Question: As shown in the figure, AB∥CD, AD intersects BC at point O. If AO=2, DO=4, BO=3, then the length of BC is ()

Answer: 9

1. Given AB is parallel to CD, according to ##Parallel Lines Axiom 2##, we can deduce ##∠OAB = ∠ODC (alternate interior angles are equal), ∠OBA = ∠OCD (alternate interior angles are equal)##.

2. In ##△OAB## and ##△ODC##, ##∠OAB = ∠ODC, ∠OBA = ∠OCD, according to the similarity criterion of triangles (AA), △OAB and △ODC are similar##.

3. According to ##the definition of similar triangles##, we get: ##BO/CO=AO/DO##.

4. Given AO=2, DO=4, BO=3, using the formula: ##3/CO=2/4##, we can solve: CO=6.

5. Therefore, the length of BC is: BC=BO+CO=3+6=9.

平行线; 内错角; 线段; 位似

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.

The original Chinese enclosed in <note></note> remains enclosed after translation.

    <note>Two parallel lines AB and CD</note> are intersected by a <note>line AD</note>, where <note>angle BAD and angle ADC</note> are located between the two parallel lines and on opposite sides of the <note>intersecting line AD</note>. Therefore, <note>angle BAD and angle ADC are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle BAD is equal to angle ADC</note>. Similarly, <note>two parallel lines AB and CD</note> are intersected by a <note>line BC</note>, where <note>angle ABC and angle BCD</note> are located between the two parallel lines and on opposite sides of the <note>intersecting line BC</note>. Therefore, <note>angle ABC and angle BCD are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle ABC is equal to angle BCD</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 diagram of this problem, <note>AB ∥ CD</note> and <note>AD intersects BC at point O</note>, according to the properties of parallel lines, we can obtain <note>∠OAB = ∠ODC</note>, <note>∠OBA = ∠OCD</note>. Specifically, AB and CD are two parallel lines, intersected by the third line <note>AD</note><note>and line DC</note>, forming the following geometric relationships: <note>alternate interior angles: ∠OAB and ∠ODC are equal, ∠OBA and ∠OCD are equal</note>.

Similarity Theorem for Triangles (AA)

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

<note>∠OAB and ∠ODC are two equal angles</note>, <note>∠OBA and ∠OCD are two equal angles</note>, thus <note>△AOB and △COD are similar</note>.

Definition of Similar Triangles

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

<note>Triangle OAB</note> and <note>Triangle ODC</note> are similar triangles. According to the definition of similar triangles: <note>∠AOB = ∠COD</note>, <note>∠BAO = ∠CDO</note>, <note>∠ABO = ∠DCO</note>; <note>AO/DO = BO/CO</note>.


Question: As shown in the figure, in quadrilateral ABCD, AD∥BC, diagonals AC and BD intersect at O. If rac{S_{	riangleADO}}{S_{	riangleDOC}}=rac{1}{3}, then the value of rac{AD}{BC} is ()

Answer: \frac{1}{3}

1. Given rac{S_{	riangle ADO}}{S_{	riangle DOC}} = rac{1}{3}, according to the ratio of areas being equal to the ratio of the corresponding heights (and the heights are perpendiculars from the same vertex to the corresponding sides), we get rac{OA}{OC} = rac{1}{3}.

2. According to the problem statement AD ∥ BC, based on the parallel postulate 2, the alternate interior angles are equal, hence ∠ADO = ∠OBC and ∠OAD = ∠OCB.

3. According to the similarity criterion for triangles (AA), the corresponding angles of △ADO and △CBO are equal, thus △ADO ∽ △CBO.

4. Since the corresponding sides of similar triangles are proportional, i.e., the proportional relationship of the sides corresponding to ∠ADO and ∠CBO, we have rac{AD}{BC} = rac{OA}{OC} = rac{1}{3}.

5. From the above proportional relationship, we finally get rac{AD}{BC} = rac{1}{3}.

平行线; 普通三角形; 内错角

<note>In the diagram of this problem, triangle ADO and triangle CBO are similar triangles</note>. According to the definition of similar triangles: <note>angle ADO = angle CBO</note>, <note>angle OAD = angle OCB</note>, <note>angle DOA = angle BOC</note>; <note>AO/OC = AD/BC = 1/3</note>.

Parallel Postulate 2 of Parallel Lines

<note>AD ∥ BC</note>, intersected by the third line <note>AC</note>, forming the following geometric relationship: alternate interior angles: <note>∠DAO = ∠OCB</note>. <note>AD ∥ BC</note>, intersected by the third line <note>BD</note>, forming the following geometric relationship: alternate interior angles: <note>∠ADO = ∠OBC</note>.

Similarity Theorem for Triangles (AA)

<note>Angle OAD is equal to angle OCB</note>, and <note>angle ODA is equal to angle OBC</note>, so <note>triangle OAD is similar to triangle OCB</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 triangle AOD, side OA is the base</note>, <note>the distance from D to AC is the height</note>. According to the area formula of a triangle, <note>the area of triangle AOD is equal to the base OA multiplied by the height and then divided by 2</note>, i.e., Area = (OA * Height) / 2. <note>In triangle DOC, side OC is the base</note>, <note>the distance from D to AC is the height</note>. According to the area formula of a triangle, <note>the area of triangle DOC is equal to the base OC multiplied by the height and then divided by 2</note>, i.e., Area = (OC * Height) / 2.


Question: As shown in the figure, D and E are points on sides AB and AC of △ABC, respectively, with DE∥BC. If AD:DB=1:3 and AE=2, then the length of AC is ()

Answer: 8

1. Given DE∥BC, according to the ##Theorem of Proportional Segments in Parallel Lines##, we get \\( \frac{AE}{AC} = \frac{AD}{AB} \\\).

2. From the given AD:DB=1:3, we can obtain \\( \frac{AD}{AB} = \frac{1}{1+3} = \frac{1}{4} \\\).

3. Substitute the value of \\( \frac{AD}{AB} \\\) into the proportion in step one, we get \\( \frac{AE}{AC} = \frac{1}{4} \\\).

4. Given AE=2, substitute it into the proportion in step three, we get \\( \frac{2}{AC} = \frac{1}{4} \\\).

5. By cross-multiplying, we can obtain AC=8.

6. Through the above reasoning, the final answer is that the length of AC is 8.

平行线; 普通三角形; 线段

Proportional Segments Theorem

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

<note>In the figure of this problem, in triangle ABC, line DE is parallel to side BC</note>, and it intersects the other two sides AB and AC at points D and E. Then, according to the Proportional Segments Theorem, we have: <note>$$ \frac{AD}{AB} = \frac{AE}{AC} $$</note>. That is, the segments intercepted are proportional to the corresponding segments of the original triangle.

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 DE and line BC are located in the same plane</note>, and <note>they have no intersection points</note>, so according to the definition of parallel lines, <note>line DE and line BC are parallel lines</note>.


Question: As shown in the figure, in the rectangle ABCD, AB=3, BC=4, point M is on the side BC, and satisfies BM=1. Draw DN perpendicular to AM, intersecting AM at point N. Then the length of DN is ()

Answer: \frac{6}{5}√{10}

1. Given that quadrilateral ABCD is a rectangle, according to the definition of a rectangle, we get ∠B=90°, and AD=BC=4.

2. In right triangle △AMB, given AB=3 and BM=1, we can use the Pythagorean theorem to find the length of AM.

3. Calculate AM, according to the Pythagorean theorem, AM=√(AB^2+BM^2)=√(3^2+1^2)=√10.

4. From the definition of a rectangle, we know ∠BAD=90°, i.e., ∠DAN+∠BAM=90°. Also, because in right triangle △AMB, ∠AMB+∠BAM=90°, we get ∠DAN=∠AMB.

5. According to the similarity criterion for triangles (AA), we can conclude that △AMB is similar to △DAN.

6. According to the definition of similar triangles, DN/AB=AD/AM.

7. Substitute the given data, DN/3=4/√10, thus we get DN=6√10/5.

8. Through the above reasoning, we finally get the answer as 6√10/5.

矩形; 垂线; 直角三角形; 线段

Definition of Rectangle

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

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

Definition of Right Triangle

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

<note>In the figure of this problem, in triangle ABM, angle ∠ABM is a right angle (90 degrees)</note>, therefore triangle ABM is a right triangle. <note>Side AB and side BM are the legs</note>, <note>side AM is the hypotenuse</note>. Similarly, <note>in triangle ADN, angle ∠AND is a right angle (90 degrees)</note>, therefore triangle ADN is a right triangle. <note>Side AN and side DN are the legs</note>, <note>side AD 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 the diagram of this problem, <note>right triangle ADM</note>, <note>angle ∠ABM is a right angle (90 degrees)</note>, <note>sides AB and BM are the legs</note>, <note>side AM is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AM^2 = AD^2 + DM^2</note>, substituting the known data gives <note>AM = √(3^2 + 1^2) = √10</note>.

