

Definition of Alternate Interior Angles

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

<note>In the figure of this problem, two parallel lines AD and BC are intersected by a line EF</note>, <note>where angle ADF and angle CBD are located between the two parallel lines and on opposite sides of the intersecting line EF</note>, therefore <note>angle ADF and angle CBD are alternate interior angles</note>. <note>Alternate interior angles are equal</note>, that is, <note>angle ADF is equal to angle CBD</note>.

Definition of Linear Pair of Angles

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

<note>Angle ADE and angle ADF have a common side DA</note>, <note>their other sides DE and DF are extensions in opposite directions</note>, so angle ADE and angle ADF are adjacent supplementary angles.


Question: As shown in the figure, the diagonals of parallelogram ABCD intersect at point O, and AB=5. The perimeter of △OCD is 23. Then the sum of the two diagonals of parallelogram ABCD is ()

Answer: 36

1. Given quadrilateral ABCD is a parallelogram, according to the properties of a parallelogram, opposite sides are equal, thus AB=CD=5.

2. Since the perimeter of triangle OCD is 23, therefore OC+OD+CD=23.

3. From step 1, it is known that CD=5, thus OC+OD=23-5=18.

4. The diagonals of a parallelogram bisect each other, hence BD=2*OD, AC=2*OC.

5. Since OC+OD=18, therefore BD+AC=2*OC+2*OD=2*(OC+OD)=2*18=36.

6. Through the above reasoning, it is concluded that the sum of the two diagonals of parallelogram ABCD is 36.

平行四边形; 线段; 中点; 普通三角形; 旋转

Definition of Parallelogram

<note>Quadrilateral ABCD is a parallelogram</note>, <note>sides AB and CD are parallel and equal</note>, <note>sides AD and BC are parallel and equal</note>, where AB=5, so <note>CD=5</note>.

Properties of Parallelogram Theorem

In this problem, in the parallelogram ABCD, <note>the opposite angles ∠BAD and ∠BCD are equal, and the opposite angles ∠ABC and ∠ADC are equal</note>; <note>the sides AB and CD are equal, and the sides AD and BC are equal</note>; <note>the diagonals AC and BD bisect each other</note>, that is, the intersection point O divides <note>the diagonal AC into two equal segments AO and OC</note>, and divides <note>the diagonal BD into two equal segments BO and OD</note>.

Perimeter Theorem of Triangle

In the figure of this problem, it is known that the three sides of triangle △OCD are <note>OC</note>, <note>OD</note>, and <note>CD</note>. According to the Perimeter Theorem of Triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, the perimeter <note>P=OC + OD + CD</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, C and D are two points on ⊙O, CD ⊥ AB, if ∠DAB = 70°, then ∠BOC = ()

Answer: 140°

1. From the figure, it can be seen that point A and point B are the endpoints of the diameter of circle O, and points C and D are on the circumference, with CD perpendicular to AB.

2. According to the problem statement, ∠DAB=70°.

3. Since CD is perpendicular to AB, let CD intersect AB at E. According to the definition of perpendicular lines, ∠AED=90°.

4. In triangle AED, according to the triangle angle sum theorem, ∠ADE=180°-∠DAE-∠AED=20°.

5. According to the inscribed angle theorem, in a circle, the inscribed angle is equal to half of the central angle that subtends the same arc, so ∠AOC=2∠ADC.

7. According to the definition of a straight angle, ∠BOC=180°-∠AOC.

9. Through the above reasoning, the final answer is 140°.

圆; 圆心角; 圆周角; 垂线; 直角三角形

Definition of Inscribed Angle

In the figure of this problem, in circle <note>O</note>, <note>the vertex of angle ADC (point D) 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>.

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>Line CD and line AB intersect to form an angle ∠AED of 90 degrees</note>, therefore according to the <note>Definition of Perpendicular Lines</note>, <note>line CD and line AB are perpendicular to each other</note>.

Triangle Angle Sum Theorem

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

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

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

Definition of Central Angle

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

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

Definition of Straight Angle

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

<note>Ray AO</note> rotates around the endpoint <note>O</note> until it forms a straight line with the initial side, creating <note>straight angle AOB</note>. According to the definition of a straight angle, <note>the measure of a straight angle is 180 degrees</note>, i.e., <note>angle AOB = 180 degrees</note>.


Question: As shown in the figure, the perpendicular bisector of the diagonal AC of parallelogram ABCD intersects sides BC and DA at points E and F respectively. Connect CF. If the perimeter of parallelogram ABCD is equal to 18cm, then the perimeter of △CDF is equal to ()

Answer: 9cm

1. Given that the perpendicular bisector of diagonal AC of parallelogram ABCD intersects sides BC and DA at points E and F, respectively.

2. The perimeter of parallelogram ABCD is 18cm. According to the properties of parallelograms, AB = CD and AD = BC, so AB + CD + AD + BC = 18cm. Let AB = CD = x and AD = BC = y, then 2x + 2y = 18, hence x + y = 9.

3. In parallelogram ABCD, connect CF. According to the properties of perpendicular bisectors, AF = CF, thus the perimeter of △CDF is equal to AD + DC.

4. Since AD = BC = y and DC = x, the perimeter of △CDF is equal to x + y, and x + y = 9.

5. Through the above reasoning, it is concluded that the perimeter of △CDF is 9cm.

平行四边形; 垂直平分线; 等腰三角形; 线段; 三角形的外角

Properties of the Perpendicular Bisector

Any point on the perpendicular bisector of a line segment is equidistant from the endpoints of the segment.

<note>The perpendicular bisector of line segment AC is line EF</note>, <note>F is on line EF</note>. According to the properties of the perpendicular bisector, <note>point F is equidistant from the endpoints A and C of line segment AC</note>, that is, <note>AF = CF</note>.

Definition of Parallelogram

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

<note>Quadrilateral ABCD is a parallelogram</note>, <note>side AB is parallel and equal to side CD</note>, <note>side AD is parallel and equal to side BC</note>. This establishes <note>AB = CD = x</note> and <note>AD = BC = y</note>.

Properties of Parallelogram Theorem

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

In the figure of this problem, in parallelogram ABCD, <note>the opposite angles ∠A and ∠C are equal</note>, <note>the opposite angles ∠B and ∠D are equal</note>; <note>sides AB and CD are equal</note>, <note>sides AD and BC are equal</note>; <note>the diagonals AC and BD bisect each other</note>.

Formula for the Perimeter of a Triangle

The perimeter of a triangle is the sum of the lengths of its three sides. Therefore, if the three sides of the triangle are denoted as \( a \), \( b \), and \( c \), then the perimeter \( P \) is given by \( P = a + b + c \).

The three sides of <note>triangle CDF</note> are <note>CD, CF, DF</note>, according to the formula for the perimeter of a triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, the perimeter <note>P=CD+CF+DF</note>.


Question: As shown in the figure, P is a point inside the parallelogram ABCD. Through point P, lines parallel to AB and AD are drawn, intersecting the parallelogram at points E, F, G, and H respectively. If S~AHPE~=3 and S~PFCG~=5, then S~△PBD~ is ()

Answer: 1

1. First, from the given conditions, it can be determined that EPGD, GPFC, EPHA, and PHBF are all parallelograms.

2. According to the area properties of parallelograms, we have ##S△DEP = S△DGP = 1/2 * S parallelogram DEPG##.

3. Similarly, through the area properties of parallelograms, we can obtain ##S△PHB = S△PBF = 1/2 * S parallelogram PHBF##.

