Question: As shown in the figure, it is known that D is a point on BC, ∠B=∠1, ∠BAC=78°, then ∠2=()

Answer: 78°

1. Given ∠B = ∠1, according to the problem, let point D be a point on BC.

3. From the figure, in triangle ABC, ∠BAC = ∠1 + ∠BAD.

4. Since ∠B = ∠1 is given, and ∠BAC = 78°.

6. In triangle ABD, according to the exterior angle theorem of triangles, we know ∠2 = ∠B + ∠BAD.

7. Because ∠B = ∠1, thus ∠2 = ∠1 + ∠BAD.

8. Combining the above conditions, we get ∠2 = ∠BAC.

9. Since ∠BAC = 78°, therefore ∠2 = 78°.

10. Through the above reasoning, the final answer is 78°.

普通三角形; 对顶角; 邻补角

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

Definition of Triangle

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

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

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

The interior angle of <note>polygon ABD</note> is <note>∠ADB</note>. The exterior angle formed by extending the adjacent sides <note>BD</note> and <note>AD</note> of this interior angle is called the exterior angle of the interior angle <note>∠ADB</note>.


Question: As shown in the figure, point C is on segment AB, point E is the midpoint of AC, and point D is the midpoint of BC. If ED = 6, then the length of segment AB is ()

Answer: 12

1. Given that point E is the midpoint of segment AC, by definition, segment AC is bisected by E, i.e., AE = CE.

2. Similarly, point D is the midpoint of segment BC, by definition, segment BC is bisected by D, i.e., BD = CD.

3. According to the conditions given in the problem, point C is on segment AB, therefore segment AB can be expressed as the sum of AC and CB.

4. From the conclusions of steps 1 and 2, AE = 1/2 AC, CE = 1/2 AC, BD = 1/2 BC, CD = 1/2 BC.

5. ##From the problem and the above formulas, we know that AB = AC + BC = 2EC + 2CD = 2(EC + CD) = 2ED.##

6. Given that the length of ED is 6, we can substitute this into the above formula to find the length of AB, i.e., AB = 2 × 6.

7. Finally, we find that the length of AB is 12.

Midpoint of a Line Segment

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

<note>Point E is the midpoint of line segment AC</note>. Therefore, according to the definition of the midpoint of a line segment, <note>point E divides line segment AC into two equal parts</note>, that is, <note>AE = CE</note>. Similarly, <note>point D is the midpoint of line segment BC</note>, therefore <note>point D divides line segment BC into two equal parts</note>, that is, <note>BD = CD</note>.

Definition of Line Segment

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

<note>Line segment AB is a part of a straight line, containing endpoint A, endpoint B, and all points between them</note>。<note>Line segment AC is a part of a straight line, containing endpoint A, endpoint C, and all points between them</note>。<note>Line segment BC is a part of a straight line, containing endpoint B, endpoint C, and all points between them</note>。


Question: As shown in the figure, in △ABC, ∠B=∠C, D is a point on side BC, E is a point on side AC, ∠ADE=∠AED, if ∠BAD=24°, then ∠CDE=()

Answer: 12°

1. Given ∠B=∠C, D is a point on side BC, point E is on side AC, ∠ADE=∠AED, ∠BAD=24°.

2. According to the exterior angle theorem of the triangle, ∠ADC is the exterior angle of △ABD, so ∠ADC=∠B+∠BAD.

4. According to the exterior angle theorem of the triangle, ∠AED is the exterior angle of △CDE, so ∠AED=∠C+∠EDC.

5. Since ∠B=∠C, and it is given that ∠ADE=∠AED, therefore ∠C+∠EDC=∠ADC-∠EDC.

7. Since it is given that ∠B=∠C, we can combine to get ##2∠EDC=24°##.

9. Through the above reasoning, the final answer is ∠CDE=12°.

等腰三角形; 普通三角形; 对称

Exterior Angle Theorem of Triangle

In the figure of this problem, <note>∠ADC is the exterior angle of △ABD</note>, according to the Exterior Angle Theorem of Triangle, <note>∠ADC=∠B+∠BAD</note>. Similarly, <note>∠AED is the exterior angle of △CDE</note>, according to the Exterior Angle Theorem of Triangle, <note>∠AED=∠C+∠EDC</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, CD is a chord, AB ⊥ CD at point E. If the radius is 5 and OE = 3, then the length of CD is ()

Answer: 8

1. Given AB is the diameter of ⊙O, CD is a chord, AB ⊥ CD at point E.

2. According to the perpendicular bisector theorem, the diameter perpendicular to the chord bisects the chord, thus CE = half of CD.

3. Connect OC, in the right triangle △OCE, OE is one of the legs, OC is the hypotenuse.

4. By the Pythagorean theorem for right triangles, OC^2 = OE^2 + CE^2.

6. Solving the equation gives CE^2 = 25 - 9 = 16, so CE = √16 = 4.

7. Since CE = half of CD, CD = 2 * CE = 2 * 4 = 8.

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

圆; 弦; 垂线; 直角三角形; 垂直平分线

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, connecting the center O and points A and B on the circumference</note>, with a length of 2 times the radius, that is, <note>AB = 2 * 5</note>.

Definition of Chord

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

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

Definition of Radius

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

<note>In circle O, point O is the center of the circle</note>, point A and <note>point B</note>, <note>point C</note> are any points on the circle, <note>line segment OA</note> and <note>line segment OB</note>, <note>line segment OC</note> are line segments from the center to any point on the circle, therefore <note>line segment OA</note> and <note>line segment OB</note>, <note>line segment OC</note> are the radii of the circle, and their length is 5.

Definition of Right Triangle

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

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

Perpendicular Diameter Theorem

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

<note>In circle O</note>, <note>diameter AB is perpendicular to chord CD at point E</note>, then according to the Perpendicular Diameter Theorem, <note>diameter AB bisects chord CD, that is, CE=ED</note>, and <note>diameter AB bisects the arcs subtended by chord CD, that is, arc CA=arc AD</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>角OEC是直角（90度）</note>, sides OE and CE are the legs, side OC is the hypotenuse, so according to the Pythagorean Theorem, <note>OC^2 = OE^2 + CE^2</note>.


Question: As shown in the figure, AB∥EF, CD⊥EF at point D, if ∠BCD=140°, then the degree of ∠ABC is ()

Answer: 50°

1. Draw a line CG through point C such that CG is parallel to AB. According to the problem, AB is parallel to EF and also parallel to CG.

2. Since AB is parallel to CG, according to the ##parallel lines axiom 2, alternate interior angles are equal##, we can conclude ∠ABC = ∠BCG.

3. According to the given conditions, CD is perpendicular to EF at point D. ##AB is parallel to EF, AB is parallel to CG, so according to the transitivity of parallel lines, EF is parallel to CG; according to the parallel lines axiom 2, same-side interior angles are supplementary, ∠GCD = 180° - ∠CDF,## therefore ∠GCD = 90°.

4. Given that ∠BCD = 140°, since ∠BCD = ∠BCG + ∠GCD, we can obtain ∠BCG = ∠BCD - ∠GCD.

5. Substituting the given conditions, we get ∠BCG = 140° - 90° = 50°.

6. ##Since ∠ABC = ∠BCG##, we can conclude ∠ABC = 50°.

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

平行线; 垂线; 内错角; 邻补角

Definition of Parallel Lines

Two lines in the same plane that do not intersect are called parallel lines.

In the figure of this problem, <note>line AB and line EF lie in the same plane and they do not intersect</note>, so according to the definition of parallel lines, <note>line AB and line EF are parallel lines</note>. Similarly, <note>line CG and line AB also lie in the same plane and they do not intersect</note>, so according to the definition of parallel lines, <note>line CG and line AB are parallel lines</note>, <note>line CG and line EF also lie in the same plane and they do not intersect</note>, so according to the definition of parallel lines, <note>line CG and line EF are parallel lines</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 AB and CG are intersected by a third line BC</note>, forming the following geometric relationship: <note>alternate interior angles ∠ABC and ∠BCG are equal</note>, <note>two parallel lines EF and CG are intersected by a third line DC</note>, forming the following geometric relationship: <note>same-side interior angles ∠GCD and ∠CDF are supplementary</note>.

Definition of Alternate Interior Angles

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

<note>Two parallel lines AB and CG</note> are intersected by a <note>line BC</note>, where <note>angle ABC</note> and <note>angle BCG</note> are located between the two parallel lines and on <note>opposite sides of the intersecting line BC</note>. Therefore, <note>angle ABC</note> and <note>angle BCG</note> are alternate interior angles. Alternate interior angles are equal, that is, <note>angle ABC</note> is equal to <note>angle BCG</note>.

Definition of Consecutive Interior Angles

When two lines are intersected by a third line, the angles that lie on the same side of the transversal and inside the two intersected lines are called consecutive interior angles. If the two lines are parallel, then the consecutive interior angles are supplementary. Conversely, if the consecutive interior angles are supplementary, then the two lines are parallel.

<note>Two lines CG and EF are intersected by a third line CD</note>, <note>angles GCD and CDF are on the same side of the transversal CD</note>, and inside the intersected lines CG and EF, so <note>angles GCD and CDF are consecutive interior angles</note>. Consecutive interior angles GCD and CDF are supplementary, that is, <note>angle GCD + angle CDF = 180 degrees</note>.

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 CG and line EF are respectively parallel to line AB</note>. According to the transitivity of parallel lines, if <note>line CG is parallel to line AB</note>, and <note>line EF is also parallel to line AB</note>, then <note>line CG and line EF are mutually parallel</note>. Therefore, <note>line CG is parallel to line EF</note>.