Similarity Theorem for Triangles (AA)

In <note>triangle ABM</note> and <note>triangle AND</note>, if <note>angle B is equal to angle AND</note>, and <note>angle DAN is equal to angle BAM</note>, then <note>triangle BAM is similar to triangle AND</note>.

Definition of Similar Triangles

In the figure of this problem, <note>triangle ADN and triangle ABM are similar triangles</note>. According to the definition of similar triangles: <note>DN/AB = AD/AM</note>.

Complementary Acute Angles in a Right Triangle

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

In the <note>right triangle ABM</note>, <note>angle B is a right angle (90 degrees)</note>, <note>angles BAM and AMB are the two acute angles other than the right angle</note>. According to the property of complementary acute angles in a right triangle, <note>the sum of angles BAM and AMB is 90 degrees</note>, that is, <note>angle BAM + angle AMB = 90°</note>.


Question: As shown in the figure, in △ABC, DE∥BC, if AB=7cm, AC=5cm, AD=3cm, then DE=()

Answer: \frac{20}{7}cm

1. Given AB=7 cm, AD=3 cm, ##therefore## BD=BA-AD=4 cm.

2. Since DE∥BC, according to ##Parallel Line Axiom 2, corresponding angles are equal, so angle A=angle BDE, and since B is the common angle, according to the similarity theorem (AA), we get ##△BDE∽△BAC.

3. According to ##the definition of similar triangles, we get BD/AB=DE/AC, i.e., 4/7=DE/5##.

4. Solving the equation ####we get DE=20/7 cm.

5. Through the above reasoning, the final answer is 20/7 cm.

平行线; 普通三角形; 线段; 位似

<note>In the diagram of this problem, triangle BDE and triangle BAC are similar triangles</note>. According to the definition of similar triangles: <note>∠BDE = ∠BAC</note>, <note>∠DBE = ∠ABC</note>, <note>∠BED = ∠ACB</note>; <note>BD/BA = DE/AC = BE/BC</note>.

Similarity Theorem for Triangles (AA)

In the figure of this problem, triangles BDE and BAC, <note>angle B is equal to angle B</note>, and <note>angle BDE is equal to angle BAC</note>, so <note>triangle BDE is similar to triangle BAC</note>.

Parallel Postulate 2 of Parallel Lines

<note>Two parallel lines DE and AC are intersected by a third line AB</note>, forming the following geometric relationship: <note>Corresponding angles: angle BDE and angle A are equal</note>.

Definition of Corresponding Angles

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

<note>Two parallel lines DE and AC</note> are intersected by a <note>line AB</note>, where <note>angle BDE and angle A are on the same side of the intersecting line AB</note>, on the <note>same side of the intersected lines DE and AC</note>. Therefore, <note>angle BDE and angle A are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle BDE is equal to angle A</note>.


Question: As shown in the figure, AP and BP are tangents to ⊙O at points A and B respectively, ∠P=60°, point C is on the major arc AB, then the measure of ∠C is ()

Answer: 60°

1. Given that AP and BP are tangent to ⊙O at points A and B respectively, according to the ##tangent properties of a circle##, we get ∠OAP=90°, ∠OBP=90°.

2. According to the interior angle sum theorem of quadrilateral APBO (the sum of the interior angles of a quadrilateral is 360°), we get ∠AOB=360°-∠P-∠OAP-∠OBP.

3. Substituting the given conditions ∠P=60°, ∠OAP=90°, ∠OBP=90°, we calculate ∠AOB=120°.

4. According to the inscribed angle theorem (the inscribed angle is half of the central angle), we get ∠ACB=0.5 * ∠AOB.

5. Substituting the value ∠AOB=120°, we calculate ∠ACB=60°.

6. Through the above reasoning, we finally get the answer as 60°.

圆; 切线; 圆周角; 等腰三角形; 弧

Definition of Tangent to a Circle

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

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

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

In the figure of this problem, <note>quadrilateral APBO</note> has four interior angles <note>∠P, ∠OAP, ∠OBP, and ∠AOB</note>. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>∠P + ∠OAP + ∠OBP + ∠AOB = 360°</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 circle O, points A, B, and C are on the circle</note>, <note>the central angle corresponding to arc AB and arc ACB is ∠AOB</note>, <note>and the inscribed angle is ∠ACB</note>. According to the Inscribed Angle Theorem, <note>∠ACB is equal to half of the central angle ∠AOB corresponding to arc AB</note>, that is, <note>∠ACB = 1/2 ∠AOB</note>.

Property of the Tangent Line to a Circle

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

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

Definition of Central Angle

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

In <note>circle O</note>, <note>point A and point B are two points on the circle</note>, and <note>the center of the circle is point O</note>. The angle formed by the lines <note>OA</note> and <note>OB</note> is called the <note>central angle ∠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 <note>circle O</note>, the <note>vertex of angle ACB (point C)</note> is on the <note>circumference</note>, and the <note>two sides of angle ACB intersect circle O at points A and B</note>. Therefore, <note>angle ACB is an inscribed angle</note>.


Question: As shown in the figure, it is known that point M is the midpoint of side AB of parallelogram ABCD, segment CM intersects BD at point E, and S△BEM=2. Then the area of the shaded part in the figure is ()

Answer: 8

1. Given that quadrilateral ABCD is a parallelogram, according to the properties of a parallelogram, we get AB = CD and AB ∥ CD.

2. According to the formula for the area of a triangle, we get the area of △DMB equals the area of △CBM, i.e., S△DMB = S△CBM.

3. Because segment CM intersects BD at point E, and point M is the midpoint of AB, we can deduce that the area of △DEM equals the area of △CEB, i.e., S△DEM = S△CEB.

4. Given that point M is the midpoint of side AB of parallelogram ABCD, we get AM = BM = 1/2 AB = 1/2 CD.

5. Since AB ∥ CD and AM = BM, △CDE and △MEB have a similarity relationship. According to the properties of similar triangles, we get CE: EM = CD: BM = 2.

6. Through the proportional relationship, we get the area of △CEB is twice the area of △BEM, i.e., S△CEB = 2S△BME.

7. Given that S△BME = 2, from the previous step we get S△CEB = 4.

8. The shaded area in the figure is the area of two △CEB, i.e., S_shaded = 4 + 4 = 8.

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

平行四边形; 中点; 线段

Definition of Parallelogram

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

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

Area Formula of a Triangle

<note>In the figure of this problem, in triangle DMB, side DM is the base, and segment BM is the height.</note> According to the area formula of a triangle, the area of triangle DMB is equal to <note>the base BM multiplied by the distance from point D to AB (which is the height of triangle DMB) multiplied by 1/2</note>.

Midpoint of a Line Segment

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

<note>In the figure of this problem, the midpoint of line segment AB is 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, the lengths of line segments AM and MB are equal. That is, <note>AM = MB</note>.

Properties of Parallelogram Theorem

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

In the parallelogram <note>ABCD</note>, the opposite angles <note>∠DAB</note> and <note>∠BCD</note> are equal, and the opposite angles <note>∠ABC</note> and <note>∠CDA</note> are equal; the sides <note>AB and CD</note> are equal, and the sides <note>AD and BC</note> are equal; the diagonals <note>AC and BD bisect each other, that is, the intersection point E divides the diagonal AC into two equal segments AE and EC, and divides the diagonal BD into two equal segments BE and ED.</note>

Definition of Similar Triangles

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

<note>Triangle CDE</note> and <note>Triangle MEB</note> are similar triangles. According to the definition of similar triangles: <note>∠CED = ∠MEB, ∠ECD = ∠EMB, ∠CDE = ∠MBE; CE/ME = ED/EB = CD/MB</note>.

Triangles with Equal Area

Two triangles that share a common base and lie between the same two parallel lines have equal areas.

In the diagram of this problem, <note>S△DMB</note> = <note>S△CBM</note>, because they share the base <note>BM</note> and are between the same parallel lines <note>(DC)</note>.


Question: As shown in the figure, in △ABC, AB=AC, ∠BA0=45°, △ABC is inscribed in ⊙O, D is a point on ⊙O, a tangent to ⊙O is drawn through point D intersecting the extension of BC at E, if DE⊥BC, AD=2√{2}, then the length of DE is ()

Answer: √{2}

1. Given AB=AC, ∠BAO=45°. Since AB=AC, it can be concluded that AH=CH, and AH passes through the center O.

2. Draw a tangent to ⊙O through point D, intersecting the extension of BC at point E, given that DE is perpendicular to BC.

3. Connect OC and OD. Since DE is a tangent to ⊙O, it can be concluded that OD is perpendicular to DE.

4. Considering the perpendicular relationship between OD and DE, and also considering that DE is perpendicular to BC, the quadrilateral ODEH is a rectangle, where ∠OHE=∠E=∠EDO=90°.