4. Calculate the area of the entire parallelogram. According to the area partition properties of parallelograms, we have ##S△ADB = S△EPD + S parallelogram AHPE + S△PHB + S△PDB##.

5. Similarly, the area on the other side can be expressed as ##S△BCD = S△PDG + S parallelogram PFCG + S△PFB - S△PDB##.

6. Subtract the above two area formulas, we get ##0 = S parallelogram AHPE - S parallelogram PFCG + 2S△PDB##.

7. Substitute the given ##S parallelogram AHPE## = 3 and ##S parallelogram PFCG## = 5, we get ##2S△PDB = 5 - 3 = 2##.

8. According to the positive nature of the area, we get ##S△PBD## = 1.

平行四边形; 平行线; 普通三角形; 点

Definition of Parallelogram

<note>Quadrilateral ABCD is a parallelogram</note>, <note>sides AB and CD are parallel and equal</note>, <note>sides AD and BC are parallel and equal</note>. Quadrilateral EPGD is a parallelogram, sides EP and DG are parallel and equal, sides ED and PG are parallel and equal. Quadrilateral GPFC is a parallelogram, sides GP and FC are parallel and equal, sides GC and PF are parallel and equal. Quadrilateral EPHA is a parallelogram, sides EP and AH are parallel and equal, sides EA and PH are parallel and equal. Quadrilateral PHBF is a parallelogram, sides PH and FB are parallel and equal, sides PF and FB are parallel and equal.

Transitivity of Parallel Lines

If two lines are each parallel to a third line, then those two lines are parallel to each other.

In the figure of this problem, <note>line DC and line EF are parallel to line AB respectively</note>. According to the transitivity of parallel lines, if <note>line DC is parallel to line AB</note>, and <note>line EF is also parallel to line AB</note>, then <note>line DC and line EF are parallel to each other</note>. Therefore, <note>line DC is parallel to line EF</note>. In the figure of this problem, <note>line DA and line GH are parallel to line CB respectively</note>. According to the transitivity of parallel lines, if <note>line DA is parallel to line CB</note>, and <note>line GH is also parallel to line CB</note>, then <note>line DA and line GH are parallel to each other</note>. Therefore, <note>line DA is parallel to line GH</note>.


Question: As shown in the figure, ⊙O is the circumcircle of △ABC. Given ∠C=60°, what is the degree measure of ∠BAO?

Answer: 30°

1. Connect OB to form △AOB.

2. According to the inscribed angle theorem, the central angle is twice the inscribed angle it subtends, i.e., ∠AOB = 2 × ∠ACB = 2 × 60° = 120°.

3. In △AOB, since OA and OB are radii, OA = OB, so △AOB is an isosceles triangle.

4. In an isosceles triangle, according to the properties of isosceles triangles, ∠BAO = ∠ABO.

5. According to the triangle angle sum theorem, ∠AOB + ∠BAO + ∠ABO = 180°. Combining with step 4, we get ∠AOB + 2 × ∠BAO = 180°.

6. Substitute ∠AOB = 120° from step 2, we get 120° + 2 × ∠BAO = 180°.

7. Solve this equation to get 2 × ∠BAO = 60°, so ∠BAO = 30°.

8. Through the above reasoning, the final answer is ∠BAO = 30°.

圆; 圆周角; 圆心角; 普通三角形; 等边三角形

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

Definition of Central Angle

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

Definition of Isosceles Triangle

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

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

Inscribed Angle Theorem

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

Properties of Isosceles Triangle

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

In the <note>isosceles triangle △AOB</note>, <note>sides OA and OB are equal</note>. Therefore, according to the <note>properties of isosceles triangles</note>, the angles opposite the equal sides are equal, i.e., <note>∠BAO = ∠ABO</note>.

Triangle Angle Sum Theorem

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


Question: As shown in the figure, it is known that AB and AD are chords of ⊙O, ∠B=20°, point C is on chord AB, connect CO and extend CO to intersect ⊙O at point D, ∠D=15°, then the degree of ∠BAD is ()

Answer: 35°

1. Connect OA. Since OA=OB and ∠ABO=20°, according to the properties of an isosceles triangle, we get ∠OAB=20° (the two base angles of an isosceles triangle are equal).

2. Since OA=OD and ∠ADO=15°, according to the properties of an isosceles triangle, we get ∠OAD=15° (the two base angles of an isosceles triangle are equal).

3. Since ∠BAD is composed of ∠OAB and ∠OAD, we can get ∠BAD=∠OAB + ∠OAD=20° + 15°=35°.

4. Through the above reasoning, we finally get the answer that ∠BAD equals 35°.

弦; 圆; 圆周角; 圆内接四边形; 内错角

Definition of Isosceles Triangle

In triangle OAB, sides OA and OB are equal, therefore triangle OAB is an isosceles triangle; similarly, in triangle ODA, sides OA and OD are equal, therefore triangle ODA is an isosceles triangle. According to the properties of an isosceles triangle, the two base angles ∠OBA and ∠OAB of triangle OAB are equal, and ∠OBA=∠OAB=20°; the two base angles ∠ODA and ∠OAD of triangle ODA are equal, and ∠ODA=∠OAD=15°.

Properties of Isosceles Triangle

In the figure of this problem, <note>in the isosceles triangle OAB, sides OA and OB are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle OAB = angle OBA</note>; <note>in the isosceles triangle OAD, sides OA and OD are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle OAD = angle ODA</note>.


Question: As shown in the figure, in rhombus ABCD, ∠BAD=80°, the perpendicular bisector of AB intersects diagonal AC at point F, E is the foot of the perpendicular, connect DF, then ∠CDF equals ()

Answer: 60°

1. Given rhombus ABCD, then ##by the property of supplementary adjacent angles of a rhombus##, we get ∠ADC=180°-∠BAD=100°.

2. The perpendicular bisector of AB intersects the diagonal AC at point F, E is the foot of the perpendicular, connect DF.

3. ##According to the property of the perpendicular bisector##, since EF is the perpendicular bisector of AB, AF=BF.

4. ##Connect the diagonals DB and AC at point O, by the property of the diagonals of a rhombus, AC is the perpendicular bisector of DB, thus FO is actually the perpendicular bisector of DB, according to the property of the perpendicular bisector,## BF=DF.

5. Based on the above conclusions, AF=BF=DF, then, ##according to the definition of an isosceles triangle, △ADF is an isosceles triangle, since ∠DFC is the exterior angle of triangle ADF, according to the property of an isosceles triangle, ∠DAF＝∠ADF, so ∠DFC=2∠DAF##.

6. ##By the definition of a rhombus and the congruence theorem (SSS), triangles ADC and ABC have corresponding sides equal, thus the two triangles are congruent##.

7. ##By the definition of a rhombus, AD is equal to DC, so by the definition of an isosceles triangle, triangle ADC is an isosceles triangle, according to the property of an isosceles triangle, ∠DAF＝∠DCF, since ∠BAD=80°, by the definition of congruent triangles, ∠DAC is equal to ∠CAB, thus ∠DAC＝∠DCF＝40°, ∠DFC=2∠DAF＝80°##.