Question: As shown in the figure, AB=8cm, AD=BC=5cm, then CD equals ()

Answer: 2cm

1. Given AB = 8cm, AD = 5cm, according to the principle of subtraction, we can derive BD = AB - AD = 3cm.

2. Given BC = 5cm, and again according to the principle of subtraction, we can derive CD = BC - BD = 2cm.

3. Therefore, we can derive that the length of CD is 2cm.

Definition of Line Segment

In the figure of this problem, <note>line segments AB, AD, BC, BD, and CD</note> are all line segments. <note>Line segment AB</note> is a part of a straight line, containing <note>endpoint A and endpoint B</note> and all points in between. <note>Line segment AD</note> is a part of a straight line, containing <note>endpoint A and endpoint D</note> and all points in between. <note>Line segment BC</note> is a part of a straight line, containing <note>endpoint B and endpoint C</note> and all points in between. <note>Line segment BD</note> is a part of a straight line, containing <note>endpoint B and endpoint D</note> and all points in between. <note>Line segment CD</note> is a part of a straight line, containing <note>endpoint C and endpoint D</note> and all points in between.


Question: As shown in the figure, point O is on the line AB. If ∠2 = 140°, then the degree of ∠1 is ()

Answer: 40°

1. Given line AB and point O, according to the diagram as shown.

2. Since point O is on line AB, ##according to the definition of a straight angle, it is known that## ∠AOB is an angle on the line, ##∠AOB = 180 degrees.##

3. According to the problem statement, ∠AOB = ∠1 + ∠2.

4. From the given conditions in the problem, ∠2 = 140°.

5. Substituting, we get ∠1 + ∠2 = 180°, that is ∠1 + 140° = 180°.

6. Solving this equation, we get: ∠1 = 40°.

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

邻补角; 直线; 对顶角

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.

<note>Angle AOB is a geometric figure formed by two rays AO and OB</note>, these two rays have a common endpoint <note>O</note>. This common endpoint <note>O is called the vertex of angle AOB</note>, and <note>rays AO and OB are called the sides of angle AOB</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, creating a straight angle AOB</note>. According to the <note>definition of a straight angle</note>, the measure of a straight angle is 180 degrees, that is, <note>angle AOB = 180 degrees</note>.


Question: As shown in the figure, place the right-angle vertex of a triangle board with a 30° angle on one side of a ruler. If ∠1=35°, then the degree of ∠2 is ()

Answer: 65°

1. Given ∠1=35°.

2. According to the problem, the set square is a right triangle, one of whose angles is 30°, i.e., ∠3=30°.

3. According to the parallel lines axiom 2, alternate interior angles are equal, and ∠4, whose vertex is at the right angle vertex of the set square, and ∠1 are alternate interior angles, so ∠4=∠1=35°.

4. Thus, let the triangle containing ∠4 and ∠3 be ABC, then ∠2 is the exterior angle of triangle ABC, so ∠2=∠4+∠3=65°.

5. Through the above reasoning, the final answer is 65°.

直角三角形; 对顶角; 邻补角

Definition of Right Triangle

In this problem, the set square is a right triangle, which includes <note>a right angle (90°)</note> and <note>a specified 30° angle</note>.

Exterior Angle Theorem of Triangle

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

In the diagram of this problem, in triangle ABC, <note>angle 2 is an exterior angle of the triangle</note>, <note>angles 3 and 4 are the two interior angles that are not adjacent to exterior angle 2</note>. According to the Exterior Angle Theorem of Triangle, <note>exterior angle 2 is equal to the sum of the two non-adjacent interior angles 3 and 4</note>, that is, <note>angle 2 = angle 3 + angle 4</note>.

Definition of Alternate Interior Angles

<note>In the diagram of this problem, two parallel lines (the two edges of the ruler)</note> are intersected by <note>a line (the right-angle side opposite the 60-degree angle of the triangle board)</note>, where <note>angle 1 and angle 4 are located between the two parallel lines and on opposite sides of the intersecting line</note>. Therefore, <note>angle 1 and angle 4 are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle 1 is equal to angle 4</note>.

Definition of Exterior Angle of a Polygon

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

<note>The angle ∠2 formed by the extension of one side adjacent to an interior angle of polygon ABC and another side</note> is called the exterior angle of the interior angle.

Parallel Postulate 2 of Parallel Lines

In the figure of this problem, <note>two parallel lines (the two edges of the ruler)</note> are intersected by <note>a line (the right-angle side opposite the 60-degree angle of the set square)</note>, forming the following geometric relationship: <note>alternate interior angles: angle 1 and angle 4 are equal</note>.


Question: As shown in the figure, in Rt△ABC, ∠C=90°, AB=10, AC=8, then sinB equals ()

Answer: \frac{4}{5}

1. In the right triangle ABC, side AB is the hypotenuse of the right triangle ABC, side AC is the opposite side of ∠ABC, AB=10, AC=8.

2. According to the definition of the sine function, for the right triangle ABC, sin(∠ABC) is equal to the opposite side AC divided by the hypotenuse AB.

3. By calculation, sin(∠ABC)=AC/AB=8/10=4/5.

4. Through the above reasoning, the final answer is 4/5.

Definition of Right Triangle

In the figure of this problem, in triangle ABC, <note>angle ACB is a right angle (90 degrees)</note>, so triangle ABC is a right triangle. <note>Side AC and side BC 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 ABC</note>, <note>angle ABC is an acute angle</note>, <note>side AC is the opposite side of angle ABC</note>, and <note>side AB is the hypotenuse</note>. According to the definition of the sine function, the <note>sine value of angle ABC</note> is equal to the <note>ratio of the opposite side AC to the hypotenuse AB</note>, that is, <note>sin(∠ABC) = AC / AB = 8 / 10 = 4 / 5</note>.


Question: As shown in the figure, extend the segment AB with a length of 8 to C, making BC=4. M and N are the midpoints of AB and BC respectively. Find the length of MN.

Answer: 6

1. Given the length of AB is 8, the length of BC is 4, and M is the midpoint of segment AB, N is the midpoint of segment BC.

2. According to the definition of midpoint, M divides AB into two equal parts, so the length of MB is MB=##1/2##AB=##1/2##×8=4.

3. Similarly, N is the midpoint of BC, N divides BC into two equal parts, so the length of BN is BN=##1/2##BC=##1/2##×4=2.

4. Since segment MN is composed of segment MB and segment BN, and MB and BN are on the same line, the length of MN is MN=MB+BN=4+2=6.

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

Midpoint of a Line Segment

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

In the figure of this problem, the midpoint of line segment AB is <note>point M</note>. According to the definition of the midpoint of a line segment, <note>point M divides line segment AB into two equal parts</note>, that is, <note>the lengths of line segments AM and MB are equal</note>. That is, <note>AM = MB</note>. Similarly, the midpoint of line segment BC is <note>point N</note>. According to the definition of the midpoint of a line segment, <note>point N divides line segment BC into two equal parts</note>, that is, <note>the lengths of line segments BN and NC are equal</note>. That is, <note>BN = NC</note>.


Question: As shown in the figure, P is a point on the chord AB of ⊙O, AB=10cm, AP=4cm, OP=5cm. Find the radius of ⊙O in cm.

Answer: 7

1. Given point P is a point on chord AB of ⊙O, AB=10cm, AP=4cm, OP=5cm.

2. Through point O, draw OD perpendicular to segment AB, D is the foot of the perpendicular, connect OB.

3. Since AB is a chord, according to the perpendicular bisector theorem, AB is perpendicularly bisected by D, i.e., BD=5cm.

4. From AP=4cm, we can obtain PB=AB-AP=10cm-4cm=6cm, thus DP=6cm-5cm=1cm.

5. In right triangle ODP, according to the Pythagorean theorem, OD=√(OP^2-DP^2)=√(5^2-1^2)=2√6.

6. Connect OB, in right triangle ODB, according to the Pythagorean theorem, OB=√(OD^2+BD^2)=√((2√6)^2+5^2)=√(24+25)=√49=7cm.

7. Through the above reasoning, it is concluded that the radius of ⊙O is 7cm.

Definition of Right Triangle

<note>In the figure of this problem, in triangle ODP, angle ODP is a right angle (90 degrees)</note>, therefore triangle ODP is a right triangle. <note>Sides OD and DP are the legs, and side OP is the hypotenuse</note>. Similarly, <note>in triangle ODB, angle ODB is a right angle (90 degrees)</note>, therefore triangle ODB is a right triangle. <note>Sides OD and DB are the legs, and side OB is the hypotenuse</note>.

Pythagorean Theorem

In this problem, <note>△OPD is a right triangle</note>, <note>∠ODP=90°</note>, <note>sides OD and DP are the legs</note>, <note>side OP is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OP^2 = OD^2 + DP^2</note>. Given <note>OP=5cm</note>, <note>DP=1cm</note>, so <note>OD=√(5^2-1^2)=2√6 cm</note>. At the same time, <note>△ODB is also a right triangle</note>, <note>∠ODB=90°</note>, <note>sides OD and BD are the legs</note>, <note>side OB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OB^2 = OD^2 + BD^2</note>, where <note>OD=2√6 cm</note>, <note>BD=5cm</note>, so <note>OB=√((2√6)^2 + 5^2)=√49=7cm</note>. <note>OB is the radius of circle O</note>.

Perpendicular Diameter Theorem

<note>Point O is the center of the circle</note>, <note>OD is perpendicular to chord AB</note>, and <note>D is the foot of the perpendicular</note>. According to the Perpendicular Diameter Theorem, diameter OD bisects chord AB, that is, <note>AD = DB = 5 cm</note>, and <note>the diameter on which OD lies</note> bisects the arcs subtended by chord AB.


Question: As shown in the figure, the circle is externally tangent to the isosceles trapezoid ABCD with the midline EF = 15cm. Then the perimeter of the isosceles trapezoid ABCD is equal to ()

Answer: 60cm

1. Given the midline EF = 15 cm, according to the property that the midline of a trapezoid is equal to half the sum of the upper base and the lower base, we get that the sum of the two bases of the trapezoid is equal to twice the midline, i.e., 2 times EF = 30 cm.

2. According to the problem, the quadrilateral ABCD is circumscribed around the circle. According to the property of a circumscribed quadrilateral, the sum of the lengths of opposite sides is equal. Therefore, the sum of the lengths of the two legs AB and CD is equal to the sum of the lengths of the two bases, i.e., AB + CD = 30 cm.

3. Since the perimeter of the trapezoid is equal to the sum of the lengths of the two bases and the two legs, the perimeter of ABCD is equal to upper base + lower base + AB + CD = 30 cm + 30 cm = 60 cm.