5. Since the quadrilateral ODEH is a rectangle, ∠AOD=90°.

6. Given AD=2√2, according to the properties of the diagonals of a rectangle, OA=OD=2.

7. Given ∠BAC=45°, since AB=AC, △ABC is an isosceles triangle, and ∠COH=45°.

8. Since O is the center of circle O, OC=radius OA=2.

9. From OC=2 and ∠COH=45°, OH=CH=√2 can be calculated.

等腰三角形; 圆; 切线; 垂线; 圆周角

Definition of Isosceles Triangle

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

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

Definition of Tangent to a Circle

The original text: <note>Circle O and line DE have only one common point D</note>, this common point is called the <note>point of tangency</note>. Therefore, <note>line DE is the tangent to circle O</note>.

Definition of Rectangle

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

In the figure of this problem, the quadrilateral ODEH is a rectangle, <note>with its interior angles ∠OHE, ∠EDO, ∠ODE, ∠HOE all being right angles (90 degrees)</note>, and <note>sides OH and DE are parallel and equal in length</note>, <note>sides OD and HE are parallel and equal in length</note>.

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

Property of Diagonals in a Rectangle

In a rectangle, the diagonals are equal in length and bisect each other.

<note>ODEH is a rectangle</note>, so the diagonals <note>OE and DH are equal and bisect each other</note>, meaning the intersection point of the diagonals <note>OE and DH is the midpoint of both diagonals</note>. Therefore, <note>segment OE is equal to segment DH</note>, <note>segment OD is equal to segment EH</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 ABC, sides AB and AC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, i.e., <note>angle ABC = angle ACB</note>.

Median Theorem of Right Triangle

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

<note>In the right triangle AOC, angle ∠AOC is a right angle (90 degrees)</note>, sides AO and OC are the legs, and side AC is the hypotenuse. <note>Point O is the midpoint of the hypotenuse AC</note>, the segment drawn from the right-angle vertex A to the midpoint O of the hypotenuse <note>segment AO is equal to half of the hypotenuse AC</note>, that is, <note>AO = 1/2 * AC</note>.

Pythagorean Theorem

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

<note>In the right triangle OHC</note>, <note>angle ∠OHC is a right angle (90 degrees)</note>, <note>sides OH and CH are the legs</note>, <note>side OC is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>OC² = OH² + CH²</note>.

Triangle Congruence Theorem (SSS)

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

In <note>right triangle ODE</note> and <note>right triangle OHE</note>, <note>OD=OE</note>, <note>OH=OE</note>, and <note>DE=HE</note>, therefore these two triangles are congruent.


Question: As shown in the figure, in △ABC, points D and E are on sides AB and AC respectively, DE ∥ BC, given EC=6, \frac{AD}{DB}=\frac{2}{3}, find the length of AE.

Answer: 4

1. Given DE∥BC, according to the ##Theorem of Proportional Segments in Parallel Lines##, we get: \frac{AD}{DB}=\frac{AE}{EC}.

2. According to the given conditions in the problem, EC=6 and \frac{AD}{DB}=\frac{2}{3}.

3. Substitute the given conditions into the proportional relationship \frac{AE}{6}=\frac{2}{3}.

4. Solve the equation \frac{AE}{6}=\frac{2}{3}, we get AE=4.

5. Through the above reasoning, we finally obtain that the length of AE is 4.

平行线; 普通三角形; 内错角

Proportional Segments Theorem

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

<note>In △ABC, line DE is parallel to side BC</note>, and it intersects the other two sides AB and AC at <note>points D and E</note>. Then, according to the <note>Proportional Segments Theorem</note>, we have: <note>\( \frac{AD}{DB} = \frac{AE}{EC} \)</note>.


Question: As shown in the figure, in △ABC, M is the midpoint of AC, E is a point on AB, AE=rac{1}{4}AB, connect EM and extend it to intersect the extension line of BC at D, then rac{BC}{CD}=()

Answer: 2

1. As shown in the figure, draw CP through point C parallel to AB, intersecting DE at point P.

2. Since CP∥AE, ##according to the parallel axiom 2 of parallel lines, alternate interior angles are equal, it can be known that ∠AEM=∠MPC, ∠EAM=∠MCP. At the same time, according to the similarity criterion theorem (AA), it can be known that## △AEM∽△CPM, according to ##the definition of similar triangles##, we have ##PC/AE=CM/AM##.

3. Since M is the midpoint of AC, and according to the ##midpoint of a line segment##, we get AM=CM, so PC=AE.

4. Given AE=##(1/4)AB##, therefore PC=##(1/4)AB##.

5. Since CP∥BE, so CP=##(1/3)BE##.

6. According to ##the parallel axiom 2 of parallel lines, corresponding angles are equal, it can be known that ∠EBD=∠PCD, ∠BED=∠CPD. At the same time, according to the similarity criterion theorem (AA), ##we get △DCP∽△DBE, so## according to the definition of similar triangles, it can be known that CP/BE=CD/BD##.

7. According to the previous conclusion, we get: ##CP/BE=1/3##, so BD=3CD.

8. Therefore BC=BD - CD = 3CD - CD = 2CD, that is ##BC/CD=2##.

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

三角形的外角; 中点; 普通三角形; 线段; 位似

Midpoint of a Line Segment

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

Definition of Similar Triangles

<note>In the figure of this problem, triangle AEM and triangle CPM are similar triangles</note>. According to the definition of similar triangles: <note>∠AEM = ∠CPM</note>, <note>∠AME = ∠CMP</note>, <note>∠EAM = ∠PCM</note>; <note>AE/PC = AM/CM = EM/PM</note>. Because <note>M is the midpoint of AC</note>, <note>AM = CM</note>, so <note>PC = AE</note>. According to <note>AE = 1/4 AB</note>, it follows that <note>PC = 1/4 AB</note>. Similarly, <note>triangle DCP and triangle DBE are similar triangles</note>. According to the definition of similar triangles: <note>∠EBD = ∠PCD</note>, <note>∠BED = ∠CPD</note>, <note>∠PDC = ∠EDB</note>; <note>PC/EB = DP/DE = DC/DB</note>.

Definition of Parallel Lines

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

The original text: <note>Line CP</note> and <note>Line AB</note> are in the same plane and do not intersect, so according to the definition of parallel lines, <note>Line CP and Line AB 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 diagram of this problem, two parallel lines <note>CP and AB</note> are intersected by a third line <note>EP</note>, forming the following geometric relationship: alternate interior angles: <note>angle AEM and angle MPC are equal</note>. Similarly, two parallel lines <note>CP and AB</note> are intersected by a third line <note>AC</note>, forming the following geometric relationship: alternate interior angles: <note>angle EAM and angle MCP are equal</note>. Similarly, two parallel lines <note>CP and AB</note> are intersected by a third line <note>BD</note>, forming the following geometric relationship: corresponding angles: <note>angle EBD and angle PCD are equal</note>. Similarly, two parallel lines <note>CP and AB</note> are intersected by a third line <note>ED</note>, forming the following geometric relationship: corresponding angles: <note>angle BED and angle CPD are equal</note>. These relationships illustrate that when two parallel lines are intersected by a third line, the alternate interior angles are equal, and the corresponding angles are equal.

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>two parallel lines CP and AB are intersected by a line AC</note>, where <note>angle EAM and angle MCP are located between the two parallel lines and on opposite sides of the intersecting line AC</note>, thus <note>angle EAM and angle MCP are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle EAM is equal to angle MCP</note>. Similarly, <note>two parallel lines CP and AB are intersected by a line EP</note>, where <note>angle AEM and angle MPC are located between the two parallel lines and on opposite sides of the intersecting line EP</note>, thus <note>angle AEM and angle MPC are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle AEM is equal to angle MPC</note>.

Similarity Theorem for Triangles (AA)

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

In the diagram of this problem, triangles <note>AEM</note> and <note>CPM</note>, if <note>angle AEM is equal to angle MPC</note>, and <note>angle EAM is equal to angle MCP</note>, then <note>triangle AEM is similar to triangle CPM.</note> Similarly, in triangles <note>DCP</note> and <note>DBE</note>, if <note>angle EBD is equal to angle PCD</note>, and <note>angle BED is equal to angle CPD</note>, then <note>triangle DCP is similar to triangle DBE.</note>

Definition of Corresponding Angles

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

<note>Two parallel lines CP and AB</note> are intersected by <note>a line BD</note>, where <note>angle EBD and angle PCD are on the same side of the intersecting line BD and on the same side of the intersected lines CP and AB</note>, thus <note>angle EBD and angle PCD are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle EBD is equal to angle PCD</note>. Similarly, <note>two parallel lines CP and AB are intersected by a line ED</note>, where <note>angle BED and angle CPD are on the same side of the intersecting line ED and on the same side of the intersected lines CP and AB</note>, thus <note>angle BED and angle CPD are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle BED is equal to angle CPD</note>.