8. ##In triangle DFC, by the triangle sum theorem, ∠CDF＝180°-∠DCF-∠DFC＝180°-40°-80°＝60°##.

菱形; 垂直平分线; 垂线; 等腰三角形; 直角三角形

Definition of Perpendicular Bisector

A line is a perpendicular bisector if and only if it passes through the midpoint of a line segment and is perpendicular to that line segment.

In the figure of this problem, <note>line EF passes through the midpoint E of segment AB, and line EF is perpendicular to segment AB</note>. Therefore, <note>line EF is the perpendicular bisector of segment AB</note>. <note>Line FO passes through the midpoint O of segment DB, and line FO is perpendicular to segment DB</note>. Therefore, <note>line FO is the perpendicular bisector of segment DB</note>.

Definition of Isosceles Triangle

In the diagram of this problem, in triangle ADF, <note>side AF and side DF are equal</note>, therefore triangle ADF is an isosceles triangle. In triangle ADC, <note>side AD and side DC are equal</note>, therefore triangle ADC is an isosceles triangle.

Definition of Rhombus

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

In the quadrilateral <note>ABCD</note>, all sides <note>AB, BC, CD, DA</note> are equal, therefore <note>the quadrilateral ABCD is a rhombus</note>.

<note>In the figure of this problem, the perpendicular bisector of segment AB is line EF</note>, <note>point F is on line EF</note>. According to the properties of the perpendicular bisector, <note>the distance from point F to the endpoints A and B of segment AB is equal</note>, that is, <note>AF = BF</note>. <note>The perpendicular bisector of segment DB is line FO</note>, <note>point F is on line FO</note>. According to the properties of the perpendicular bisector, <note>the distance from point F to the endpoints D and B of segment DB is equal</note>, that is, <note>DF = FB</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 DAF, angle ∠DFC is an exterior angle of the triangle</note>, <note>angles ∠FAD and ∠ADF are the two interior angles that are not adjacent to the exterior angle ∠DFC</note>, according to the Exterior Angle Theorem of Triangle, <note>the exterior angle ∠DFC is equal to the sum of the two non-adjacent interior angles ∠FAD and ∠ADF</note>, that is, <note>angle ∠DFC = angle ∠FAD + angle ∠ADF</note>.

Adjacent Angles of Rhombus are Supplementary

Any pair of adjacent angles in a rhombus are supplementary.

In the figure of this problem, in rhombus ABCD, <note>angle ADC and angle DAB are a pair of adjacent angles</note>, according to the properties of the rhombus, <note>any pair of adjacent angles of a rhombus are supplementary</note>, that is, <note>angle ADC + angle DAB = 180 degrees</note>. Similarly, <note>angle DCB and angle CBA are also a pair of adjacent angles</note>, satisfying <note>angle DCB + angle CBA = 180 degrees</note>.

Properties of the Diagonals of a Rhombus

In a rhombus, the diagonals bisect each other and are perpendicular to each other.

In the figure of this problem, in rhombus ABCD, <note>diagonals AC and BD bisect each other and are perpendicular to each other</note>. Specifically, <note>point O is the intersection of diagonals AC and BD</note>, and <note>AO=OC and DO=OB</note>. At the same time, <note>angle DOA and angle DOC are both right angles (90 degrees)</note>, so <note>diagonals AC and BD are perpendicular to each other</note>.

Properties of Isosceles Triangle

In the <note>isosceles triangle ADF</note>, sides <note>DF and AF are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>angle ADF = angle DAF</note>. In the <note>isosceles triangle ADC</note>, sides <note>AD and DC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>angle DAC = angle DCA</note>.

Definition of Congruent Triangles

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

<note>Triangle ADC and triangle ABC are congruent triangles</note>, the <note>corresponding sides and corresponding angles of triangle ADC are equal to those of triangle ABC</note>, that is:
<note>side AD = side AB</note>
<note>side DC = side BC</note>
side AC = side AC
At the same time, the corresponding angles are also equal:
<note>angle DAC = angle CAB</note>
<note>angle DCA = angle BCA</note>
<note>angle ADC = angle ABC</note>

Triangle Congruence Theorem (SSS)

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

In the diagram of this problem, in triangles ADC and ABC, <note>side AD is equal to side AB</note>, <note>side DC is equal to side BC</note>, <note>side AC is equal to side AC</note>, therefore, according to the Triangle Congruence Theorem (SSS), these two triangles are congruent.

Triangle Angle Sum Theorem

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

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

Definition of Exterior Angle of a Polygon

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

In the diagram of this problem, <note>an interior angle of polygon ADF is ∠AFD</note>, <note>the angle formed by extending the adjacent sides DF and AF</note> is called <note>the exterior angle ∠DFC of the interior angle ∠AFD</note>.


Question: As shown in the figure, the diagonals AC and BD of parallelogram ABCD intersect at point O, and point E is the midpoint of CD. The perimeter of △ABD is 16 cm. Then the perimeter of △DOE is ()

Answer: 8cm

1. Given that quadrilateral ABCD is a parallelogram, according to the ##theorem of parallelogram properties##, the diagonals of a parallelogram bisect each other, so point O is the midpoint of BD.

2. ##According to the theorem of parallelogram properties, opposite sides are equal, AD=BC, AB=CD, and since side BD is shared, according to the congruent triangles theorem (SSS), triangle ABD is congruent to triangle CDB##.

3. ##Since point O is the midpoint of BD and E is the midpoint of CD, according to the midline theorem of triangles, segment OE is the midline of triangle CDB##.

4. According to the midline theorem of triangles, ##in a triangle, if a segment connects the midpoints of two sides, then this segment is parallel to and equal to half of the third side##, so OE=rac{1}{2}BC.

5. ##Since point O is the midpoint of BD, E is the midpoint of CD, and OE=rac{1}{2}BC##, the perimeter of △DOE is equal to rac{1}{2} the perimeter of △BCD.

6. ##Since triangle ABD is congruent to triangle CDB##, the perimeter of △ABD is equal to the perimeter of △CDB.

7. Thus, it is derived that the perimeter of △DOE=rac{1}{2}×16=8cm.

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

平行四边形; 中点; 线段; 普通三角形; 点

Formula for the Perimeter of a Triangle

<note>The three sides of triangle DEO are DO, DE, and OE</note>, according to the formula for the perimeter of a triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, <note>Perimeter P=DO+DE+OE</note>, in <note>triangle ABD, the three sides are AB, AD, and DB</note>, according to the formula for the perimeter of a triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, <note>Perimeter P=AB+AD+DB</note>, in <note>triangle BCD, the three sides are BC, DC, and BD</note>, according to the formula for the perimeter of a triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, <note>Perimeter P=BC+DC+BD</note>.

Triangle Midline Theorem

In a triangle, if a line segment connects the midpoints of two sides, then this line segment is parallel to the third side and is equal to half the length of the third side.