4. The final answer is 60 cm.

圆; 等腰三角形; 切线; 梯形; 平行线

Properties of a Tangential Quadrilateral

A quadrilateral is called a tangential quadrilateral if each of its four sides touches (is tangent to) a circle. For a tangential quadrilateral, the sums of the lengths of its opposite sides are equal.

<note>Quadrilateral ABCD is a tangential quadrilateral</note>, with its four sides tangent to the circle. According to the properties of a tangential quadrilateral, <note>the sums of the lengths of the opposite sides of quadrilateral ABCD are equal</note>, that is, <note>side AB + side CD = side AD + side BC</note>.

Definition of Trapezoid

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

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

Median Line Theorem of Trapezoid

The median line of a trapezoid is the line segment that connects the midpoints of the non-parallel sides. This line segment is parallel to the bases (the parallel sides of the trapezoid) and its length is equal to half the sum of the lengths of the two bases.

In the diagram of this problem, in the trapezoid <note>ABCD</note>, ##<note>side AD</note>## and ##<note>side BC</note>## are the two bases of the trapezoid, <note>point E</note> and <note>point F</note> are the midpoints of the two legs of the trapezoid, and <note>segment EF</note> is the median line connecting the midpoints of the legs. According to the Median Line Theorem of Trapezoid, <note>segment EF</note> is parallel to ##<note>side AD</note>## and ##<note>side BC</note>##, and the length of <note>segment EF</note> is equal to half the sum of the lengths of ##<note>side AD</note>## and ##<note>side BC</note>##, that is, <note>EF = (##AD## + ##BC##) / 2</note>. Since it is known that <note>EF = 15 cm</note>, therefore ##<note>AD</note> + <note>BC</note>## = 2 * <note>15</note> = <note>30 cm</note>.


Question: As shown in the figure, points A, B, C, and D are all on ⊙O. If ∠BOD = 110°, then the degree of ∠BCD is ()

Answer: 125°

1. According to the problem, points A, B, C, and D are all on ⊙O, and it is known that ∠BOD=110°.

2. According to the inscribed angle theorem, ∠BAD (i.e., the opposite angle of ∠BCD) is equal to half of the central angle it subtends, that is, ∠BAD=∠BOD/2=55°.

3. Since quadrilateral ABCD is a cyclic quadrilateral, and the opposite angles of a cyclic quadrilateral are supplementary, ∠BCD=180°-∠BAD.

4. Thus, ∠BCD=180°-55°=125°.

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

圆; 圆心角; 圆周角; 圆内接四边形

Definition of Inscribed Angle

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

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

Definition of Central Angle

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

<note>In the diagram of this problem, in circle O, points B and D are two points on the circle, and the center of the circle is point O</note>. <note>The angle ∠BOD formed by the lines OB and OD is called the central 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>In the given figure, in circle O, ##points A, B, D## are on the circle</note>, <note>the central angle corresponding to ##arc BD## is ∠BOD, and the inscribed angle is ∠BAD</note>. According to the Inscribed Angle Theorem, <note>∠BAD is equal to half of the central angle ∠BOD corresponding to the ##arc BD##</note>, that is, <note>∠BAD = 1/2 ∠BOD</note>.

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

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

<note>Quadrilateral ABCD is a cyclic quadrilateral of circle O</note>, and the vertices A, B, C, D are all on the circle. According to the Diagonal Supplementary Theorem for Cyclic Quadrilateral, the sum of each pair of opposite angles of quadrilateral ABCD is equal to 180°. Specifically, <note>∠BAD + ∠BCD = 180°</note>; <note>∠ABC + ∠CDA = 180°</note>. Therefore, <note>∠BAD and ∠BCD are supplementary</note>, that is, <note>∠BCD=180°-∠BAD=180°-55°=125°</note>.


Question: As shown in the figure, given △ABC, point D is on the extension line of BC, ∠ACD=140°, ∠ABC=50°, then the measure of ∠A is ()

Answer: 90°

1. Observe the problem. As shown in the figure, it is known that point D is on the extension line of BC, the measure of ∠ACD is 140°, and the measure of ∠ABC is 50°.

####

##2. According to the definition of the exterior angle of a polygon, ∠ACD is the exterior angle of triangle ACB. According to the exterior angle theorem of triangles, the exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles, i.e., ∠ACD = ∠BAC + ∠ABC.

##3##. Substitute the given conditions into ##∠ACD = ∠BAC + ∠ABC##, we get 140° = ∠BAC + 50°.

##4##. Rearrange the equation to get: ∠BAC = 140° - 50°.

##5##. Calculate to get: ∠BAC = 90°.

##6##. Through the above reasoning, the final answer is 90°.

普通三角形; 三角形的外角; 邻补角

In the figure of this problem, an interior angle of triangle ACB is <note>∠ACB</note>. The angle formed by extending the adjacent sides <note>BC</note> and <note>AC</note> of this interior angle is called the exterior angle <note>∠ACD</note> of the interior angle <note>∠ACB</note>.

Exterior Angle Theorem of Triangle

In the figure of this problem, in triangle ABC, <note>angle ACD is an exterior angle of the triangle</note>, <note>angle BAC and angle ABC are the two non-adjacent interior angles to exterior angle ACD</note>. According to the Exterior Angle Theorem of Triangle, <note>the exterior angle ACD is equal to the sum of the two non-adjacent interior angles BAC and ABC</note>, i.e., <note>angle ACD = angle BAC + angle ABC</note>.


Question: As shown in the figure, quadrilateral ABCD is an inscribed quadrilateral of ⊙O, ∠BCD=110°, then the degree of ∠BOD is ()

Answer: 140°

1. Given that quadrilateral ABCD is an inscribed quadrilateral of ⊙O, and it is known that ∠BCD=110°.

2. According to the properties of an inscribed quadrilateral, the opposite angles are supplementary, that is, ∠A and ∠C in quadrilateral ABCD are supplementary. Based on this property, we get ∠A = 180° - ∠BCD.

3. According to the calculation, we get ∠A = 180° - 110° = 70°.

4. By the inscribed angle theorem, the central angle is twice the inscribed angle subtended by the same arc. Therefore, ∠BOD = 2∠A.

5. According to the calculation, we get ∠BOD = 2 x 70° = 140°.

6. Therefore, the final answer is 140°.

圆内接四边形; 圆周角; 圆心角; 圆

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.

In the figure of this problem, <note>the four vertices A, B, C, and D of quadrilateral ABCD are all on the same circle O</note>. <note>This circle is called the circumcircle of quadrilateral ABCD</note>. Therefore, <note>quadrilateral ABCD is a cyclic quadrilateral</note>. According to the properties of cyclic quadrilaterals, the sum of opposite angles is equal to 180 degrees, that is, <note>∠A + ∠C = 180 degrees</note>.

In the cyclic quadrilateral ABCD, the vertices A, B, C, and D are all on the circle. According to the Diagonal Supplementary Theorem for Cyclic Quadrilateral, the sum of each pair of opposite angles of quadrilateral ABCD is equal to 180°. Specifically, ∠A + ∠C = 180°.

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 BD and arc BAD is ∠BOD</note>, <note>the inscribed angle is ∠BAD</note>. According to the Inscribed Angle Theorem, <note>∠BAD is equal to half of the central angle ∠BOD corresponding to arc BD</note>, that is, <note>∠BAD = 1/2 ∠BOD</note>.


Question: As shown in the figure, BD and CE are respectively the altitude and angle bisector of △ABC, and they intersect at point O. If ∠BCA=70°, then the measure of ∠BOE is ()

Answer: 55°

1. Given BD is the altitude of △ABC. According to the definition of altitude, BD is perpendicular to AC, i.e., ∠BDC=90°.

2. Given CE is the angle bisector of △ABC, according to the definition of angle bisector, CE bisects ∠ACB. Given ∠ACB=70°, we can obtain ∠ECB=∠ECA=35°.

3. In the right triangle COD, according to the triangle angle sum theorem, we can obtain ∠COD=180°-90°-35°=55°.

4. According to the definition of vertical angles, ∠BOE=∠COD=55°.

垂线; 普通三角形; 垂直平分线; 对顶角

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).

<note>The line segment BD from vertex B perpendicular to the opposite side AC</note> is the altitude of vertex B. <note>The line segment BD forms a right angle (90 degrees) with side AC</note>, which indicates that the line segment BD is the perpendicular distance from vertex B to the opposite side AC.

Definition of Angle Bisector

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

<note>The vertex of angle ACB is point C</note>, from <note>point C a line CE is drawn</note>, this line <note>divides angle ACB into two equal angles</note>, that is, <note>∠ECB and ∠ECA are equal</note>. Therefore, <note>line CE is the angle bisector of angle ACB</note>.

Triangle Angle Sum Theorem

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

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

Definition of Vertical Angles

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

<note>In the figure of this problem, two intersecting lines CE and BD intersect at point O</note>, forming four angles: <note>angle EOD, angle COB, angle BOE, and angle COD</note>. According to the definition of vertical angles, <note>angle EOD and angle COB are vertical angles, and angle BOE and angle COD are vertical angles</note>. Since vertical angles are equal, <note>angle EOD = angle COB, and angle BOE = angle COD</note>.


Question: As shown in the figure, in △ABC, AB=10cm, AC=18cm, point M starts from point A and moves towards point B at a speed of 2cm per second, point N starts from point C and moves towards point A at a speed of 3cm per second. When one moving point reaches the endpoint, the other moving point also stops. When △AMN is an isosceles triangle with MN as the base, the moving time is ()

Answer: 3.6秒

1. Let the time of motion be x seconds. In △ABC, it is known that AB=10cm, AC=18cm.

2. Point M starts from point A and moves towards point B at a speed of 2cm per second, so AM=2x cm.

3. Point N starts from point C and moves towards point A at a speed of 3cm per second, so AN=18 - 3x cm.

4. When △AMN is an isosceles triangle, AM=AN.

5. Therefore, we get the equation 2x = 18 - 3x.

6. By solving the equation 2x = 18 - 3x, we get: 5x = 18, so x = 3.6.

7. Through the above reasoning, when △AMN is an isosceles triangle with MN as the base, the time of motion is 3.6 seconds.