Question: As shown in the figure, in △ABC, D and E are points on AB and AC respectively, satisfying AD=3, AE=2, EC=1, DE∥BC. Find AB=()

Answer: 4.5

1. Given DE∥BC, according to the ##theorem of proportional segments divided by parallel lines##, we get ##AD/DB = AE/EC##.

2. Since AD=3, AE=2, EC=1, ##therefore, 3/DB = 2/1##.

3. Solving the equation ##3/DB = 2##, we can find ##DB = 3/2 = 1.5##.

4. Since AB = AD + DB, we have AB = 3 + 1.5 = 4.5.

5. Through the above reasoning, the final answer is AB=4.5.

平行线; 普通三角形; 线段; 内错角; 等腰三角形

Proportional Segments Theorem

<note>In triangle ABC, line DE is parallel to side BC and intersects the other two sides AB and AC at points D and E</note>, according to the Proportional Segments Theorem, <note>we have: AD/DB = AE/EC, that is 3/DB = 2/1</note>, by solving the equation we get DB = 3/2=1.5, <note>so AB=AD+DB=3+1.5=4.5</note>.


Question: As shown in the figure, lines l_{1}∥l_{2}∥l_{3}, given: AB=4, BC=6, DE=3, find EF=()

Answer: 4.5

1. Given that the lines ## L1∥ L2 ∥ L3##, according to the theorem of proportional segments formed by parallel lines, let the length of EF be x, we get ##AB/BC = DE/EF##.

##2##. Substitute the given conditions AB = 4, BC = 6, DE = 3 into the proportion, we get ##4/6 = 3/x##.

##3##. By cross-multiplying both sides of the equation, we get 4x = 18.

##4##. Solving this equation, we get x = ##18/4 = 4.5##.

##5##. Therefore, we can conclude that the length of EF is ##4.5##.

<note>Line ## L1∥ L2 ∥ L3##</note>, and the segments between them are divided into proportional parts by parallel lines. According to the Proportional Segments Theorem, we have:
\[
\frac{<note>AB}{BC}</note> = \frac{<note>DE}{EF}</note>
\]
That is, the intercepted segments are proportional to the corresponding segments of the original triangle.


Question: As shown in the figure, in △ABC, AB=15, AC=12, BC=9, a moving circle passing through point C and tangent to side AB intersects CB and CA at points E and F respectively. Then the minimum length of segment EF is ()

Answer: \frac{36}{5}

1. Given AB=15, AC=12, BC=9, according to the ##converse of the Pythagorean theorem##, △ABC is a right triangle, ∠ACB=90°.

2. By constructing an auxiliary line, draw a circle passing through point C and tangent to side AB.

3. Let the tangent point of the circle with AB be D, ##also let the center of the circle be O, connect OD, OD is the radius##.

4. According to the ##properties of the tangent to the circle, OD is perpendicular to AB, ∠ODA=90°, connect CO##.

5. When CD is the diameter of the circle, the length of EF is minimized ##, so points C, O, D are on the same line, thus ∠CDA=90°##.

6. ##According to the area of the right triangle, the area of △ABC=1/2*AC*BC=1/2*12*9=54. Additionally, there is another way to calculate the area of the triangle. According to the triangle area formula, since CD is perpendicular to AB, CD is also the height of △ABC, therefore the area of △ABC=1/2*AB*CD=1/2*15*CD. Since it is the same triangle, 1/2*15*CD=54, finally obtaining CD=EF=36/5##.

7. Through the above reasoning, the final answer is ##36/5##.

切线; 圆; 线段; 普通三角形; 弦

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

Definition of Tangent to a Circle

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

The original text: <note>The circle and the line AB have only one common point D</note>, this common point is called the point of tangency. Therefore, <note>the line AB is the tangent to the circle</note>.

Converse of the Pythagorean Theorem

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

<note>The three sides of triangle △ABC are AB, AC, and BC</note>, and satisfy <note>AB² = AC² + BC²</note>. According to the converse of the Pythagorean Theorem, <note>△ABC is a right triangle</note>, and the angle opposite the longest side <note>AB, angle ACB, is a right angle</note>.

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

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

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

Definition of 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>CD is the diameter</note>, connecting <note>the center O</note> and <note>points C and D on the circumference</note>, with a length equal to twice the radius, i.e., <note>CD = 2OD</note>.

Area of Right Triangle

The area of a right triangle is equal to half the product of the two legs that form the right angle, i.e., Area = 1/2 * base * height.

<note>In the right triangle ABC, angle ACB is a right angle (90 degrees)</note>, <note>sides AC and BC are the legs of the right angle</note>, one leg is used as the base, and the other leg is used as the height, so the area of the right triangle is equal to half the product of these two legs, i.e., area = 1/2 * side AC * side BC.

Area Formula of a Triangle

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

In <note>triangle ABC</note>, <note>side AC is the base</note>, and <note>segment CD is the height</note>. According to the area formula of a triangle, the area of triangle ABC is equal to <note>base AC</note> multiplied by <note>height CD</note> and then divided by 2, i.e., area = <note>(AC * CD) / 2</note>.


Question: As shown in the figure, it is known that AB∥CD∥EF, AD:AF=3:5, BE=15, then the length of CE is equal to ()

Answer: 6

1. Given AB∥CD∥EF, according to the theorem of proportional segments intercepted by parallel lines, which states that the corresponding segments intercepted by two parallel lines on a third line are proportional, we can derive the proportional relationship between BC and BE.

2. According to the problem condition AD:AF=3:5, combining this ratio with the above ratio, we can obtain \frac{BC}{BE} = \frac{AD}{AF}.

3. Substituting the known data, we get \frac{BC}{15} = \frac{3}{5}.

4. By solving the proportion equation \frac{BC}{15} = \frac{3}{5}, we get BC=9.

5. Since CE=BE-BC, we get CE = 15 - 9 = 6.

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

平行线; 内错角; 同位角; 线段; 位似

Proportional Segments Theorem

Original text: <note>AB∥CD∥EF</note>, and <note>AD:AF=3:5</note>. According to the Proportional Segments Theorem, we can deduce that <note>the ratio of BC to BE is 3:5</note>, which means <note>\(\frac{BC}{BE} = \frac{AD}{AF}\)</note>. Specifically, <note>line AB is parallel to EF</note> and intersects the other two sides <note>AD and AF at points B and E</note>. Therefore, according to the Proportional Segments Theorem, we have <note>\(\frac{BC}{BE} = \frac{AD}{AF}\)</note>.


Question: As shown in the figure, in the parallelogram ABCD, point E is on side AD, CE intersects BD at point F. If EF = \frac{1}{3}FC, then \frac{AE}{ED} = ()

Answer: 2

1. Given in parallelogram ABCD, point E is on side AD, CE intersects BD at point F, and EF = \frac{1}{3}FC.

2. In parallelogram ABCD, AD∥BC, AD=BC. Therefore, according to the parallel axiom 2 and the similarity theorem of triangles (AA), we can obtain △DEF∽△BCF.

3. According to the definition of similar triangles, we get \frac{DE}{BC} = \frac{EF}{FC} = \frac{1}{3}.

4. Since AD = BC, it can be written as \frac{DE}{AD} = \frac{1}{3}.

5. Thus, we get DE:AD = 1:3.

6. Since AD = AE + ED, it can be written as DE:(AE+ED) = 1:3.

7. Solving this, we get \frac{AE}{ED} = 2.

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

平行四边形; 线段; 中点; 普通三角形; 位似

Definition of Parallelogram

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

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

Definition of Similar Triangles

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

In the figure of this problem, <note>triangle DEF and triangle BCF are similar triangles</note>. According to the <note>definition of similar triangles</note>, we have: <note>∠EDF = ∠CBF</note>, <note>∠DFE = ∠BFC</note>, <note>∠DEF = ∠BCF</note>; <note>DE/BC = EF/FC = DF/BF</note>.

Similarity Theorem for Triangles (AA)

<note>Angle FED is equal to angle FCB</note>, and <note>angle EDF is equal to angle FBC</note>, so <note>triangle FED is similar to triangle BFC</note>.

Parallel Postulate 2 of Parallel Lines

<note>Two parallel lines AD and BC</note> are intersected by <note>the third line EC</note>, forming the following geometric relationship: <note>alternate interior angles: angle DEF = angle FCB</note>. <note>Two parallel lines AD and BC</note> are intersected by <note>the third line BD</note>, forming the following geometric relationship: <note>alternate interior angles: angle EDF = angle DBC</note>.

Definition of Alternate Interior Angles

In the diagram of this problem, two <note>parallel lines AD and BC</note> are intersected by <note>two lines EC and BD</note>, where <note>angle EDF and angle DBC are located between the two parallel lines and on opposite sides of the intersecting line BD, thus angle EDF and angle DBC are alternate interior angles</note>. Alternate interior angles are equal, i.e., <note>angle EDF = angle DBC</note>. <note>Angle DEF and angle FCB are located between the two parallel lines and on opposite sides of the intersecting line EC, thus angle DEF and angle FCB are alternate interior angles</note>. Alternate interior angles are equal, i.e., <note>angle DEF = angle FCB</note>.