<note>Point E is the midpoint of side CD</note>, <note>Point O is the midpoint of side BD</note>, <note>Line segment OE connects these two midpoints</note>. According to the Triangle Midline Theorem, <note>Line segment OE is parallel to the third side BC</note> and <note>equals half of the third side BC</note>, that is, <note>OE || BC</note>, and <note>OE = 1/2 * BC</note>.

Properties of Parallelogram Theorem

<note>In parallelogram ABCD</note>, <note>sides AB and CD are equal</note>, <note>sides AD and BC are equal</note>; <note>the diagonals AC and BD bisect each other</note>, that is, <note>the intersection point O divides diagonal AC into two equal segments AO and OC</note>, <note>and divides diagonal BD into two equal segments BO and OD</note>.

Triangle Congruence Theorem (SSS)

<note>In triangles ABD and CDB</note>, <note>AB=CD</note>, <note>AD=BC</note>, and <note>BD is the common side</note>. According to the Side-Side-Side (SSS) Congruence Theorem, <note>triangle ABD is congruent to triangle CDB</note>.

Definition of Congruent Triangles

<note>Triangles ABD and CDB are congruent triangles</note>, the <note>corresponding sides and corresponding angles of triangle ABD are equal to those of triangle CDB</note>, that is: <note>side AB = side CD</note>, <note>side BD = side BD</note>, <note>side AD = side CB</note>, and the corresponding angles are also equal: <note>angle BAD = angle CDB</note>, <note>angle ABD = angle BDC</note>, <note>angle ADB = angle DCB</note>.


Question: As shown in the figure, AB is a chord of ⊙O, and a tangent AC is drawn through point A. If ∠BAC=55°, then ∠AOB equals ()

Answer: 110°

1. Given that AC is the tangent to ⊙O, according to the property of the tangent, OA⊥AC.

2. From this, we can deduce that ∠OAC=90°.

3. Given that ∠BAC=55°, according to the definition of angles, ∠BAC and ∠OAB sum up to ∠OAC.

5. According to the definition of radius, we know that OA=OB. According to the definition of an isosceles triangle, ΔOAB is an isosceles triangle. According to the property of an isosceles triangle, in ΔOAB, ∠OAB=∠OBA.

7. In ΔOAB, according to the triangle sum theorem, ∠AOB=180°-∠OAB-∠OBA=180°-2×35°.

圆; 切线; 圆心角; 弧; 圆周角

Definition of Tangent to a Circle

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

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

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.

In the figure of this problem, in circle O, point A is the point of tangency of line AC with the circle, and segment OA is the radius of the circle. According to the property of the tangent line to a circle, <note>the tangent line AC is perpendicular to the radius OA at the point of tangency A</note>, <note>i.e., ∠OAC = 90°</note>.

Properties of Isosceles Triangle

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

Triangle Angle Sum Theorem

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

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.

In the figure of this problem, <note>angle OAC is a geometric figure formed by two rays AO and AC</note>, these two rays <note>have a common endpoint A</note>. <note>This common endpoint A is called the vertex of angle OAC</note>, and <note>rays AO and AC are called the sides of angle OAC</note>.

Definition of Radius

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

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

Definition of Isosceles Triangle

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


Question: As shown in the figure, in the parallelogram ABCD, the line CE passing through point C is perpendicular to AB, with E as the foot of the perpendicular. If ∠EAD = 54°, then the degree of ∠BCE is ()

Answer: 36°

1. Given that line CE is perpendicular to segment AB, with the foot of the perpendicular being point E, we have ∠AEC=90°.

2. Given ∠EAD=54°, let the intersection of AD and CE be point F. According to the triangle angle sum theorem, we have ∠AFE = 180°-90°-54°=36°.

3. Quadrilateral ABCD is a parallelogram. By the properties of parallelograms, side AD is parallel to side BC.

4. According to the parallel line axiom 2, when AD∥BC, ∠ECB = ∠AFE.

5. From the above reasoning, we obtain ∠BCE = 36°.

平行四边形; 垂线; 内错角; 邻补角; 直角三角形

Triangle Angle Sum Theorem

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

Properties of Parallelogram Theorem

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

In the figure of this problem, in parallelogram ABCD, <note>the opposite angles ∠BAD and ∠BCD are equal</note>, <note>the opposite angles ∠B and ∠D are equal</note>; <note>side AB and side CD are equal</note>, <note>side AD and side BC are equal</note>; <note>the diagonals AC and BD bisect each other</note>.

Parallel Postulate 2 of Parallel Lines

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

In the figure of this problem, <note>two parallel lines AD and BC are intersected by a third line CE</note>, forming the following geometric relationships: 1. <note>Corresponding angles: angle AFE and angle BCE are equal</note>. 2. <note>Alternate interior angles: angle DFC and angle BCF are equal</note>. 3. <note>Consecutive interior angles: angle AFC and angle BCF are supplementary, that is, angle AFC + angle BCF = 180 degrees</note>. These relationships demonstrate that when two parallel lines are intersected by a third line, corresponding angles are equal, alternate interior angles are equal, and consecutive interior angles are supplementary.

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.

<note>Two parallel lines AD and BC</note> are intersected by <note>a line EC</note>, where <note>angle AFE and angle ECB are on the same side of the intersecting line EC</note>, on the <note>same side of the two intersected lines AD and BC</note>, therefore <note>angle AFE and angle ECB are corresponding angles</note>. Corresponding angles are equal, that is <note>angle AFE is equal to angle ECB</note>.


Question: As shown in the figure, ∠MON=90°, moving points A and B are located on rays OM and ON respectively, the side AB of rectangle ABCD is 6, and BC is 4. Then the maximum length of segment OC is ()

Answer: 8

1. Given ∠MON=90°, the sides of rectangle ABCD are AB=6 and BC=4.

2. Take the midpoint E of AB and connect OE, CE.

3. Since E is the midpoint of AB, ##according to the midpoint theorem,## BE=##1/2##AB=3.

4. According to the given ∠MON=90°, by ##the median theorem of the hypotenuse in a right triangle,## OE=##1/2##AB=3.

5. In the right triangle BCE, according to the Pythagorean theorem, CE=##√(BE^2 + BC^2) = √(3^2 + 4^2)##= 5.

6. In △OCE, according to the triangle inequality theorem, CE + OE > OC.

7. When points O, E, and C are collinear, the maximum value of OC is OE + CE = 3 + 5 = 8.

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

射线; 矩形; 直角三角形; 点; 线段

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 AB is point E</note>. According to the definition of the midpoint of a line segment, <note>point E divides line segment AB into two equal parts</note>, that is, <note>the lengths of line segments AE and EB are equal</note>. That is, <note>AE = EB = 3</note>.

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 (≅).

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

Pythagorean Theorem

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

<note>In the diagram of this problem, in the right triangle BCE, ##∠CBE## is a right angle (90 degrees)</note>, the sides <note>BE and BC are the legs</note>, and the side <note>CE is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>##CE^2 = BE^2 + BC^2##</note>, therefore <note>CE = √(BE^2 + BC^2) = √(3^2 + 4^2) ##= 5.</note>

Median Theorem of Right Triangle

In a right triangle, the line segment drawn from the vertex of the right angle to the midpoint of the hypotenuse is equal to half the length of the hypotenuse.