等腰三角形; 线段; 普通三角形; 点

Definition of Isosceles Triangle

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

<note>△AMN is an isosceles triangle</note>, with <note>MN as the base</note>. According to the problem, when <note>△AMN is an isosceles triangle</note>, it must satisfy <note>AM=AN</note>, that is, <note>in △AMN, side AM and side AN are equal</note>. Also, AM=2x cm, AN=18 - 3x cm.


Question: Given a horizontally placed cylindrical drainage pipe, the radius of the pipe's cross-section is 1m, and the water level is 0.2m high. Find the width of the water surface in the cross-section of the drainage pipe.

Answer: 1.2m

1. Let the chord inside the circle be AB (water surface width), draw OC perpendicular to AB at C, OC intersects the circle O at D, connect OB, as shown in the figure.

2. Since OB and OD are the radii of the circle, OB = OD = 1 meter.

3. Given CD = 0.2 meters, therefore OC = OD - CD = 1 - 0.2 = 0.8 meters.

4. In the right triangle OBC, using the Pythagorean theorem, we get BC = √(OB^2 - OC^2) = √(1^2 - 0.8^2) = √(1 - 0.64) = √0.36 = 0.6 meters.

5. According to the perpendicular bisector theorem, the line OC is perpendicular to and bisects the chord AB, that is, AC = BC = 1/2 AB.

6. Therefore, AB = 2 * 0.6 meters = 1.2 meters.

7. Through the above reasoning, the final width of the water surface in the cross-section of the drainage pipe is 1.2 meters.

圆; 弦; 圆周角; 正弦; 余弦

Definition of Radius

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

The original Chinese enclosed in <note>点O是圆心</note>, <note>点D是圆上的任意一点</note>, <note>线段OD是从圆心到圆上任意一点的线段</note>, therefore <note>线段OD是圆的半径</note>.

Definition of Circle

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

In the figure of this problem, in circle O, <note>point O is the center of the circle</note>, <note>the radius is 1 meter</note>. In the figure, <note>all points that are 1 meter away from point O are on circle O</note>.

Definition of Right Triangle

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

In the figure of this problem, in triangle OBC, <note>angle OCB is a right angle (90 degrees)</note>, so <note>triangle OBC is a right triangle</note>. <note>Side OC and side BC are the legs</note>, <note>side OB is the hypotenuse</note>.

Pythagorean Theorem

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

In this problem, <note>in the right triangle OBC, ∠OCB is a right angle (90 degrees)</note>, <note>sides OC and BC are the legs, and side OB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OB^2 = OC^2 + BC^2</note>, that is, <note>1^2 = 0.8^2 + BC^2</note>. By calculation, <note>BC = √(1 - 0.64) = √0.36 = 0.6 meters</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, radius OD is perpendicular to chord AB</note>, then according to the Perpendicular Diameter Theorem, <note>radius OD bisects chord AB</note>, that is, <note>AC = BC</note>, and <note>radius OC bisects the two arcs subtended by chord AB</note>, that is, <note>arc AD = arc BD</note>.


Question: As shown in the figure, BP bisects ∠ABC and intersects CD at point F, DP bisects ∠ADC and intersects AB at point E. If ∠A = 40°, ∠P = 38°, then the degree of ∠C is ()

Answer: 36°

1. According to the problem statement, BP bisects ∠ABC, DP bisects ∠ADC.

2. According to the definition of angle bisector, ∠ADP = ∠PDC, ∠CBP = ∠PBA.

3. From the given conditions, ∠A = 40°, ∠P = 38°.

6. According to the definition of exterior angles of a polygon, the exterior angle of ∠AED is ∠DEB, the exterior angle of ∠PEB is ∠DEB. Similarly, the exterior angle of ∠CFB is ∠BFD, the exterior angle of ∠PFD is ∠BFD. Therefore, according to the exterior angle theorem of a triangle, we can obtain two equations ∠A + ∠ADP = ∠P + ∠ABP and ∠C + ∠CBP = ∠P + ∠PDC.

7. Substituting the given conditions, we get 40° + y = 38° + x and ∠C + x = 38° + y.

8. Since BP and DP bisect the angles, we can solve for ∠A + ∠C = 2∠P by summing the respective angles equal to the sum of the exterior angles.

10. Simplifying the above expression, we get ∠C = 2 × 38° - 40°.

12. Through the above reasoning, we finally obtain the answer as 36°.

普通三角形; 三角形的外角; 对顶角; 邻补角; 射线

Definition of Angle Bisector

<note>The vertex of angle ABC is point B</note>, from <note>point B, a line BP is drawn</note>, this line divides angle ABC into two equal angles, that is, <note>∠CBP and ∠PBA are equal</note>. Therefore, <note>line BP is the angle bisector of angle ABC</note>. Similarly, <note>the vertex of angle ADC is point D</note>, from <note>point D, a line DP is drawn</note>, this line divides angle ADC into two equal angles, that is, <note>∠ADP and ∠PDC are equal</note>. Therefore, <note>line DP is the angle bisector of angle ADC</note>.

Exterior Angle Theorem of Triangle

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

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

Similarly, in triangle AED, <note>angle DEB</note> is an exterior angle of the triangle, <note>angle EAD</note> and <note>angle EDA</note> are the two interior angles that are not adjacent to the exterior angle <note>DEB</note>. According to the Exterior Angle Theorem of Triangle, the exterior angle <note>DEB</note> is equal to the sum of the two non-adjacent interior angles <note>EAD</note> and <note>EDA</note>, that is, <note>angle DEB = angle EAD + angle EDA</note>.

Similarly, in triangle PFD, <note>angle BFD</note> is an exterior angle of the triangle,

Definition of Triangle

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

<note>Triangle AED</note> is a geometric figure formed by three non-collinear points <note>A, E, D</note> and their connecting line segments <note>AE, AD, DE</note>. <note>Points A, E, D</note> are the three vertices of the triangle, and the line segments <note>AE, AD, DE</note> are the three sides of the triangle. Similarly, <note>Triangle PEB</note> is a geometric figure formed by three non-collinear points <note>P, E, B</note> and their connecting line segments <note>PE, PB, BE</note>. <note>Points P, E, B</note> are the three vertices of the triangle, and the line segments <note>PE, PB, BE</note> are the three sides of the triangle. Similarly, <note>Triangle PFD</note> is a geometric figure formed by three non-collinear points <note>P, F, D</note> and their connecting line segments <note>PF, PD, FD</note>. <note>Points P, F, D</note> are the three vertices of the triangle, and the line segments <note>PF, PD, FD</note> are the three sides of the triangle. Similarly, <note>Triangle CF


Question: As shown in the figure, in △ABC, ∠A=90°, AB=AC, BD bisects ∠ABE, DE⊥BC, if BC=10cm, then the perimeter of △DEC is ()

Answer: 10cm

1. Given ∠A=90° and AB=AC, combined with the definition of an isosceles right triangle, △ABC is an isosceles right triangle.

2. According to the Pythagorean theorem, AB^2 + AC^2 = BC^2, since AB=AC, we have 2AB^2 = BC^2. Taking the square root of both sides, we get √2AB = √2AC = BC, i.e., AB=AC=10cm/√2=5√2cm.

3. Since BD bisects ∠ABE, and DE⊥BC.

4. According to BD=BD, ∠ABD=∠EBD, ∠BAD=∠DEB, by the triangle angle sum theorem, ∠ADB=∠EDB=180°-∠ABD-90°=180°-∠EBD-90°, we conclude △ABD≌△EBD (ASA, i.e., angle-side-angle theorem).

5. Since △ABD≌△EBD, by the definition of congruent triangles, AB=BE.

6. Since BC=BE+EC, BE=AB=5√2, we can find EC equals 10-5√2.

7. Since △ABC is an isosceles right triangle, ∠C=45°, since DE⊥BC, ∠DEC is 90°. By the triangle angle sum theorem, ∠C=∠EDC=180°-90°-45°=45°. By the properties of isosceles triangles, the two legs DE and CE of the right triangle DEC are equal. According to the definition of an isosceles right triangle, △DEC is also an isosceles right triangle.

8. By the Pythagorean theorem, we can derive the side ratio of an isosceles right triangle as 1:1:√2, and find DC=10√2-10.

9. According to the triangle perimeter formula, the perimeter of △DEC is DC+2EC=10.

直角三角形; 垂线; 等腰三角形; 线段; 三角形的外角

<note>Triangle ABC</note> is a geometric figure composed of <note>three non-collinear points A, B, C</note> and their connecting line segments <note>AB, AC, BC</note>. Points <note>A, B, C</note> are the three vertices of the triangle, and line segments <note>AB, AC, BC</note> are the three sides of the triangle. ##<note>Triangle ABD</note> is a geometric figure composed of <note>three non-collinear points A, B, D</note> and their connecting line segments <note>AB, AD, BD</note>. Points <note>A, B, D</note> are the three vertices of the triangle, and line segments <note>AB, AD, BD</note> are the three sides of the triangle. <note>Triangle EBD</note> is a geometric figure composed of <note>three non-collinear points E, B, D</note> and their connecting line segments <note>BE, DE, BD</note>. Points <note>E, B, D</note> are the three vertices of the triangle, and line segments <note>BE, DE, BD</note> are the three sides of the triangle. <note>Triangle DEC</note> is composed of

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>Triangle ABC is an isosceles right triangle</note>, where <note>angle BAC is a right angle (90 degrees)</note>, <note>sides AB and AC are equal legs</note>. ##<note>Triangle DEC is an isosceles right triangle</note>, where <note>angle DEC is a right angle (90 degrees)</note>, <note>sides DE and EC are equal legs</note>.##

Definition of Angle Bisector

In the figure of this problem, <note>the vertex of angle ABE is point B</note>, <note>a line BD is drawn from point B</note>, this line divides angle ABE into two equal angles, that is, <note>angle ABD and angle EBD are equal</note>. Therefore, <note>line BD is the angle bisector of angle ABE</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.

The original text: <note>Line DE and line BC intersect to form a 90-degree angle ∠DEC</note>, so according to the definition of perpendicular lines, <note>line DE and line BC are perpendicular to each other</note>.