Question: As shown in the figure, ⊙O is the circumcircle of △ABC, AD is the diameter of ⊙O, and EA is the tangent to ⊙O. If ∠EAC=120°, then the measure of ∠ABC is ()

Answer: 60°

1. Given EA is the tangent line of ⊙O, #OA is the radius of ⊙O. According to the property of the tangent line to a circle, the tangent line is perpendicular to the radius at the point of tangency##, we know ∠EAD=90°.

2. From ∠EAC=120° and ∠EAD=90°, we can deduce ∠DAC=∠EAC-∠EAD=30°.

3. Since AD is the diameter of ⊙O, according to ##(Corollary 2 of the Inscribed Angle Theorem) the inscribed angle subtended by the diameter is a right angle##, we can conclude ∠ACD=90°.

4. ##According to the Triangle Sum Theorem, we know## ∠ADC=180°-∠ACD-∠DAC, ####we can get ∠ADC=60°.

5. From ∠ADC=60°, according to ##Corollary 1 of the Inscribed Angle Theorem##, ∠ABC is also equal to ∠ADC=60°.

##6##. Through the above reasoning, the final answer is 60°.

圆周角; 切线; 圆; 直角三角形; 圆内接四边形

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

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

Definition of Inscribed Angle

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

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

Definition of Diameter

<note>Line segment AD is the diameter</note>, connecting <note>the center O of the circle</note> and <note>points A and D on the circumference</note>, with a length equal to 2 times the radius, i.e., <note>AD = 2r</note>.

Definition of Tangent to a Circle

The original text: <note>圆⊙O与直线EA有且只有一个公共点A</note>, this common point is called the point of tangency. Therefore, <note>直线EA是圆⊙O的切线</note>.

<note>Point A is the tangent point of line EA and the circle</note>, <note>segment OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent EA is perpendicular to the radius OA at the tangent point A</note>, that is, <note>∠EAD = 90 degrees</note>.

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

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

In circle O, the inscribed angle subtended by the diameter AD is a right angle (90 degrees).

Triangle Angle Sum Theorem

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

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

Definition of Radius

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


Question: As shown in the figure, △ABC∽△DEF, the similarity ratio is 1:2. If EF=2, then the length of BC is ()

Answer: 1

1. Given △ABC is similar to △DEF, the similarity ratio is 1:2.

2. According to the properties of similar triangles, the similarity ratio is equal to the ratio of corresponding sides, i.e., \\(rac{BC}{EF} = rac{1}{2}\\).

3. Given EF = 2.

4. From the previous steps, we get \\(rac{BC}{EF} = rac{1}{2} 	imes EF = rac{1}{2} 	imes 2 = 1\\).

5. Through the above reasoning, the final answer is 1.

普通三角形; 线段; 位似

Definition of Similar Triangles

<note>△ABC and △DEF are similar triangles</note>. According to the definition of similar triangles: <note>∠A = ∠D, ∠B = ∠E, ∠C = ∠F</note>; <note> $\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = \frac{1}{2}$ </note>.


Question: As shown in the figure, △ABC∽△AED, ∠ADE=80°, ∠A=60°, then ∠B equals ()

Answer: 40°

1. Given ∠ADE=80°, ∠A=60°, according to the triangle angle sum theorem, we get ∠ADE+∠A+∠AED=180°.

2. From ∠ADE=80° and ∠A=60°, based on the triangle angle sum theorem, we get ∠AED=180°-∠ADE-∠A=180°-80°-60°=40°.

3. Also, since △ABC∽△AED, according to the corresponding angles of similar triangles theorem, we get ∠B=∠AED.

4. Based on the previous conclusion, we get ∠B=40°.

5. Through the above reasoning, we finally get the answer ∠B=40°.

等腰三角形; 内错角; 普通三角形

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

Triangle Angle Sum Theorem

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

Definition of Similar Triangles

<note>Triangles ABC and AED are similar triangles</note>. According to the definition of similar triangles: <note>∠BAC = ∠DAE</note>, ∠ABC = <note>∠AED</note>, ∠BCA = <note>∠EDA</note>; AB/<note>AE</note> = BC/DE = AC/<note>AD</note>.


Question: As shown in the figure, E is a point on side AD of parallelogram ABCD. Draw EF ∥ AB through point E, intersecting BD at F. If DE:EA=2:3 and EF=4, find the length of CD.

Answer: 10

1. Given DE:EA=2:3, ##therefore## DE:DA=2:5.

2. Since EF is parallel to AB and intersects line BD at F, according to ##parallel lines parallel axiom 2, corresponding angles are equal, it can be known## that ∠DEF=∠DAB, ∠EFD=∠ABD, ##based on the similarity triangle determination theorem (AA), so triangle DEF is similar to triangle DAB,## we have DE/DA=EF/AB.

3. Substitute the given conditions, EF=4, and DE:DA=2:5, i.e., DE/DA=2/5.

4. From the above similarity triangle ratio relationship 2/5=4/AB, solving this ratio gives AB=10.

5. Due to the property of equal opposite sides of a parallelogram, the opposite sides AB and CD of parallelogram ABCD are equal, therefore CD=AB.

6. Finally, the length of CD is 10.

平行四边形; 平行线; 线段

<note>Triangle DEF and triangle DAB are similar triangles</note>. According to the <note>definition of similar triangles</note>, we have: <note>∠DEF=∠DAB</note>, <note>∠EFD=∠ABD</note>, <note>∠EDF=∠ADB</note>; <note>DE/DA=EF/AB=DF/DB</note>.

Similarity Theorem for Triangles (AA)

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

Given that <note>angle DEF is equal to angle DAB</note> and <note>angle DFE is equal to angle DBA</note>, therefore <note>triangle DEF is similar to triangle DAB</note>.

Properties of Parallelogram Theorem

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

In the figure of this problem, <note>in parallelogram ABCD</note>, <note>the opposite angles ∠DAB and ∠BCD are equal</note>, <note>the opposite angles ∠ABC and ∠CDA are equal</note>; <note>sides AB and CD are equal</note>, <note>sides AD and BC are equal</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 diagram of this problem, <note>the two parallel lines EF and AB</note> are intersected by <note>the third line DA</note>, forming <note>corresponding angles ∠DEF=∠DAB</note>. Similarly, <note>EF and AB are intersected by the third line DB</note>, forming <note>corresponding angles ∠EFD=∠ABD</note>.

Definition of Corresponding Angles

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

In the diagram of this problem, two parallel lines <note>EF and AB</note> are intersected by a line <note>AD</note>, where <note>angle DEF and angle DAB are on the same side of the intersecting line AD and on the same side of the intersected lines EF and AB</note>, thus <note>angle DEF and angle DAB are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle DEF is equal to angle DAB</note>. Similarly, the lines <note>EF and AB</note> are intersected by a line <note>DB</note>, where <note>angle DFE and angle DBA are on the same side of the intersecting line DB and on the same side of the intersected lines EF and AB</note>, thus <note>angle DFE and angle DBA are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle DFE is equal to angle DBA</note>.


Question: As shown in the figure, given ∠1 + ∠2 = 100°, then ∠3 = ( ).

Answer: 130°

1. Given ∠1 + ∠2 = 100°.

2. By the definition of vertical angles, ∠1 = ∠2, so the equation from the previous step can be rewritten as 2 ∠1 = 100°.

3. Therefore, ∠1 = 50°.

4. According to the definition of adjacent supplementary angles, we can write the equation ∠1 + ∠3 = 180°.

5. Substitute the value of ∠1 into the equation to get 50° + ∠3 = 180°.

6. Rearrange the equation from the previous step to get ∠3 = 180° - 50° = 130°.

7. Through the above reasoning, the final answer is ∠3 = 130°.

Definition of Linear Pair of Angles

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

In the diagram of this problem, ∠1 and ∠3 have a common side, and their other sides are extensions in opposite directions, so ∠1 and ∠3 are adjacent supplementary angles.

Definition of Vertical Angles

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

<note>Two intersecting lines intersect at the center</note>, forming four angles. According to the definition of vertical angles, <note>angle 1 and angle 2 are vertical angles</note>. Since vertical angles are equal, <note>angle 1 = angle 2</note>.


Question: As shown in the figure, AB is the diameter of ⊙O, PA is tangent to ⊙O at point A, segment PO intersects ⊙O at point C, connect BC, if ∠P = 36°, then ∠B equals ()

Answer: 27°

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

 2. From ∠P=36° and the conclusion in step 1 ∠OAP=90°, in △AOP, according to the sum of the angles in a triangle theorem, we get ∠AOP=180°-∠OAP-∠P=180°-90°- 36° = 54°.