<note>In the right triangle BOA, angle BOA is a right angle (90 degrees)</note>, sides OB and OA are the legs, and side AB is the hypotenuse. <note>Point E is the midpoint of the hypotenuse AB</note>, and the segment drawn from the right angle vertex O to the midpoint E of the hypotenuse <note>OE is equal to half of the hypotenuse AB, that is, OE = 1/2 * AB</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 OE</note>, <note>side CE</note>, and <note>side OC form a triangle</note>. According to the <note>Theorem of Triangle Inequality</note>, the sum of any two sides is greater than the third side, that is, <note>side CE + side OE > side OC</note>, <note>side CE + side OC > side OE</note>, <note>side OC + side OE > side CE</note>.

Definition of Right Triangle

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

<note>In the diagram of this problem, in triangle BCE, angle CBE is a right angle (90 degrees)</note>, therefore triangle BCE is a right triangle. <note>Side BC and side BE are the legs, side CE is the hypotenuse</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, point C is on ⊙O, if ∠ABC=30°, then ∠CAB is ()

Answer: 60°

1. Given AB is the diameter of circle O, and point C is on circle O. According to the ##(Corollary 2 of the Inscribed Angle Theorem) the inscribed angle subtended by the diameter is a right angle##, it follows that ∠ACB is a right angle, i.e., ∠ACB=90°.

 2. From ∠ABC=30° and ∠ACB=90°, ##according to the Triangle Angle Sum Theorem##, which states that the sum of the three interior angles of a triangle is equal to 180°, it is calculated that ∠CAB=180°-∠ACB-∠ABC=180°-90°-30°=60°.

圆; 圆周角; 直角三角形

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

Triangle Angle Sum Theorem

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

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

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

<note>In the diagram of this problem, in circle O, the angle subtended by the diameter AB at the circumference, angle ACB, is a right angle (90 degrees)</note>. (Or <note>In the diagram of this problem, the angle subtended by the circumference, angle ACB, is 90 degrees, so the chord AB subtended by it is the diameter</note>.)


Question: As shown in the figure, AB is the diameter of ⊙O, points C and D are on ⊙O, connect AC, BC, AD, CD. If ∠BAC=50°, then the degree of ∠ADC is equal to ()

Answer: 40°

1. Given that AB is the diameter of ⊙O, according to the theorem 'The angle subtended by a diameter is a right angle', we get ∠ACB=90°.

2. From ∠ACB=90° and the given condition ∠BAC=50°, according to the 'Sum of angles in a triangle theorem', we obtain ∠ABC=40°.

4. According to the corollary of the Inscribed Angle Theorem, in a circle, any two inscribed angles corresponding to the same arc are equal, i.e., ∠ADC=∠ABC. Based on the above reasoning, we get ∠ABC=40°, so ∠ADC=40°.

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

圆; 圆周角; 圆内接四边形; 直角三角形

Definition of Inscribed Angle

Original: <note>在本题图中，##∠ABC、∠ADC##都是圆周角</note>。以<note>##∠ABC##为例，##角ABC##的##顶点B##在圆周上</note>，<note>##角ABC##的两边##AB和BC##分别与圆O相交于##点A和点C##</note>。因此，<note>##角ABC##是一个圆周角</note>。

Translation: <note>In the diagram of this problem, ##∠ABC, ∠ADC## are both inscribed angles</note>. Taking <note>##∠ABC## as an example, the vertex B of ##∠ABC## is on the circumference</note>, <note>the two sides of ##∠ABC##, AB and BC, intersect the circle O at points A and C respectively</note>. Therefore, <note>##∠ABC## is an inscribed angle</note>.

<note>In the diagram of this problem, in circle O, the inscribed angle ACB subtended by the diameter AB is a right angle (90 degrees)</note>.

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

<note>In circle O, the inscribed angles ∠ABC and ∠ADC corresponding to arc AC are equal</note>. According to <note>Corollary 1 of the Inscribed Angle Theorem</note>, the inscribed angles <note>∠ABC and ∠ADC corresponding to the same arc AC are equal</note>, that is, <note>∠ABC = ∠ADC</note>.

Triangle Angle Sum Theorem

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


Question: As shown in the figure, line segment AB passes through the center O of the circle and intersects ⊙O at points A and C. ∠B = 30°, line BD is tangent to ⊙O at point D. Then, the measure of ∠ADB is ()

Answer: 120°

1. Given that the line segment AB passes through the center of the circle O, intersecting ⊙O at points A and C, and the line BD is tangent to ⊙O at point D. Connect OD.

2. Since BD is the tangent to ⊙O at point D, according to the properties of tangents, we get ∠ODB=90°.

3. According to the triangle angle sum theorem, because ∠B=∠OBD=30°, in △ODB, ∠DOB=180°-∠ODB-∠OBD=180°-90°-30°=60°.

4. According to the inscribed angle theorem, ∠DOC=∠DOB=2∠DAC.

5. The inscribed angle ∠DAC=1/2∠DOC=1/2 * 60°=30°.

6. According to the triangle angle sum theorem, in △ADB, ∠ADB=180°-∠DAB-∠DBA=180°-30°-30°=120°.

7. Through the above reasoning, the final answer is 120°.

圆; 切线; 圆心角; 圆周角; 等腰三角形

Definition of Circle

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

In the <note>circle ⊙O</note>, <note>point O is the center</note>, and the <note>radius is OA (or OC)</note>. All points in the figure that are equidistant from point O, equal to the distance of <note>OA (or OC)</note>, lie on the <note>circle ⊙O</note>.

Definition of Tangent to a Circle

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

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

Definition of Radius

In the figure of this problem, in circle ⊙O, point O is the center of the circle, point D is any point on the circle, the line segment OD is the line segment from the center of the circle to any point on the circle, therefore the line segment OD is the radius of the circle.

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

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

Triangle Angle Sum Theorem

<note>Angle ODB</note> + <note>Angle DOB</note> + <note>Angle OBD</note> = 180°. In triangle ADB, <note>Angle ADB</note>, <note>Angle BAD</note> and <note>Angle DBA</note> are the three interior angles of triangle ADB. According to the Triangle Angle Sum Theorem, <note>Angle ADB</note> + <note>Angle BAD</note> + <note>Angle DBA</note> = 180°.

Inscribed Angle Theorem

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

<note>In circle O, points A, D, and C are on the circle</note>, <note>the central angle corresponding to arc CD is ∠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 arc CD, i.e., ∠DAC = 1/2 ∠DOC</note>.

Definition of Inscribed Angle

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

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


Question: As shown in the figure, in parallelogram ABCD, it is known that ∠AOB=90°, AC=8cm, AD=5cm, then the length of BD is ()

Answer: 6cm

1. Given that quadrilateral ABCD is a parallelogram, according to the ##properties of parallelograms theorem##, the diagonals bisect each other. Therefore, OA = ##1/2##AC = ##1/2## × 8 = 4 (cm).

2. ∵ ∠AOB = 90°, according to the ##definition of adjacent supplementary angles##, we have ∠AOD = 180° - ∠AOB = 90°.

3. In △ADO, according to the Pythagorean theorem ##and the definition of right triangles##, since ∠AOD = 90°, we get OD = √##(AD^2 - OA^2)## = √##(5^2 - 4^2)## = √##(25 - 16)##= √9 = 3 (cm).