Pythagorean Theorem

<note>In the figure of this problem, in the right triangle ABC, ##∠BAC## is a right angle (90 degrees)</note>, sides AB and AC are the legs, and side BC is the hypotenuse, so according to the Pythagorean Theorem, <note>##BC^2 = AB^2 + AC^2##</note>.

If two angles and the included side of one triangle are respectively equal to two angles and the included side of another triangle, then the two triangles are congruent.

<note>∠ABD=∠EBD=45°</note>, <note>∠BAD=∠BED=90°</note>, and <note>BD=BD</note>. Since the two triangles have two angles and the included side equal respectively, according to the Angle-Side-Angle (ASA) Criterion for Congruence of Triangles, it can be concluded that <note>triangle ABD is congruent to triangle EBD</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 ABD and triangle EBD are congruent triangles</note>, the corresponding sides and corresponding angles of triangle ABD are equal to those of triangle EBD, namely:
side <note>BD = side BD</note>
side <note>AD = side ED</note>
side <note>AB = side EB</note>, and the corresponding angles are also equal:
angle <note>EBD = angle DBA</note>
angle <note>DEB = angle DAB</note>
angle <note>ADB = angle EDB</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 EDC</note> are <note>EC, DC, ED</note> respectively. 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, i.e., the perimeter <note>P=EC+DC+ED</note>.

Triangle Angle Sum Theorem

In the figure of this problem, in triangle ABD, <note>angle ABD, angle BAD, and angle ADB are the three interior angles of triangle ABD</note>, according to the Triangle Angle Sum Theorem, <note>angle ABD + angle BAD + angle ADB = 180°</note>. In triangle BDE, <note>angle BDE, angle BED, and angle DBE are the three interior angles of triangle BDE</note>, according to the Triangle Angle Sum Theorem, <note>angle BDE + angle BED + angle DBE = 180°</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 DEC</note>, <note>angle EDC and angle ECD are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the sides opposite the equal angles are equal, that is, side ED = side EC</note>.


Question: As shown in the figure, quadrilateral ABCD is a tangential quadrilateral of ⊙O, and AB=10, CD=12. Then the perimeter of quadrilateral ABCD is ()

Answer: 44

1. Given quadrilateral ABCD is a circumscribed quadrilateral of ⊙O, according to ##the property of circumscribed quadrilateral, the sum of the two pairs of opposite sides of a circumscribed quadrilateral are equal##.

2. From the given conditions AB=10, CD=12, we get AB+CD=10+12=22.

3. According to ##the property of circumscribed quadrilateral##, we know AD+BC=22.

4. ##According to the formula for the perimeter of a polygon, we know## the perimeter of quadrilateral ABCD = AB+BC+CD+DA.

5. Substituting the given conditions into the perimeter formula, we get the perimeter of quadrilateral ABCD = 10+BC+12+AD.

6. From the given AD+BC=22, we get 10+12+AD+BC=10+12+22.

7. Therefore, the perimeter of quadrilateral ABCD = 44.

圆内接四边形; 切线; 圆

A quadrilateral is called a tangential quadrilateral if each of its four sides touches (is tangent to) a circle. For a tangential quadrilateral, the sums of the lengths of its opposite sides are equal.

In the figure of this problem, the quadrilateral <note>ABCD is a tangential quadrilateral</note>, with its four sides tangent to the circle. According to the properties of a tangential quadrilateral, <note>the sums of the lengths of the opposite sides of quadrilateral ABCD are equal</note>, that is, <note>side AB + side CD = side AD + side BC</note>.

Perimeter Formula for Polygons

The perimeter of a polygon is equal to the sum of the lengths of all its sides. For an n-sided polygon, if the lengths of the sides are \( a_1, a_2, \ldots, a_n \), then the perimeter \( P \) is given by \( P = a_1 + a_2 + \ldots + a_n \).

In the figure of this problem, it is known that the sides of the quadrilateral <note>ABCD</note> are <note>AB, BC, CD, DA</note> respectively. According to the perimeter formula for polygons, the perimeter of the quadrilateral is equal to the sum of the lengths of all its sides, i.e., perimeter <note>P=AB+BC+CD+DA</note>.


Question: As shown in the figure, the diagonals AC and BD of quadrilateral ABCD are perpendicular to each other at point O, and AC = 12, BD = 9. What is the area of quadrilateral ABCD?

Answer: 54

1. Given quadrilateral ABCD, AC is perpendicular to BD with the foot of the perpendicular being O, and AC=12, BD=9.

2. According to the definition of height, OA is the height of △ABD, OB is the height of △ABC, OC is the height of △BCD, and OD is the height of △ACD.

3. The area of △AOD is \\( \frac{1}{2} \\times AO \\times OD \\), the area of △COD is \\( \frac{1}{2} \\times OD \\times OC \\). The area of △AOB is \\( \frac{1}{2} \\times AO \\times OB \\), the area of △COB is \\( \frac{1}{2} \\times OC \\times OB \\).

4. Adding the areas of all triangles, we get the area of quadrilateral ABCD: S_{quadrilateral ABCD} = S_{△AOD} + S_{△COD} + S_{△BOC} + S_{△AOB}.

5. Substituting the areas from steps 3 and 4, we get: S_{quadrilateral ABCD} = \\( \frac{1}{2} \\times AO \\times OD \\) + \\( \frac{1}{2} \\times OD \\times OC \\) + \\( \frac{1}{2} \\times AO \\times OB \\) + \\( \frac{1}{2} \\times OC \\times OB \\).

6. Combining common terms: S_{quadrilateral ABCD} = \\( \frac{1}{2} \\times (OC + OA) \\times (OD + OB) \\).

7. Since OA + OC = AC, OB + OD = BD, substituting these into the equation, we get: S_{quadrilateral ABCD} = \\( \frac{1}{2} AC \\times BD \\).

8. Given AC=12 and BD=9, substituting these values, we get: S_{quadrilateral ABCD} = \\( \frac{1}{2} \\times 12 \\times 9 = 54 \\).

9. Through the above reasoning, the final answer is 54.

垂线; 直角三角形; 普通四边形

Definition of Altitude

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

<note>The line segment AO perpendicular to the opposite side BD from vertex A is the altitude of vertex A in △ABD</note>, <note>The line segment BO perpendicular to the opposite side AC from vertex B is the altitude of vertex B in △ABC</note>, <note>The line segment CO perpendicular to the opposite side BD from vertex C is the altitude of vertex C in △BCD</note>, <note>The line segment DO perpendicular to the opposite side AC from vertex D is the altitude of vertex D in △ACD</note>.

Area Formula of a Triangle

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

<note>In the figure of this problem, in triangle AOD, side AO is the base, and segment OD is the height</note>. According to the area formula of a triangle, the area of triangle AOD is equal to the base <note>AO multiplied by the height OD and then divided by 2</note>, that is, area = <note>(AO * OD) / 2</note>. Similarly, <note>the area of triangle COD</note> is equal to the base <note>OC multiplied by the height OD and then divided by 2</note>, that is, area = <note>(OC * OD) / 2</note>. <note>The area of triangle AOB is equal to the base AO multiplied by the height OB and then divided by 2</note>, that is, area = <note>(AO * OB) / 2</note>. <note>The area of triangle BOC</note> is equal to the base <note>OC multiplied by the height OB and then divided by 2</note>, that is, area = <note>(OC * OB) / 2</note>.


Question: As shown in the figure, in parallelogram ABCD, points E and F are on AD and AB respectively. Connect EB, EC, FC, and FD in sequence. The areas of the shaded regions are S~1~, S~2~, S~3~, and S~4~ respectively. Given S~1~=2, S~2~=12, S~3~=3, find the value of S~4~.

Answer: 7

1. Given parallelogram ABCD, points E and F are on AD and AB respectively. Define the areas of four triangles as S1, S2, S3, and S4, as shown in the figure.

2. Let the area of parallelogram ABCD be S, then the areas of triangles △CBE and △CDF are S/2 respectively.

3. Since the entire figure is composed of several parts, the total area of ABCD can be expressed by adding the areas of each part: S = area of triangle △CBE + area of triangle △CDF + (shaded parts S1 + S4 + S3) - S2.

4. According to the given conditions, the area of shaded part S1 is 2, the area of S2 is 12, and the area of S3 is 3, therefore: S = (1/2)S + (1/2)S + 2 + S4 + 3 - 12.

5. Simplifying the equation: S = S + 2 + S4 + 3 - 12, further simplification: 0 = 2 + S4 + 3 - 12.

6. Simplifying the equation, we get: S4 = 7.

7. Through the above reasoning, the final answer is S4 = 7.

平行四边形; 普通三角形; 线段; 平行线; 对称

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

Area Formula of a Triangle

In the figure of this problem, according to the area formula of a triangle, <note>the area of triangle BCE</note> is equal to <note>the base BC</note> multiplied by <note>the height on the side BC (i.e., the height on the side BC of the parallelogram)</note> and then divided by 2, <note>the area of triangle CDF</note> is equal to <note>the base ED</note> multiplied by <note>the height on the side CD (i.e., the height on the side CD of the parallelogram)</note> and then divided by 2.

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 \( A = \text{base length} \times \text{height} \).

<note>In the figure of this problem</note>, in the parallelogram xxx, <note>side xx is the base</note>, and the corresponding <note>height is the vertical distance from the base xx to the opposite side, denoted as xx</note>. Therefore, according to the area formula of a parallelogram, the area of parallelogram ABCD <note>equals the length of base BC multiplied by the corresponding height</note>, and also <note>equals the length of base CD multiplied by the corresponding height</note>.


Question: As shown in the figure, in △ABC, AB=10, AC=6, line DE passing through point A is parallel to CB, the bisectors of ∠ABC and ∠ACB intersect DE at E and D respectively, then the length of DE is ()

Answer: 16

1. According to the problem, BE is the bisector of ∠ABC. Based on the definition of angle bisector, we get ∠ABE=∠EBC. CD is the bisector of ∠ACB. Based on the definition of angle bisector, we get ∠ACD=∠BCD.