####

3. In circle O, according to the inscribed angle theorem, the inscribed angle ∠ABC corresponding to the minor arc AC is half of the central angle ∠AOC, which means ∠AOC= 2∠ABC.

4. According to the conclusion in step 3 and the calculation in step 2 ∠AOC=∠AOP=54°, we can get ∠ABC=1/2 ∠AOC.

5. Substituting the angle calculation result into the formula, we can get ∠B = 54°/2 = 27°.

6. Through the above reasoning, the final answer is 27°.

圆; 切线; 圆周角; 直角三角形

Definition of Diameter

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

Definition of Chord

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

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

Definition of Tangent to a Circle

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

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

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

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

Triangle Angle Sum Theorem

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

Inscribed Angle Theorem

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

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

Definition of Central Angle

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

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

Definition of Inscribed Angle

In the problem diagram, in circle O, the vertex of angle ABC (point B) is on the circumference, and the two sides of angle ABC intersect circle O at points A and C respectively. Therefore, angle ABC is an inscribed angle.


Question: As shown in the figure, in Rt△ABC, ∠C=90°, BC=3, AC=4, if △ABC∽△BDC, then CD=()

Answer: \frac{9}{4}

1. Given Rt△ABC, ∠C=90°, BC=3, AC=4, according to ##Pythagorean theorem##, we can find the length of AB. By the Pythagorean theorem, AB = √(##AC^2## + ##BC^2##) = √(##4^2## + ##3^2##) = √(16 + 9) = √25 = 5.

2. Also, according to the problem, △ABC ∽ △BDC, based on the ##definition of similar triangles##, the corresponding sides of similar triangles are proportional. Therefore, ##BC/CD=AC/BC##.

3. Substitute the known lengths into the proportion ##BC/CD=AC/BC##, i.e., 3/CD=4/3.

4. Solve for CD by cross-multiplying, obtaining CD = ##(3×3)/4 = 9/4##.

5. Through the above reasoning, the final answer is CD = 9/4.

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

Definition of Similar Triangles

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

In the figure of this problem, <note>triangle ABC and triangle BDC are similar triangles</note>. According to the definition of similar triangles: <note>∠BAC = ∠DBC</note>, ∠ABC = ∠BDC, ∠ACB = ∠DCB; <note>BC/CD = AC/BC = AB/BD</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 figure of this problem, in the right triangle ABC, angle ∠ACB is a right angle (90 degrees)</note>, <note>sides AC and BC are the legs</note>, <note>side AB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AB^2 = AC^2 + BC^2</note>.


Question: As shown in the figure, C and M are two points on the line segment AB, and point M is the midpoint of the line segment AC. If AB = 8cm and BC = 2cm, then the length of AM is ()

Answer: 3cm

1. Given AB=8cm, BC=2cm.

2. From the figure, it can be seen that AC=AB-BC=8cm-2cm=6cm.

3. Since M is the midpoint of segment AC, according to the definition of midpoint, AM is equal to half of AC.

4. Because AC=6cm, therefore AM=6cm/2=3cm.

5. Through the above reasoning, the final answer is 3cm.

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


Question: As shown in the figure, AD∥BE∥CF, lines l1 and l2 intersect these three parallel lines at points A, B, C, D, E, F respectively, rac{AB}{BC}=rac{2}{3}, DE=6, then the value of EF is ()

Answer: 9

1. Given AD∥BE∥CF, lines l1 and l2 intersect these three parallel lines at points A, B, C, D, E, F respectively.

2. According to the theorem of proportional segments formed by parallel lines, we can obtain: rac{AB}{BC} = rac{DE}{EF}.

3. It is known from the problem that rac{AB}{BC} = rac{2}{3} and DE=6.

4. Substitute the known conditions into the proportion: rac{6}{EF} = rac{2}{3}.

5. Solve this proportion to get: EF = 9.

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

平行线; 线段; 内错角; 位似

Definition of Parallel Lines

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

The original text: <note>AD∥BE∥CF are in the same plane</note>, and they do not intersect, so according to the definition of parallel lines, <note>AD, BE, and CF are parallel lines</note>.

Proportional Segments Theorem

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

In the figure of this problem, <note>AD∥BE∥CF</note>, <note>lines l1, l2 intersect at points A, B, C, D, E, F respectively</note>. According to the Proportional Segments Theorem, we can obtain <note>\(\frac{AB}{BC}=\frac{DE}{EF}\)</note>.


Question: As shown in the figure, it is known that AD is the median of △ABC, and the perimeter of △ABD is 3cm larger than the perimeter of △ACD. Then the difference between AB and AC is ()

Answer: 3cm

1. Given that AD is the median of △ABC, according to the ##definition of the median of a triangle##, we have BD=DC.

2. Since BD=DC, ##according to the theorem of the perimeter of a triangle##, the difference in the perimeter of △ABD and △ACD is (AB+AD+BD)-(AC+AD+DC)=AB-AC.

3. According to the given condition, the perimeter of △ABD is 3cm greater than the perimeter of △ACD, so we get AB-AC=3.

4. Through the above reasoning, we finally conclude that the difference between AB and AC is 3.

普通三角形; 中点; 线段

Definition of Median of a Triangle

A median of a triangle is a line segment drawn from one vertex of the triangle to the midpoint of the opposite side.

<note>Vertex A is a vertex of the triangle</note>, <note>the opposite side BC is the side opposite to vertex A</note>. <note>Point D is the midpoint of side BC</note>, <note>line segment AD is the segment from vertex A to midpoint D of side BC</note>, therefore <note>AD is a median of triangle ABC</note>.

Perimeter Theorem of Triangle

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

<note>The three sides of triangle △ABD are AB, AD, and BD</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., the perimeter P = <note>AB</note> + <note>AD</note> + <note>BD</note>. In <note>the three sides of triangle △ADC are AC, AD, and DC</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., the perimeter P = <note>AC</note> + <note>AD</note> + <note>DC</note>.


Question: As shown in the figure, AB is the tangent to ⊙O at point B, AO intersects ⊙O at point C. If ∠BAO=40°, then the degree of ∠OCB is ()

Answer: 65°

1. Given that AB is the tangent to ⊙O and B is the point of tangency, according to the ##property of the tangent to a circle##, OB is perpendicular to AB, i.e., ∠OBA=90°.

2. Given ∠BAO=40°, and ∠OBA=90°, according to the triangle angle sum theorem, we get ##∠BOA##=50°.

3. ##According to the definition of radius and the definition of isosceles triangle, we get## OB=OC, both are radii of ⊙O, and triangle OBC is an isosceles triangle.

####

##4.## According to the property of isosceles triangles, we get ∠OCB=∠OBC, #according to the triangle angle sum theorem, we know# 2∠OCB= ##180°-∠BOA##

##5.## Calculating, we get ∠OCB=1/2 ×130°=65°.

##6.## Through the above reasoning, we finally get the answer ∠OCB=65°.

圆; 切线; 直角三角形; 圆心角; 圆周角

Definition of Tangent to a Circle

The circle <note>O</note> and the line <note>AB</note> have exactly one common point <note>B</note>, which is called the point of tangency. Therefore, the line <note>AB</note> is the tangent to the circle <note>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>Point O is the center of the circle</note>, point B and point C are any points on the circle, <note>line segment OB and line segment OC are segments from the center of the circle to any point on the circle</note>, therefore <note>line segment OB and line segment OC are the radii of circle O</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 OBC, side OB and side OC are equal</note>, therefore triangle OBC is an isosceles triangle.

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

Triangle Angle Sum Theorem

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

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

Properties of Isosceles Triangle

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

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


Question: As shown in the figure, D and E are points on sides AB and AC of △ABC, and DE∥BC. If DE:BC=3:5 and AD=6, then AB=()

Answer: 10

1. Given that D and E are points on sides AB and AC of △ABC, and DE∥BC, ##according to the parallel axiom 2 of parallel lines, corresponding angles are equal, thus ∠C=∠AED. Since ∠A is common, △ADE∽△ACB##, we have: DE/BC=AD/AB.

2. The problem states DE:BC=3:5, substituting into the above equation, we get: 3/5 = AD/AB.

3. The problem also states AD=6, based on the above proportion, we get: 3/5 = 6/AB.

4. By cross-multiplying, we obtain: 3 * AB = 5 * 6, i.e., 3AB = 30.

5. Solving the equation, we get: AB = 30 / 3, i.e., AB = 10.

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

平行线; 普通三角形; 位似; 线段

Parallel Postulate 2 of Parallel Lines

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

<note>Two parallel lines DE and BC are intersected by a third line AC</note>, forming the following geometric relationship: <note>corresponding angles: angle C and angle AED are equal</note>

Similarity Theorem for Triangles (AA)

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

In the diagram of this problem, triangles ADE and ABC, <note>∠C=∠AED</note> and <note>∠A is shared</note>, so triangles ADE and ABC are similar.