4. By the ##properties of parallelograms theorem##, we get BD = 2 × OD = 2 × 3 = 6 (cm).

##5##. Through the above reasoning, the final answer is 6 cm.

平行四边形; 垂线; 直角三角形; 线段

Definition of Parallelogram

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

<note>Quadrilateral ABCD is a parallelogram</note>, <note>sides AB and CD are parallel and equal</note>, <note>sides AD and BC are parallel and equal</note>.

Definition of Diagonal

A diagonal is a line segment connecting one vertex of a polygon to another vertex that is not adjacent to it.

In the figure of this problem, in parallelogram ABCD, <note>the diagonal is the line segment AC connecting vertex A and the non-adjacent vertex C</note>, and <note>the line segment BD connecting vertex B and the non-adjacent vertex D</note>. Therefore, <note>the line segments AC and BD are the diagonals of parallelogram ABCD</note>.

Properties of Parallelogram Theorem

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

In the figure of this problem, in parallelogram ABCD, <note>angles ∠A and ∠C are equal</note>, <note>angles ∠B and ∠D are equal</note>; <note>sides AB and CD are equal</note>, <note>sides AD and BC are equal</note>; the diagonals <note>AC and BD bisect each other</note>, that is, <note>the intersection point O divides diagonal AC into two equal segments OA and OC</note>, <note>and divides diagonal BD into two equal segments OB and OD</note>.

Definition of Linear Pair of Angles

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

<note>In the figure of this problem, angle AOD and angle AOB have a common side AO, and their other sides OD and OB are extensions of each other in opposite directions</note>, so angle AOD and angle AOB are adjacent supplementary angles.

Pythagorean Theorem

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

<note>In △ADO, ∠AOD is a right angle (90 degrees)</note>, <note>sides OD and AO are the legs</note>, <note>side AD is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OD^2 = AD^2 - AO^2</note>.

Definition of Right Triangle

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


Question: As shown in the figure, AB is the diameter of ⊙O, CD is a chord of ⊙O, and CD ⊥ AB at E. Connect AD and BC respectively. Given ∠D = 65°, find ∠OCD = ()

Answer: 40°

1. Given that AB is the diameter of ⊙O, CD is a chord of ⊙O, and CD ⊥ AB at E, according to the problem statement and given ∠D=65°.

2. According to the Corollary 1 of the Inscribed Angle Theorem, given ∠D=65°, then ∠ABC=∠EBC=65°.

3. Since CD ⊥ AB, and E is the intersection point of AB and CD, therefore ∠CEB=90°.

4. In ΔCEB, according to the Interior Angle Sum Theorem, ∠ECB=180°-∠CEB-∠EBC=180°-90°-65°=25°.

5. Since OB and OC are the radii of circle O, OB=OC, according to the definition of an isosceles triangle, △OBC is an isosceles triangle.

6. According to the properties of an isosceles triangle, in the isosceles triangle △OBC, ∠OBC=∠OCB=65°.

7. Since in △COB, ∠OCB=∠OCE+∠ECB, then ∠OCE=∠OCB-∠ECB.

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

圆; 垂线; 直角三角形; 圆周角; 弧

In <note>circle O</note>, the <note>circumferential angle ∠ADC corresponding to arc AC</note> is equal to <note>∠ABC</note>. According to <note>Corollary 1 of the Inscribed Angle Theorem</note>, the <note>circumferential angles ∠ADC and ∠ABC corresponding to the same arc AB</note> are equal, that is, <note>∠ADC = ∠ABC</note>.

Definition of Inscribed Angle

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

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

Triangle Angle Sum Theorem

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

Properties of Isosceles Triangle

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

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

Definition of Radius

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

<note>Point O is the center of the circle, point A is any point on the circle, segment OA is the segment from the center to any point on the circle, therefore segment OA is the radius of the circle</note>。<note>In circle O, point O is the center, point B is any point on the circle, segment OB is the segment from the center to any point on the circle, therefore segment OB is the radius of the circle</note>。<note>In circle O, point O is the center, point C is any point on the circle, segment OC is the segment from the center to any point on the circle, therefore segment OC is the radius of the circle</note>。


Question: As shown in the figure, CD is the diameter of ⊙O, and chord DE ∥ OA. If the measure of ∠D is 50°, then the measure of ∠C is ()

Answer: 25°

1. Given CD is the diameter of ⊙O, chord DE is parallel to OA, and the measure of ∠D is 50°.

2. Since DE is parallel to OA, according to the parallel line axiom 2, ∠AOD is equal to ∠D, thus the measure of ∠AOD is also 50°.

3. ∠AOD is the central angle of minor arc AD, ∠ACD is the inscribed angle of minor arc AD.

4. According to the inscribed angle theorem, the inscribed angle is half of the central angle. Therefore, ∠C is half of ∠AOD.

5. From this calculation, ∠C = 50° / 2 = 25°.

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

圆; 圆周角; 平行线; 弦; 直角三角形

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.

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

Definition of Central Angle

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

Parallel Postulate 2 of Parallel Lines

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

<note>In the figure of this problem, two parallel lines DE and OA are intersected by a third line DO</note>, forming the following geometric relationships: 1. <note>Alternate interior angles: ∠AOD and ∠D are equal</note>. Therefore, <note>∠AOD = ∠D = 50°</note>.

Inscribed Angle Theorem

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

Definition of Alternate Interior Angles

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

In the figure of this problem, two parallel lines <note>OA and DE</note> are intersected by a line <note>OD</note>, where <note>angle D and angle AOD are located between the two parallel lines and on opposite sides of the intersecting line OD</note>, thus <note>angle D and angle AOD are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle D is equal to angle AOD</note>.


Question: As shown in the figure, given that points A, B, and C are on ⊙O, AC⊥BO at D, and ∠B=50°, then the degree of ∠BOC is ()

Answer: 80°

1. Given that points A, B, and C are on ⊙O, AC is perpendicular to BO and intersects at point D.

2. According to the definition of ##perpendicular lines##, it is concluded that ∠ADB=90°.

3. From the given condition ##∠B=∠ABD=50°##, ##according to the triangle angle sum theorem, it is concluded that ∠BAD=180°-∠ABD-∠ADB=180°-90°-50°=40°, and ∠BAD=∠BAC##.

4. ##According to the inscribed angle theorem, ∠BAC is equal to half of the central angle ∠BOC that subtends arc BC##.

5. Therefore, ∠BOC=2##∠BAC##.

6. Substituting ##∠BAC##=40°, we get ∠BOC=2*40°=80°.

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

圆; 圆周角; 圆心角; 垂线

Definition of Perpendicular Lines

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

The lines <note>AC and BO</note> intersect to form an angle <note>∠ADB is 90 degrees</note>, so according to the <note>definition of perpendicular lines</note>, <note>lines AC and BO are perpendicular to each other</note>.

Definition of Central Angle

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

Definition of Inscribed Angle

In the figure of this problem, <note>circle O</note>, <note>the vertex A of angle BAC is on the circumference</note>, and the two sides of angle BAC <note>intersect circle O at points B and C respectively</note>. Therefore, angle BAC is an <note>inscribed angle</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 BC is ∠BOC</note>, and the inscribed angle is <note>∠BAC</note>. According to the Inscribed Angle Theorem, <note>∠BOC is equal to twice the inscribed angle ∠BAC corresponding to arc AC, that is, ∠BOC = 2 * ∠BAC</note>.