2. Since DE∥BC, according to the parallel axiom 2 of parallel lines, alternate interior angles are equal. Therefore, we have: ∠E=∠ABE=∠EBC, ∠D=∠ACD=∠BCD.

3. Therefore, triangles ABE and ADC are isosceles triangles, AD=AC, AE=AB.

4. Therefore, DE=AE + AD=AB + AC= 10 + 6 =16.

平行线; 普通三角形; 内错角; 线段; 对称

Definition of Angle Bisector

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

In the figure of this problem, <note>the vertex of angle ABC is point B</note>, a line BE is drawn from <note>point B</note>, <note>this line divides angle ABC into two equal angles, i.e., ∠ABE and ∠EBC are equal</note>. Therefore, <note>line BE is the angle bisector of angle ABC</note>. Similarly, <note>the vertex of angle ACB is point C</note>, a line CD is drawn from <note>point C</note>, <note>this line divides angle ACB into two equal angles, i.e., ∠ACD and ∠DCB are equal</note>. Therefore, <note>line CD is the angle bisector of angle ACB</note>.

Parallel Postulate 2 of Parallel Lines

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

<note>Two parallel lines DE and BC are intersected by a third line CD</note>, forming the following geometric relationship: <note>Alternate interior angles: angle D and angle DCB are equal</note>. <note>Two parallel lines DE and BC are intersected by a third line BE</note>, forming the following geometric relationship: <note>Alternate interior angles: angle E and angle EBC are equal</note>.

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

Properties of Isosceles Triangle

<note>In the isosceles triangle DAC, angle D and angle ACD are equal</note>. Therefore, according to the properties of isosceles triangles, the sides opposite the equal angles are equal, that is, <note>side AD = side AC</note>. <note>In the isosceles triangle EAB, angle E and angle ABE are equal</note>. Therefore, according to the properties of isosceles triangles, the sides opposite the equal angles are equal, that is, <note>side AE = side AB</note>.


Question: As shown in the figure, in △ABC, ∠ACB=90°, △CBD is folded along CD, making point B fall exactly on point E on side AC. If ∠A=24°, then the degree of ∠BDC is ()

Answer: 69°

1. After folding, △CBD makes point B fall at point E. According to the properties of folding, ∠ACB=2∠BCD.

2. In △ABC, it is known that ∠ACB=90°, ∠A=24°. Using the triangle angle sum theorem, calculate the sum of the interior angles of △ABC to be 180°, and obtain ∠B=180°-∠ACB-∠A=180°-90°-24°=66°.

3. According to the conclusion in step 1, obtain ∠BCD=∠ACB/2=90°/2=45°.

4. In △BDC, using the triangle angle sum theorem, calculate the sum of the interior angles of △BDC to be 180°, and obtain ∠BDC=180°-∠BCD-∠B=180°-45°-66°=69°.

5. Through the above reasoning, the final answer is 69°.

直角三角形; 对称; 余弦; 正弦; 垂线

Triangle Angle Sum Theorem

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

In the figure of this problem, <note>in triangle ABC, angle ACB, angle CAB, and angle ABC are the three internal angles of triangle ABC</note>, according to the Triangle Angle Sum Theorem, <note>angle ACB + angle CAB + angle ABC = 180°</note>. For <note>△ABC</note>, it is known that <note>∠ACB=90°</note>, <note>∠CAB=24°</note>, then it can be obtained: <note>∠ABC=180°-∠ACB-∠CAB=180°-90°-24°=66°</note>. For <note>△BDC</note>, <note>angle BDC, angle BCD, and angle CBD are the three internal angles of triangle BDC</note>, it is known that <note>∠BCD=45°</note>, <note>∠CBD=66°</note>, then it can be obtained: <note>∠BDC=180°-∠BCD-∠CBD=180°-45°-66°=69°</note>.


Question: As shown in the figure, PA and PB are tangents to ⊙O at points A and B respectively. Point C is on ⊙O, and ∠P=36°. Find ∠ACB.

Answer: 72°

1. As shown in the figure, connect OA and OB, OA and OB are radii of circle O.

2. According to the given conditions, PA and PB are tangents to circle O. Since the tangent is perpendicular to the radius, we have ∠PAO = ∠PBO = 90°.

3. Given ∠P = 36°, we have ∠AOB = 360° - ∠PAO - ∠PBO - ∠P = 360° - 90° - 90° - 36° = 144°.

4. According to the inscribed angle theorem, the inscribed angle ∠ACB of arc AB is equal to half of the central angle ∠AOB, so we have ∠ACB = 1/2 * ∠AOB = 1/2 * 144° = 72°.

5. Through the above reasoning, the final answer is ∠ACB = 72°.

圆; 切线; 圆周角; 圆内接四边形; 普通三角形

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

Definition of 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>In the diagram of this problem, circle O</note> and <note>lines PA and PB</note> each have only <note>one common point A and B</note>, respectively. These two common points are called points of tangency. Therefore, <note>lines PA and PB are tangents 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.

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

Sum of Interior Angles of a Quadrilateral Theorem

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

<note>∠PAO = ∠PBO = 90°</note>, so ∠APB + <note>∠AOB</note> + 2 * 90° = 360°, hence <note>∠AOB = 144°</note>.

Inscribed Angle Theorem

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

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


Question: As shown in the figure, C and D are two points on the line segment AB. If CB=4cm, DB=7cm, and D is the midpoint of AC, then the length of AC is equal to ()

Answer: 6cm

1. Given that point C and point D are two points on segment AB, the length of CB is 4cm, and the length of DB is 7cm.

2. From the relationship of segment addition and subtraction, we can get the length of DC as DC = DB - CB = 7cm - 4cm = 3cm.

3. Given that point D is the midpoint of AC, according to the definition of midpoint, the midpoint divides the distance between two points equally, so the length of AC should be twice the length of DC, i.e., AC = 2 * DC.

4. Calculate the length of AC, and get AC = 2 * 3cm = 6cm.

5. Through the above reasoning, we finally get the length of AC as 6cm.

Midpoint of a Line Segment

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

In this problem, <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, <note>the lengths of line segments AD and DC are equal</note>. That is, <note>AD = DC</note>.

Definition of Line Segment

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

<note>Line segment AB is a part of a straight line, containing endpoint A and endpoint B and all points between them</note>. Line segment AB has two endpoints, which are <note>A and B</note>, and <note>every point on line segment AB is located between endpoint A and endpoint B</note>. <note>Line segments AC, AD, DC, CB, and DB are also line segments</note>, each containing their corresponding endpoints and all points between them.


Question: As shown in the figure, in circle O with a radius of 5, the length of chord AB is 8. Find the distance from the center O to the chord AB.

Answer: 3

1. Given that the radius of circle O is 5, and the length of chord AB is 8.

2. Connect segments OA and OB, ##let OC be perpendicular to AB at C##.

3. According to the ##perpendicular bisector theorem##, OC is perpendicular to AB, and AC is equal to CB.

4. Therefore, AC = CB = ##(1/2)## AB = 4.

5. In the right triangle △OAC, according to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the two legs.

6. According to the Pythagorean theorem, OC = ##√(OA^2 - AC^2)##.

7. Substituting the given conditions, OC = ##√(5^2 - 4^2)##.

8. Calculating, OC = ##√(25 - 16) = √(9)## = 3.

9. Therefore, the distance from the center O to the chord AB is 3.

圆; 弦; 垂线; 直角三角形

Definition of Radius

<note>Point O is the center of the circle</note>, <note>Point A and Point B are any two points on the circle</note>, <note>Line segments OA and OB are segments from the center of the circle to any point on the circle</note>, therefore <note>Line segments OA and OB are the radii of the circle</note>, with a length of 5.

Definition of Chord

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

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

In <note>circle O</note>, <note>line segment OC is perpendicular to chord AB</note>, then according to the Perpendicular Diameter Theorem, <note>line segment OC bisects chord AB</note>, that is, <note>AC=CB=##(1/2)##AB=4####</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 OAC, ∠OCA is a right angle (90 degrees)</note>, the sides <note>OC and AC are the legs</note>, the side <note>OA is the hypotenuse</note>, so according to the Pythagorean Theorem, ##<note>OA^2= OC^2 + AC^2</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 OAC, angle OCA is a right angle (90 degrees)</note>, therefore <note>triangle OAC is a right triangle</note>. <note>Side OC and side AC are the legs</note>, <note>side OA is the hypotenuse</note>.


Question: As shown in the figure, in △ABC, the line DE passing through vertex A is parallel to BC. The angle bisectors of ∠ABC and ∠ACB intersect DE at points E and D, respectively. If AC = 3 and AB = 4, then the length of DE is ()

Answer: 7

1. Given that line DE is parallel to BC and passes through vertex A, and the bisectors of ∠ABC and ∠ACB intersect DE at points E and D, respectively.

2. Since the bisectors of ∠ABC and ∠ACB intersect DE at points E and D, ∠EBC = ∠ABE and ∠DCB = ∠ACD.

3. According to the parallel line axiom 2, corresponding angles are equal, we get ∠DCB = ∠CDE and ∠EBC = ∠BED.

4. Since ∠DCB = ∠CDE, ∠ADC = ∠ACD, similarly, since ∠EBC = ∠BED, ∠ABE = ∠AEB.

5. In triangles ABE and ADC, according to the properties of isosceles triangles, since ∠ADC = ∠ACD, AD = AC. Similarly, since ∠ABE = ∠AEB, AB = AE.

6. We get the length of DE equals AD plus AE, i.e., DE = AD + AE.

7. From step 5, we know AD = AC = 3, AE = AB = 4, so DE = 3 + 4 = 7.

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

平行线; 三角形的外角; 线段; 普通三角形; 平行四边形

Definition of Angle Bisector

In the diagram of this problem, <note>the vertex of angle ABC is point B</note>, <note>a line BE is drawn from point B</note>, and this line divides angle ABC into two equal angles, that is, <note>angle EBC and angle ABE are equal</note>. Therefore, <note>line BE is the angle bisector of angle ABC</note>. Similarly, <note>the vertex of angle ACB is point C</note>, <note>a line CD is drawn from point C</note>, and this line divides angle ACB into two equal angles, that is, <note>angle DCB and angle ACD are equal</note>. Therefore, <note>line CD is the angle bisector of angle ACB</note>.