Definition of Similar Triangles

<note>Triangles ADE and ABC are similar triangles</note>. According to the definition of similar triangles: <note>angle A = angle A</note>, <note>angle B = angle ADE</note>, <note>angle C = angle AED</note>; <note>AD/AB = AE/AC = DE/BC</note>.

Definition of Corresponding Angles

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

<note>Two parallel lines DE and BC</note> are intersected by <note>a line AC</note>, where <note>angle C and angle AED are on the same side of the intersecting line AC and on the same side of the intersected lines DE and BC</note>, therefore, <note>angle C and angle AED are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle C is equal to angle AED</note>.


Question: As shown in the figure, BC is tangent to ⊙O at point C, the extension of BO intersects ⊙O at point A, connect AC, if ∠ACB=120°, then the degree of ∠A equals ()

Answer: 30°

1. Connect OC.

2. Since point A is on the extension of BO, OA=OC, according to the properties of an isosceles triangle, ∠OAC=∠ACO.

3. Since BC is tangent to ⊙O at point C, OC is the radius, so OC is perpendicular to the tangent BC, i.e., ∠OCB=90°.

4. Given ∠ACB=120°, so ∠ACO=∠ACB-∠OCB=120°-90°=30°.

5. Since ∠OAC=∠ACO, ∠A=∠ACO=30°.

6. Through the above reasoning, the final answer is 30°.

圆; 切线; 圆周角; 三角形的外角; 同旁内角

Definition of Tangent to a Circle

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

The circle <note>⊙O</note> and the line <note>BC</note> have only one common point <note>C</note>, which is called the point of tangency. Therefore, the line <note>BC</note> is the tangent to the circle <note>⊙O</note>.

Definition of Radius

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

Definition of Isosceles Triangle

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

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

In <note>circle ⊙O</note>, point <note>C is the tangent point of line BC with the circle</note>, and <note>line segment OC is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>tangent line BC is perpendicular to the radius OC passing through the tangent point C</note>, i.e., <note>∠OCB=90°</note>.

Properties of Isosceles Triangle

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


Question: In the figure above, what is the value of y?

Answer: 60

1. Given that 4x degrees and 2x degrees are adjacent supplementary angles, we get 180 degrees = 4x + 2x.

2. Solving the equation, we get x = 30 degrees.

3. Since y degrees and 4x degrees are adjacent supplementary angles, we have y + 4x = 180 degrees. Substituting x = 30 degrees, we find the value of y.

5. Through the above reasoning, the final answer is answer = 60.

Definition of Linear Pair of Angles

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

In the figure of this problem, <note>angle 4x and angle 2x share a common side, and their other sides are extensions in opposite directions</note>, so <note>angle 4x and angle 2x are supplementary adjacent angles</note>. Similarly, <note>angle 4x and angle y share a common side, and their other sides are extensions in opposite directions</note>, so <note>angle 4x and angle y are supplementary adjacent angles</note>.


Question: As shown in the figure, BC=rac{1}{2}AB, D is the midpoint of AC, if DC=3, then the length of AB is ()

Answer: 4

1. Given that D is the midpoint of AC, according to the definition of the midpoint of a line segment, the length of AC is twice that of DC, i.e., AC=2DC.

2. According to the given DC=3, substituting into the relationship derived in the previous step AC=2DC, we get AC=2×3=6.

3. Given that BC is half of AB, according to the problem statement BC=1/2 AB, we can deduce that the length of AB is 2/3 of AC.

4. Expressing the above relationship with an equation, we get AB=2/3 AC.

5. Substituting the previously calculated AC=6, we get AB=2/3×6=4.

6. Through the above reasoning, the final answer is 4.

Midpoint of a Line Segment

In the figure of 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>. Therefore, <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.

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


Question: As shown in the figure, in △ABC, DE∥BC, if rac{AD}{AB}=rac{1}{3}, then rac{AE}{AC}=()

Answer: \frac{1}{3}

1. Given in △ABC, according to the problem statement, DE∥BC.

2. According to the theorem of parallel lines dividing segments proportionally, we can deduce: rac{AD}{AB} = rac{AE}{AC}.

3. The problem states rac{AD}{AB} = rac{1}{3}, substituting into the conclusion ∴ rac{AE}{AC} = rac{1}{3}.

4. Therefore, through the above reasoning, the final answer is rac{1}{3}.

普通三角形; 平行线; 位似

Proportional Segments Theorem

In the figure of this problem, in triangle ABC, <note>line DE is parallel to side BC</note>, and <note>it intersects the other two sides AB and AC at points D and E</note>, then according to the Proportional Segments Theorem, we have: <note>AD/AB = AE/EC</note>, that is, the intercepted segments are proportional to the corresponding segments of the original triangle.


Question: As shown in the figure, in △ABC, points D and E are on sides AB and AC respectively, DE∥BC, and AE=1, AC=5, AB=6. Then the length of AD is ()

Answer: 1.2

1. Given that points D and E are on sides AB and AC respectively, and DE∥BC. ##Therefore, according to the parallel axiom 2 of parallel lines, we have ∠ADE=∠ABC (corresponding angles are equal), ∠AED=∠ACB (corresponding angles are equal).##

2. According to ##the theorem of similar triangles (AA)##, △ADE ∽ △ABC. ##According to the definition of similar triangles, \rac{AE}{AC} = \rac{AD}{AB}##.

####

##3##. Given the conditions AE=1, AC=5, AB=6, substitute into the proportion: \rac{1}{5} = \rac{AD}{6}.

##4##. Solve this proportion equation: \rac{AD}{6} = \rac{6}{5}.

##5##. Therefore, we obtain the length of AD as \rac{6}{5} or 1.2.

平行线; 普通三角形; 线段; 位似

Definition of Corresponding Angles

In the diagram of this problem, <note>two parallel lines DE and BC</note> are intersected by <note>a line AB</note>, where <note>angle ADE and angle ABC are on the same side of the intersecting line AB</note>, on the same side of the intersected <note>two lines DE and BC</note>. Therefore, <note>angle ADE and angle ABC are corresponding angles</note>. Corresponding angles are equal, i.e., <note>angle ADE is equal to angle ABC</note>. Similarly, <note>two parallel lines DE and BC</note> are intersected by <note>a line AC</note>, where <note>angle AED and angle ACB are on the same side of the intersecting line AC</note>, on the same side of the intersected <note>two lines DE and BC</note>. Therefore, <note>angle AED and angle ACB are corresponding angles</note>. Corresponding angles are equal, i.e., <note>angle AED is equal to angle ACB</note>.

Parallel Postulate 2 of Parallel Lines

<note>Two parallel lines DE and BC</note> are intersected by <note>the third line AB</note>, forming the following geometric relationship:
Corresponding angles: <note>angle ADE and angle ABC are equal</note>.
Similarly, <note>two parallel lines DE and BC</note> are intersected by <note>the third line AC</note>, forming the following geometric relationship:
1. Corresponding angles: <note>angle AED and angle ACB are equal</note>.
These relationships illustrate that when <note>two parallel lines are intersected by a third line</note>, the corresponding angles are equal.

Similarity Theorem for Triangles (AA)

<note>In the diagram of this problem, in triangles ADE and ABC, angle ADE is equal to angle ABC, and angle AED is equal to angle ACB</note>, so triangle ADE is similar to triangle ABC.

Definition of Similar Triangles

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

<note>Triangles ADE and ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠DAE = ∠BAC</note>, <note>∠ADE = ∠ABC</note>, <note>∠AED = ∠ACB</note>; <note>AE/AC = AD/AB = DE/BC</note>.


Question: As shown in the figure, AB is tangent to ⊙O at B, and the secant line ACD passes through the center O. If ∠BCD=70°, then the degree of ∠A is ()

Answer: 50°

1. Given AB is tangent to ⊙O at B, ACD is a secant line and passes through the center O, according to the given condition ∠BCD=70°.

2. Draw auxiliary lines OB and OC through the center O, ##according to the definition of radius,## the two segments OB and OC are equal radii, i.e., OB=OC.

3. ##Since OB=OC, according to the definition of an isosceles triangle, △OBC is an isosceles triangle.## Since ∠BCD=70°, ##according to the properties of an isosceles triangle##, we get ∠OBC=∠BCD=70°.

4. In △OBC, according to the triangle angle sum theorem, we get ∠BOC=180°-∠BCD-∠OBC = 180°-70°-70°=40°.

5. Since AB is tangent to ⊙O at B, according to ##the properties of a tangent to a circle##, we know OB⊥AB, i.e., ∠ABO=90°. ##According to the definition of a right triangle, we know △AOB is a right triangle.##

6. Finally, using ##the complementary property of acute angles in a right triangle##, in △AOB, we get ∠A=90°-∠BOC=90°-40°=50°.