Triangle Angle Sum Theorem

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

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


Question: As shown in the figure, in ⊙O, it is known that ∠AOB=110°, C is a point on the circumference, then ∠ACB is ()

Answer: 125°

1. Take a point D on the major arc AB, and connect AD and BD.

2. According to the inscribed angle theorem, the inscribed angle is equal to half of the central angle it subtends, thus ∠ADB=∠AOB/2=110°/2=55°.

3. Since the four vertices of quadrilateral ADBC are all on circle O, according to the definition of a cyclic quadrilateral, quadrilateral ADBC is a cyclic quadrilateral.

4. In a cyclic quadrilateral, the sum of the measures of opposite angles is 180°, so ∠ACB+∠ADB=180°.

5. Substitute the known ∠ADB=55°, then ∠ACB=180°-55°=125°.

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

圆; 圆心角; 圆周角; 弦

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 D of angle ∠ADB is on the circumference</note>, <note>the two sides of angle ∠ADB intersect circle O at points A and B respectively</note>. Therefore, <note>angle ∠ADB is an inscribed angle</note>.##

Definition of Central Angle

In the diagram of this problem, in circle O, <note>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>.

Inscribed Angle Theorem

<note>In the figure of this problem, in circle O, points A, B, and D are on the circle</note>, <note>the central angle corresponding to arc AB is ∠AOB=110°</note>, <note>the inscribed angle is ∠ADB</note>. According to the Inscribed Angle Theorem, <note>∠ADB is equal to half of the central angle ∠AOB corresponding to arc AB</note>, that is, <note>∠ADB = 1/2 ∠AOB</note>.

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.

Original: <note>The four vertices A, D, B, and C of quadrilateral ADBC are all on the same circle</note>. This <note>circle is called the circumcircle of quadrilateral ADBC</note>. Therefore, quadrilateral ADBC is a cyclic quadrilateral. According to <note>the properties of cyclic quadrilaterals</note>, we can conclude that <note>the sum of opposite angles is equal to 180 degrees</note>, that is, <note>angle ACB + angle ADB = 180 degrees</note>, <note>angle DAC + angle DBC = 180 degrees</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>In the cyclic quadrilateral ADBC, the vertices ADBC are 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 ADBC is equal to 180°</note>. Specifically, <note>∠ADB + ∠ACB = 180°</note>; <note>∠DAC + ∠ABC = 180°</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, chord CD intersects AB, and ∠ABC=32°, then the degree of ∠CDB is ()

Answer: 58°

1. Given that AB is the diameter of ⊙O, ##according to (corollary 2 of the inscribed angle theorem) the inscribed angle subtended by the diameter is a right angle##, we get ∠ACB=90°.

2. ##According to the theorem of the sum of the angles in a triangle##, and with ∠ACB=90° and ∠ABC=32°, we get ∠BAC=180°-∠ACB-∠ABC=58°.

3. ##According to corollary 1 of the inscribed angle theorem##, ∠CDB and ∠BAC are both inscribed angles subtended by arc BC, hence ∠CDB=∠BAC.

4. Substituting the value of ∠BAC calculated earlier, we get ∠CDB=58°.

5. Through the above reasoning, the final answer is 58°.

圆; 圆周角; 直角三角形; 弦

Definition of Diameter

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

In the figure of this problem, <note>AB is the diameter</note>, <note>connecting the center O of the circle with points A and B on the circumference</note>, with a length equal to 2 times the radius, that is, <note>AB = 2r</note>.

Definition of Inscribed Angle

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

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

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

Triangle Angle Sum Theorem

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

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

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


Question: As shown in the figure, point C is on line segment AB, point D is the midpoint of AC. If CD = 3cm and AB = 10cm, then the length of BC is ()

Answer: 4cm

1. Given that point D is the midpoint of segment AC, according to the definition of midpoint, the length of AC is twice that of CD, i.e., AC = 2 × CD.

2. Given that the length of CD is 3cm, substitute it into the conclusion from the previous step to obtain AC = 2 × 3 = 6cm.

3. From the relationship between segments AC and AB, we know that BC = AB - AC.

4. Substitute the given lengths of AB (10cm) and AC (6cm) to obtain BC = 10 - 6 = 4cm.

5. Through the above reasoning, we finally conclude that the length of BC is 4cm.

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


Question: As shown in the figure, ⊙O is the circumcircle of △ABC, ∠AOB=60°, then the degree of ∠C is ()

Answer: 30°

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

3. In this problem, the inscribed angle ACB is equal to half of the central angle AOB that subtends the arc AB.

4. Therefore, ∠ACB = 1/2 × ∠AOB = 1/2 × 60° = 30°.

5. From the above reasoning, the final answer is ∠ACB=30°.

圆; 圆心角; 圆周角; 普通三角形; 弧

Definition of Inscribed Angle

<note>In the figure of this problem, in circle O, 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>.

Inscribed Angle Theorem

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

<note>Arc AB</note> and <note>Arc ACB</note> correspond to the central angle ∠AOB and the inscribed angle ∠ACB. According to the Inscribed Angle Theorem, ∠ACB is equal to half of the central angle ∠AOB that subtends <note>Arc AB</note>, i.e., ∠ACB = 1/2 ∠AOB.

Definition of Central Angle

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

Original text: <note>In the figure 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>.


Question: As shown in the figure, ⊙O is the circumcircle of △ABC, ∠BOC=3∠AOB. If ∠ACB=20°, then the degree of ∠BAC is ()

Answer: 60°

1. Given ∠ACB=20°, according to the inscribed angle theorem, #### we get ∠AOB=2×∠ACB=2×20°=40°.

2. According to the condition of the problem ∠BOC=3×∠AOB, we get ∠BOC=3×40°=120°.

3. ##According to the inscribed angle theorem, we get ∠BOC=2×∠BAC.##

4. Therefore ∠BAC=##120°/2=60°.##

##5##. Due to the above reasoning, we finally get the degree of ∠BAC as 60°.

圆; 圆心角; 圆周角; 普通三角形; 普通多边形

Definition of Inscribed Angle

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

Inscribed Angle Theorem

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

Definition of Central Angle

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


Question: As shown in the figure, AB is the diameter of ⊙O, CD is the chord of ⊙O, ∠ABD=53°, then ∠BCD is ()

Answer: 37°

1. Given AB is the diameter of ⊙O, CD is the chord of ⊙O, ∠ABD=53°.

2. Connect AD, according to (corollary 2 of the inscribed angle theorem) the inscribed angle subtended by the diameter is a right angle, we get ∠BDA=90°.

3. Since ∠ABD=53°, ∠BDA=90°, by the triangle angle sum theorem, we know ∠DAB is 180°-90°-53°=37°.

4. Because ∠DAB and ∠BCD correspond to the same arc DB, according to corollary 1 of the inscribed angle theorem, they are equal.