Parallel Postulate 2 of Parallel Lines

<note>DE∥BC</note>, and it is intersected by <note>lines CD and BE</note>, forming the following geometric relationships: <note>alternate interior angles: ∠EBC and ∠BED are equal, ∠BCD and ∠CDE are equal</note>.

Properties of Isosceles Triangle

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

<note>In isosceles triangle ADC, angle ADC = angle ACD</note>. Therefore, according to the properties of isosceles triangles, the sides opposite the equal angles are equal, that is, <note>side AD and side AC are equal</note>. Similarly, <note>in isosceles triangle ABE, angle ABE = angle AEB</note>. Therefore, according to the properties of isosceles triangles, the sides opposite the equal angles are equal, that is, <note>side AB and side AE 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 figure of this problem, <note>the two parallel lines BC and DE</note> are intersected by <note>lines CD and BE</note>. Among them, <note>angle DCB and angle CDE</note>, <note>angle EBC and angle BED</note> are located between the two parallel lines and are on opposite sides of the intersecting lines <note>CD and BE</note>. Therefore, <note>angle DCB and angle CDE are alternate interior angles</note>, <note>angle EBC and angle BED are alternate interior angles</note>. <note>Alternate interior angles are equal, i.e., ∠DCB=∠CDE, ∠EBC=∠BED</note>.


Question: As shown in the figure, PA and PB are tangent to ⊙O at points A and B respectively, ∠C=55°, then ∠P equals ()

Answer: 70°

1. Given PA is tangent to ⊙O at A, PB is tangent to ⊙O at B, connect OA and OB.

2. According to the properties of the tangent to a circle: OA is perpendicular to PA, OB is perpendicular to PB.

4. Since quadrilateral POAB is a quadrilateral, according to the theorem of the sum of interior angles of a quadrilateral, the sum of the interior angles of the quadrilateral is 360°, so ∠P = 360° - ∠AOB - ∠OAP - ∠OBP.

5. According to the problem, ∠C = 55°, then according to the inscribed angle theorem, the central angle is twice the inscribed angle corresponding to the arc, thus ∠AOB = 2∠C = 110°.

6. Therefore, ∠P = 360° - ∠AOB - ∠OAP - ∠OBP = 360° - 90° - 90° - 110° = 70°.

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

圆; 切线; 等腰三角形; 圆周角; 三角形的外角

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

<note>In quadrilateral AOBP, angles PAO, AOB, OBP, and APB are the four interior angles of the quadrilateral</note>. According to the sum of interior angles of a quadrilateral theorem, the sum of these four interior angles is 360°, that is, <note>angle PAO + angle AOB + angle OBP + angle APB = 360°</note>.

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 <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>, that is, <note>∠ACB = 1/2 ∠AOB</note>.

Definition of Inscribed Angle

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

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

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


Question: A ship departs from point A on the sea surface and travels 40 nautical miles in the direction of 40° south-west to reach point B. Then, from point B, it travels 40 nautical miles in the direction of 20° north-west to reach point C. Find the distance between points A and C.

Answer: 40海里

1. Given: A ship departs from point A on the sea surface and travels 40 nautical miles in the direction of 40° south-west to reach point B, then travels 40 nautical miles in the direction of 20° north-west from point B to reach point C.

2. According to the problem, auxiliary lines can be constructed: Draw a line perpendicular to the sea surface from point A towards the south and another line perpendicular to the sea surface from point B towards the north.

3. Since the ship travels 40° south-west, ∠BAA' = 40° (where A' is a point on the line extended south from point A).

4. Similarly, since the ship travels 20° north-west, ∠CBB' = 20° (where B' is a point on the line extended north from point B).

5. Since these two lines are in the north-south direction, ##according to the parallel line axiom 2, BB' | | AA', ∠B'BA and ∠BAA' are alternate interior angles, so ∠B'BA=∠BAA',## thus ##∠ABC## = ∠BAA' + ∠CBB' = 40° + 20° = 60°.

6. The actual travel path from point A to point B, AB = 40 nautical miles, and the actual travel path from point B to point C, BC = 40 nautical miles.

7. Therefore, according to the given ##∠ABC## = 60° and AB = 40 nautical miles, BC = 40 nautical miles, it can be concluded that △ABC is an equilateral triangle (##determination theorem of equilateral triangle (60-degree angle determination of isosceles triangle)##).

8. Therefore, AC = AB = 40 nautical miles.

9. Through the above reasoning, the final answer is 40 nautical miles.

直角三角形; 方向角; 线段; 等腰三角形; 平移

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.

<note>∠BAA' is a geometric figure composed of rays BA and AA'</note>, these two rays have a common endpoint A. This common endpoint A is called <note>the vertex of ∠BAA'</note>, and rays BA and AA' are called <note>the sides of ∠BAA'</note>. Similarly, <note>∠CBB' is a geometric figure composed of rays CB and BB'</note>, these two rays have a common endpoint B. This common endpoint B is called <note>the vertex of ∠CBB'</note>, and rays CB and BB' are called <note>the sides of ∠CBB'</note>.

Definition of Triangle

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

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

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

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

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

Parallel Postulate 2 of Parallel Lines

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

<note>Two parallel lines BB' and AA'</note> are intersected by <note>a third line AB</note>, forming the following geometric relationship: <note>Alternate interior angles: angle B'BA and angle BAA' are equal</note>.

Definition of Alternate Interior Angles

In the figure of this problem, <note>two parallel lines AA' and BB'</note> are intersected by <note>a line AB</note>, where <note>angle BAA'</note> and <note>angle ABB'</note> are located between the two parallel lines and on opposite sides of the intersecting line AB. Therefore, <note>angle BAA'</note> and <note>angle ABB'</note> are alternate interior angles. Alternate interior angles are equal, i.e., <note>angle BAA' is equal to angle ABB'</note>.


Question: As shown in the figure, AB∥CD, BE is the perpendicular bisector of AD, DC=BC, if ∠A=70°, then ∠C=()

Answer: 100°

1. Given BE is the perpendicular bisector of AD, according to the definition of perpendicular bisector, we get AB = BD.

2. According to the properties of an isosceles triangle, in △ABD, ∠ADB = ∠A = 70°.

3. Since AB∥CD, according to the parallel postulate 2 of parallel lines, the interior angles on the same side are supplementary, ∠ADC= 180° -∠A=110°.

4. Therefore, ∠CDB = ∠ADC - ∠ADB=110°-70°=40°.

5. Since CD=CB, triangle CBD is an isosceles triangle, so ∠CDB=∠CBD=40°.

6. In △BCD, according to the triangle sum theorem, ∠C = 180° - 2×∠CDB=100°.

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

平行线; 垂直平分线; 等腰三角形; 同位角; 内错角

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.

<note>Line BE passes through the midpoint E of segment AD</note>, and <note>line BE is perpendicular to segment AD</note>. Therefore, <note>line BE is the perpendicular bisector of segment AD</note>.

Properties of Isosceles Triangle

In the isosceles triangle <note>ΔBCD</note>, <note>sides BC and DC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠DBC = ∠DCB</note>.

Parallel Postulate 2 of Parallel Lines

<note>In the diagram of this problem, two parallel lines AB and CD are intersected by a third line AD</note>, forming the following geometric relationship: Consecutive interior angles: <note>∠A and ∠ADC are supplementary, i.e., ∠A + ∠ADC = 180 degrees</note>.

Triangle Angle Sum Theorem

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

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

When two lines are intersected by a third line, the angles that lie on the same side of the transversal and inside the two intersected lines are called consecutive interior angles. If the two lines are parallel, then the consecutive interior angles are supplementary. Conversely, if the consecutive interior angles are supplementary, then the two lines are parallel.

<note>Two lines AB and CD are intersected by a third line AD</note>, <note>angles A and ADC are on the same side of the intersecting line AD</note>, and <note>within the intersected lines CD and AB</note>, so <note>angles A and ADC are consecutive interior angles</note>. <note>Consecutive interior angles A and ADC are supplementary</note>, that is, <note>angle A + angle ADC = 180 degrees</note>.


Question: As shown in the figure, in quadrilateral ABCD, AD∥BC, BF bisects ∠ABC and intersects AD at point F, CE bisects ∠BCD and intersects AD at point E, AB=8, CD=6, EF=2. Find the length of AD.

Answer: 12

1. Given that line segment AD is parallel to line segment BC, and line segment BF bisects ∠ABC, we get ∠ABF equal to ∠CBF.

2. Since line segment CE bisects ∠BCD, we get ∠DCE equal to ∠BCE.

3. Because line segment AD is parallel to line segment BC, ##according to the parallel line postulate 2##, we get ∠CBF equal to ∠AFB, and ∠DEC equal to ∠BCE.

4. Combining the results of the above two steps, we know that ∠ABF is equal to ∠AFB, and ∠DCE is equal to ∠DEC.

5. Because ∠ABF is equal to ∠AFB, according to ##the properties of an isosceles triangle##, we can get line segment AB equal to line segment AF, thus the length of AF is 8.

6. Similarly, since ∠DCE is equal to ∠DEC, according to ##the properties of an isosceles triangle##, we can get line segment CD equal to line segment DE, thus the length of DE is 6.

7. Given that the length of line segment EF is 2, according to the relationship of segment addition and subtraction, the length of line segment AD is equal to the length of line segment AF plus the length of line segment DE minus the length of line segment EF.

8. Calculating, we get: AD = AF + DE - EF = 8 + 6 - 2 = 12.

9. Through the above reasoning, the final answer is: 12.

平行线; 平行四边形; 内错角; 平移; 线段

Definition of Angle Bisector

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

In the diagram of this problem, the vertex of <note>∠ABC is point B</note>, and a line BF is drawn from <note>point B</note>, which divides <note>∠ABC into two equal angles</note>, namely <note>∠ABF and ∠CBF are equal</note>. Therefore, <note>line BF is the angle bisector of ∠ABC</note>. Additionally, the vertex of <note>∠BCD is point C</note>, and a line CE is drawn from <note>point C</note>, which divides <note>∠BCD into two equal angles</note>, namely <note>∠DCE and ∠BCE are equal</note>. Therefore, <note>line CE is the angle bisector of ∠BCD</note>.

Definition of Parallel Lines

Two lines in the same plane that do not intersect are called parallel lines.

<note>Line segment AD and line segment BC are located in the same plane</note>, and they do not intersect, so according to the definition of parallel lines, <note>line segment AD and line segment BC are parallel lines</note>.

Parallel Postulate 2 of Parallel Lines

<note>In the diagram of this problem, two parallel lines AD and BC are intersected by a third line BF</note>, forming the following geometric relationships:
1. Alternate interior angles: <note>∠FBC</note> and <note>∠AFB</note> are equal.
2. Alternate interior angles: <note>∠BCE</note> and <note>∠DEC</note> are equal.
<note>These relationships demonstrate that when two parallel lines are intersected by a third line, the alternate interior angles are equal.</note>

Properties of Isosceles Triangle

In the figure of this problem, <note>in the isosceles triangle ABD</note>, <note>angle ABD = angle ADB</note>. Therefore, according to the properties of the isosceles triangle, the sides opposite to the equal angles are equal, <note>i.e., side AB and side AD are equal</note>. <note>In the isosceles triangle EDC</note>, <note>angle DEC and angle DCE are equal</note>. Therefore, according to the properties of the isosceles triangle, the sides opposite to the equal angles are equal, <note>i.e., side DE = side DC</note>.

Definition of Alternate Interior Angles

<note>In the diagram of this problem, two parallel lines AD and BC</note> are <note>intersected by lines BF and CE</note>, among which <note>∠CBF and ∠AFB</note>, <note>angles ∠DEC and ∠BCE</note> are located between the two parallel lines, and on opposite sides of the intersecting lines <note>BF and CE</note>, therefore <note>∠CBF and ∠AFB are alternate interior angles</note>, <note>angles ∠DEC and ∠BCE are alternate interior angles</note>. <note>Alternate interior angles are equal</note>, that is, <note>angle ∠CBF is equal to angle ∠AFB</note>, and <note>angle ∠DEC is equal to angle ∠BCE</note>.

Definition of Isosceles Triangle

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

<note>In triangle AFB, side AB and side AF are equal</note>, therefore <note>triangle AFB is an isosceles triangle</note>, <note>in triangle DEC, side DE and side DC are equal</note>, therefore <note>triangle DEC is an isosceles triangle</note>.


Question: As shown in the figure, through a point inside the triangle, parallel lines are drawn to each of the three sides. If the perimeter of the triangle is 6cm, then the sum of the perimeters of the three shaded triangles in the figure is ()

Answer: 6cm

1. Given a point inside a triangle, draw lines parallel to the three sides of the triangle. Let the three vertices of the triangle be A, B, and C. Draw EN parallel to BC, PM parallel to AB, and DQ parallel to AC. Assume EN, PM, and DQ are related to the point F inside triangle ABC.

2. By the definition of a parallelogram, quadrilaterals EFBP, FQCN, and ADFM are parallelograms.

3. In the parallelogram EFBP, EF is equal to BP, and PF is equal to BE; in the parallelogram FQCN, FQ is equal to NC, and FN is equal to CQ; in the parallelogram ADFM, DF is equal to AM, and FM is equal to AD.

4. Calculate the sum of the perimeters of the three shaded triangles: DE + EF + FD + FM + FN + MN + FP + PQ + FQ.

5. Substitute the results from step 3, the sum of the perimeters of the three shaded triangles = DE + BP + AM + AD + QC + MN + BE + PQ + NC.

6. Further combine the parts, (AD + DE + BE) + (BP + PQ + CQ) + (NC + MN + AM), which equals AB + BC + AC.

7. According to the problem, the perimeter of the original triangle is 6 cm, so the sum of the perimeters of the three shaded triangles is 6 cm.

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

普通三角形; 平行线; 点

Properties of Parallelogram Theorem

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

<note>In quadrilateral EFBP, EF = BP, PF = BE</note>; <note>In quadrilateral FQCN, FQ = NC, FN = CQ</note>; <note>In quadrilateral ADFM, DF = AM, FM = AD</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 EFBP</note>, <note>Quadrilateral FQCN</note>, <note>Quadrilateral ADFM</note> are parallelograms. In the parallelogram <note>EFBP</note>, <note>EF is equal to BP</note>, <note>PF is equal to BE</note>; in the parallelogram <note>FQCN</note>, <note>FQ is equal to NC</note>, <note>FN is equal to CQ</note>; in the parallelogram <note>ADFM</note>, <note>DF is equal to AM</note>, <note>FM is equal to AD</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.

In the figure of this problem, it is known that the three sides of triangle <note>△DEF</note> are <note>DE</note>, <note>DF</note>, and <note>EF</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, i.e., perimeter <note>P=DE+DF+EF</note>. It is known that the three sides of triangle <note>△MNF</note> are <note>MN</note>, <note>MF</note>, and <note>NF</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, i.e., perimeter <note>P=MN+MF+NF</note>. It is known that the three sides of triangle <note>△PQF</note> are <note>PQ</note>, <note>PF</note>, and <note>QF</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, i.e., perimeter <note>P=PQ+PF+QF</note>.


Question: As shown in the figure, in △ABC, points D and E are the midpoints of AB and AC respectively. If DE=1.5, then the length of BC is ()

Answer: 3

1. Given that points D and E are the midpoints of segments AB and AC respectively, according to the Midline Theorem, segment DE is the midline of △ABC parallel to BC.

2. According to the Midline Theorem, the length of the midline of a triangle is equal to half of the third side of the triangle, in this case, DE = 1/2 * BC.

3. Given DE = 1.5, based on the conclusion from the previous step, we can deduce BC = 2 * DE.

4. Substituting the given value of DE, we calculate BC = 2 * 1.5 = 3.

5. Therefore, through the above reasoning, we can finally conclude that the length of segment BC is 3.

中点; 平行线; 普通三角形; 线段

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.

In the diagram of this problem, in triangle ABC, <note>point D is the midpoint of side AB</note>, <note>point E is the midpoint of side AC</note>, <note>line segment DE connects these two midpoints</note>. According to the Triangle Midline Theorem, <note>line segment DE is parallel to the third side BC and equals half of the third side BC</note>, i.e., <note>DE || BC</note>, and <note>DE = 1/2 * BC</note>. Given that DE=1.5, therefore <note>BC=2 * DE=3</note>.


Question: As shown in the figure, in parallelogram ABCD, AB=10cm, AD=15cm, AC and BD intersect at point O. OE⊥BD intersects AD at E. Then the perimeter of △ABE is ()

Answer: 25cm

1. Given that in parallelogram ABCD, AB=10cm, AD=15cm, and AC, BD intersect at point O.

2. Since point O is the midpoint of BD, and OE is perpendicular to BD, according to the definition of the perpendicular bisector, segment EO is the perpendicular bisector of segment BD.

3. Because EO is the perpendicular bisector of BD, it can be concluded that BE is equal to ED.

4. In △ABE, the perimeter is AB + AE + BE = AB + AE + DE = AB + AD.

5. Also, since AD and BC are equal in parallelogram ABCD.

6. Therefore, the perimeter of △ABE is AB + BC.

7. According to the given information, AB is equal to 10cm, BC is equal to 15cm, substituting these values, the perimeter of △ABE is 10cm + 15cm = 25cm.

平行四边形; 线段; 垂线; 普通三角形; 直角三角形

Definition of Parallelogram

<note>Quadrilateral ABCD is a parallelogram</note>, in which <note>AB∥DC and AB=DC=10cm</note>, <note>AD∥BC and AD=BC=15cm</note>.

Definition of Perpendicular Bisector

<note>Line segment EO is the perpendicular bisector of line segment BD</note>, <note>midpoint O is the midpoint of BD</note>, and <note>EO is perpendicular to BD</note>.

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 segment BD is line OE</note>, <note>point E is on line OE</note>. According to the properties of the perpendicular bisector, <note>the distance from point E to the endpoints B and D of segment BD is equal</note>, that is, <note>BE = ED</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 \( P = a + b + c \).

<note>The three sides of triangle BAE are AB, AE, and BE</note>, according to the formula for the perimeter of a triangle, the <note>perimeter of a triangle is equal to the sum of the lengths of its three sides</note>, that is, <note>perimeter P=AB+AE+BE</note>.


Question: As shown in the figure, the height of the bottom plate of a truck compartment from the ground is rac{3}{2} meters. To facilitate unloading, a wooden plank is often used to form an inclined plane. To ensure that the angle between the inclined plane and the horizontal ground does not exceed 30°, the length of the wooden plank should be at least ()

Answer: 3米

1. As shown in the figure, let AC be the height of the carriage floor from the ground, i.e., AC=##3/2## meters, and AB be the length of the plank.

2. Since the plank needs to form a slope, making the angle between the slope and the horizontal ground not greater than 30°, therefore in the right triangle ABC, ∠B=30°.

3. According to the properties of a ##30°-60°-90°## triangle, in a ##30°-60°-90°## triangle, the side opposite the 30° angle is half of the hypotenuse. The hypotenuse AB is equal to 2 times the opposite side AC.##

4. Therefore, AB=2*AC.

5. Substituting AC=##3/2## meters, we get AB=2×##3/2##=3 meters.

6. Hence, in order to make the angle between the slope and the horizontal ground not greater than 30°, the length of the plank must be at least 3 meters.

直角三角形; 仰角; 正弦

Definition of Right Triangle

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

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

Properties of 30°-60°-90° Triangle

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

In the figure of this problem, in the <note>30°-60°-90° triangle ABC</note>, angle <note>ABC is 30 degrees</note>, angle <note>BAC is 60 degrees</note>, angle <note>ACB is 90 degrees</note>. Side <note>AB is the hypotenuse</note>, side <note>AC is the side opposite the 30-degree angle</note>, side <note>BC is the side opposite the 60-degree angle</note>. According to the properties of the 30°-60°-90° triangle, side <note>AC is equal to half of side AB</note>, side <note>BC is equal to √3 times side AC</note>. That is: <note>AB = 1/2 * AC</note>, <note>BC = AC * √3</note>.