圆; 圆周角; 弧; 弦; 直角三角形

Definition of Diameter

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

Definition of Chord

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

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

Definition of Central Angle

<note>In the figure 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 formed by the lines <note>OA</note> and <note>OB</note>, <note>∠AOB</note>, is called the central angle. Similarly, <note>points C and B are two points on the circle</note>, and the angle formed by the lines <note>OC</note> and <note>OB</note>, <note>∠COB</note>, is also a central angle.

Inscribed Angle Theorem

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

In <note>circle O</note>, <note>the angles subtended by the diameter AB at the circumference, ∠BDA and ∠BCA, are right angles (90 degrees)</note>.

Triangle Angle Sum Theorem

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

<note>In circle O, the inscribed angles ∠ACD and ∠ABD corresponding to arc AD are equal</note>. According to Corollary 1 of the Inscribed Angle Theorem, <note>the inscribed angles ∠ACD and ∠ABD corresponding to the same arc AD are equal</note>, i.e., <note>∠ACD = ∠ABD</note>.

Definition of Inscribed Angle

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


Question: As shown in the figure, in △ABC, AB=AC, ∠A=40°, with C as the center and the length of CB as the radius, draw an arc intersecting AB at point D. Connect CD, then ∠ACD equals ()

Answer: 30°

1. Given AB=AC and ∠A=40°, ##according to the definition of an isosceles triangle, we can conclude that △ABC is an isosceles triangle. According to the properties of an isosceles triangle and the triangle angle sum theorem##, we get ∠ABC=∠ACB=##(180°-40°)/2=70°##.

2. With point C as the center, draw an arc with the radius equal to the length of BC, intersecting AB at point D, then connect CD.

3. ##According to the definition of radius, we get## BC=CD. According to the properties of an isosceles triangle, we get ##∠CBD##=∠CDB.

4. In ΔBCD, ##∠CDB=∠ABC##=70°, according to the exterior angle theorem, we get ##∠ACD=∠CDB-∠A=70°-40°=30°##.

####

##5##. Through the above reasoning, we finally get the answer as 30°.

等腰三角形; 圆; 弧; 圆周角; 内错角

Definition of Isosceles Triangle

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

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

Properties of Isosceles Triangle

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

Original text: <note>In isosceles triangle BCD, sides BC and CD are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle BCD = angle CDB</note>. Similarly, <note>in isosceles triangle ABC, sides AB and AC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle ABC = angle ACB</note>.

Exterior Angle Theorem of Triangle

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

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

Definition of Radius

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

In the figure of this problem, <note>in circle C, point C is the center of the circle</note>, <note>point B is any point on the circle</note>, <note>line segment CB is the line segment from the center to any point on the circle</note>, therefore <note>line segment CB is the radius of the circle</note>.


Question: As shown in the figure, in ⊙O, the length of chord AB is 2, OC ⊥ AB at C, OC = 1. If two tangents to ⊙O are drawn from a point P outside ⊙O, with points of tangency at A and B respectively, then the measure of ∠APB is ()

Answer: 90°

1. Given that the length of chord AB is 2, OC is perpendicular to AB and intersects at point C, OC equals 1.

2. Connect OA and OB, PA and PB are tangents, then according to the ##properties of the tangent to a circle##, ∠OAP equals ∠OBP equals 90 degrees.

3. ##Since OC is perpendicular to AB at point C, according to the perpendicular bisector theorem, AC equals BC equals 1##, therefore OC=AC=BC.

4. From this, it can be concluded that triangle ACO and triangle BCO are isosceles right triangles.

5. ##Therefore, combining the triangle angle sum theorem, we can deduce ∠AOC=∠OAC=(180°-90°)/2=45°, ∠BOC=∠OBC=(180°-90°)/2=45°, and then calculate ∠AOB=∠AOC+∠BOC=45°+45°=90°##

6. ##According to the quadrilateral angle sum theorem, we can know that ∠APB=360°-∠AOB-∠OAP-∠OBP=360°-90°-90°-90°=90°##

7. Through the above reasoning, the final answer is 90 degrees.

圆; 切线; 弦; 圆周角; 等腰三角形

Definition of Chord

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

Definition of Tangent to a Circle

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

In the figure of this problem, <note>circle O and lines PA and PB have only one common point A and B, respectively, and these two common points are called points of tangency</note>. Therefore, <note>lines PA and PB are tangents to circle O</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, OC is perpendicular to chord AB at C</note>, then according to the <note>Perpendicular Diameter Theorem</note>, OC bisects chord AB, that is <note>AC=BC</note>, and <note>OC bisects the two arcs subtended by chord AB</note>.

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

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

Definition of Isosceles Right Triangle

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

<note>In the figure of this problem, triangle ACO and triangle BCO are isosceles right triangles</note>, <note>where angle ACO and angle BCO are right angles (90 degrees)</note>, <note>side AC and side CO are equal</note>, <note>side BC and side CO are equal</note>.

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

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

Triangle Angle Sum Theorem

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

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


Question: As shown in the figure, it is known that AD intersects BC at point O, AB∥CD. If ∠B=40°, ∠D=30°, then the measure of ∠AOC is ()

Answer: 70°

1. Given AB∥CD, ##according to the parallel axiom 2 of parallel lines##, we get ∠DAB = ∠CDA = 30°.

2. ##According to the exterior angle theorem of triangles##, that is, the exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles, we get ∠AOC = ∠DAB + ∠ABC.

3. From the given condition, ∠ABC = 40°, combining the above steps, we get ∠AOC = 30° + 40°.

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

平行线; 内错角; 对顶角

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>Two parallel lines CD and AB</note> are intersected by a third <note>line AD</note>, forming the following geometric relationship: alternate interior angles: <note>angle DAB and angle CDA are equal</note>. These relationships illustrate that <note>when two parallel lines are intersected by a third line, the alternate interior angles are equal</note>.

Definition of Alternate Interior Angles

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

In the diagram of this problem, two parallel lines <note>AB and CD</note> are intersected by a line <note>AD</note>. Among them, <note>angle DAB and angle CDA</note> are located between the two parallel lines and on opposite sides of <note>line AD</note>. Therefore, <note>angle DAB and angle CDA</note> are alternate interior angles. Alternate interior angles are equal, that is, <note>angle DAB is equal to angle CDA</note>.

Exterior Angle Theorem of Triangle

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

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

The original: <note>One of the interior angles of triangle AOB is ∠AOB</note>, extending the adjacent side of this interior angle <note>BO to form the angle ∠COA</note> is called the exterior angle of the interior angle <note>∠AOB</note>.


Question: As shown in the figure, AB is the diameter of O, ∠AOC=110°, then ∠D=()

Answer: 35°

2. According to the definition of a straight angle, ∠AOB is a straight angle, ∠AOB=180°, thus ∠BOC=180°-∠AOC.

4. According to the inscribed angle theorem, the inscribed angle ∠BDC is equal to half of the central angle corresponding to the arc it intercepts.

6. Substituting the value, we get ∠BDC=1/2*70°=35°.

7. Through the above reasoning, the final answer is 35°.

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

Definition of Central Angle

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

Definition of Straight Angle

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

<note>Ray OA rotates around endpoint O to form a straight line with the initial side, forming straight angle AOB</note>. According to the definition of a straight angle, <note>a straight angle measures 180 degrees</note>, i.e., <note>angle AOB = 180 degrees</note>.

Inscribed Angle Theorem

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

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