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

圆; 切线; 圆周角; 线段; 三角形的外角

Definition of Tangent to a Circle

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

Secant Line

A straight line that intersects a circle at two distinct points is called a secant line of the circle.

The line <note>ACD</note> intersects the <note>circle O</note> at <note>two points</note>, namely <note>point C</note> and <note>point D</note>. According to the definition of a secant line, the line ACD intersects the circle O at two distinct points, so the <note>line ACD</note> is a <note>secant line of circle O</note>.

Definition of Radius

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

Definition of Isosceles Triangle

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

Triangle Angle Sum Theorem

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

Properties of Isosceles Triangle

<note>In the isosceles triangle OBC, sides OB and OC are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the angles opposite the equal sides are equal</note>, that is, <note>angle OBC = angle OCB = 70°</note>.

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

Complementary Property of Acute Angles in Right Triangle

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

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

Definition of Right Triangle

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

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


Question: Place a pair of right-angled triangles as shown in the figure, so that the right-angled side of the triangle containing the 30° angle and one right-angled side of the triangle containing the 45° angle are on the same line. Then the degree of ∠1 is ()

Answer: 75°

1. In the figure, since the right-angle side of the triangle board containing a 30° angle and one right-angle side of the triangle board containing a 45° angle are on the same line, #let the triangle board containing a 30° angle be triangle ABC, with the 30° angle corresponding to ∠A, the 90° angle corresponding to ∠B, and the 60° angle corresponding to ∠C; the triangle board containing a 45° angle be triangle DEF, with the 45° angle corresponding to ∠E, the 90° angle corresponding to ∠D, and the remaining angle corresponding to ∠F, the intersection point where ∠1 is located is O, thus the two triangles intersect to form a new triangle COF##.

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##2##. ##According to the definition of vertical angles, we can obtain ∠1=∠COF,## to solve ∠1, we can calculate the other two interior angles of the triangle where its vertical angle is located respectively.

##3##. For the triangle board containing a 30° angle, ##according to the triangle interior angle sum theorem, we can deduce## its angles are 30° and 60° respectively.

##4##. For the triangle board containing a 45° angle, ##according to the triangle interior angle sum theorem, we can deduce## its angles are 45° and 45° respectively.

##5##. ##The triangle where ∠COF## is located####, its two interior angles are 60° and 45° respectively.

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

##7##. Thus ##∠COF## can be expressed as 180° - (60° + 45°) = 75°.

##8##. Therefore, the angle of ∠1 is 75°.

直角三角形; 邻补角; 对顶角

Definition of Vertical Angles

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

<note>Two intersecting lines form four angles: ∠1, ##∠EOC##, ##∠COF##, and ##∠AOF##</note>. According to the definition of vertical angles, <note>∠1 and ##∠COF## are vertical angles</note>, <note>##∠EOC## and ##∠AOF## are vertical angles</note>. Since vertical angles are equal, <note>∠1=##∠COF##</note>, <note>##∠EOC##=##∠AOF##</note>.

Definition of Right Triangle

<note>Triangles with a 30° angle and triangles with a 45° angle are both right triangles</note>. The angles of the triangle with a 30° angle are <note>30°, 60°, and 90°</note>, and its right-angle sides are <note>the two sides adjacent to the 30° angle and the 60° angle</note>. The angles of the triangle with a 45° angle are <note>45°, 45°, and 90°</note>, and its right-angle sides are <note>the two sides adjacent to the two 45° angles</note>.

Triangle Angle Sum Theorem

<note>Angle 60°</note>, angle 45°, and <note>the vertical angle of ∠1</note> are the three interior angles of the triangle. According to the Triangle Angle Sum Theorem, <note>Angle 60°</note> + angle 45° + <note>the vertical angle of ∠1</note> = 180°, so <note>the vertical angle of ∠1</note> = 180° - (<note>60°</note> + 45°) = <note>75°</note>.


Question: As shown in the figure, a∥b, point B is on line b, and AB⊥BC, ∠1=35°, then ∠2=()

Answer: 55°

1. Given AB is perpendicular to BC, and ∠1=35°, according to the definition of a straight angle, we get ∠ABb=90°-35°=55°.

2. By the parallel postulate 2 of parallel lines, corresponding angles are equal, ∠2 is equal to ∠ABb.

3. Therefore, ∠2 is equal to 55°.

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

平行线; 同位角; 直角三角形

Definition of Perpendicular Lines

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

<note>Line AB and line BC intersect to form an angle ∠ABC of 90 degrees</note>, therefore according to the definition of perpendicular lines, <note>line AB and line BC are perpendicular to each other</note>.

Definition of Corresponding Angles

In the figure of this problem, <note>two parallel lines a and b are intersected by a line AB</note>, where <note>angle 2 and angle ABb are on the same side of the intersecting line AB, on the same side of the intersected lines a and b</note>, therefore <note>angle 2 and angle ABb are corresponding angles</note>. <note>Corresponding angles are equal, i.e., angle 2 = angle ABb</note>.

Parallel Postulate 2 of Parallel Lines

<note>Two parallel lines a and b</note> are intersected by a third <note>line AB</note>, forming the following geometric relationship: 1. Corresponding angles: <note>angle 2</note> and <note>angle ABb</note> are equal.

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>A ray rotates around endpoint B to form a straight line b with the initial side</note>. According to the definition of a straight angle, <note>a straight angle measures 180 degrees</note>, that is, <note>angle 1 + 90 degrees + angle ABb = 180 degrees</note>.


Question: As shown in the figure, if a∥b, ∠1=115°, then ∠2=()

Answer: 65°

1. Let line a be AB, line b be CD. Given that line a is parallel to line b, according to the parallel lines axiom 2, it is known that the interior angles on the same side are supplementary, thus ∠1 + ∠2 = 180°.

2. From ∠1 + ∠2 = 180° and the given ∠1 = 115°, it follows that ∠2 = 180° - 115°.

3. After calculation, it is obtained that ∠2 = 65°.

Definition of Parallel Lines

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

The lines <note>line a and line b</note> are in the same plane and do not intersect, so according to the definition of parallel lines, <note>line a and line b are parallel lines</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.

In the diagram of this problem, two straight lines <note>AB</note> and <note>CD</note> are intersected by a third straight line <note>AC</note>. Angles <note>1</note> and <note>2</note> are on the same side of the transversal <note>AC</note> and within the intersected lines <note>AB</note> and <note>CD</note>, so angles <note>1</note> and <note>2</note> are consecutive interior angles. Consecutive interior angles <note>1 and 2 are supplementary</note>, that is, <note>angle 1 + angle 2 = 180 degrees</note>.

Parallel Postulate 2 of Parallel Lines

In the figure of this problem, <note>line a and line b are intersected by a line</note>, thus forming <note>same-side interior angles ∠1 and ∠2</note>. According to the <note>Supplementary Angles Theorem for Parallel Lines</note>, we can conclude that <note>∠1 + ∠2 = 180°</note>. Since <note>∠1 = 115°</note>, therefore <note>∠2 = 180° - 115° = 65°</note>, finally obtaining the answer <note>∠2 = 65°</note>.


Question: As shown in the figure, parallel lines a and b are intersected by line c. If ∠1 = 50°, then the degree of ∠2 is ()

Answer: 130°

1. Given that line a and line b are parallel and intersected by line c, ∠1 = 50°.

2. Let line a and line c intersect at point O, line a be AB, line c be CD. According to the definition of vertical angles, the vertical angle of ∠1 is ∠3, thus ∠1 = ∠3 = 50°.

####

3. Next, we can conclude that ∠2 and ∠3 are same-side interior angles. According to the parallel postulate 2, we get ∠2 + ∠3 = 180°.

4. Substituting the given condition, ∠2 + 50° = 180°, we get ∠2 = 130°.

5. Through the above reasoning, we finally obtain the answer ∠2 is 130°.

Definition of Vertical Angles

In the figure of this problem, two intersecting lines <note>a and c intersect at point x</note>, forming four angles: angle <note>AOC</note>, angle <note>1</note>, angle <note>2</note>, and angle <note>DOB</note>. According to the definition of vertical angles, angle <note>1 and angle 2 are vertical angles</note>, angle <note>AOC and angle DOB are vertical angles</note>. Since vertical angles are equal in measure, <note>angle 1 = angle 2</note>, <note>angle AOC = angle DOB</note>.

Parallel Postulate 2 of Parallel Lines

In the diagram of this problem, <note>two parallel lines a and b</note> are intersected by <note>a third line c</note>, forming the following geometric relationships: 1. Corresponding angles: <note>angle AOC and angle 2 are equal</note>. 2. Alternate interior angles: <note>angle BOD and angle 2 are equal</note>. 3. Consecutive interior angles: <note>angle 2 and angle 3 are supplementary, that is, angle 2 + angle 3 = 180 degrees</note>. These relationships illustrate that when two parallel lines are intersected by a third line, corresponding angles are equal, alternate interior angles are equal, and consecutive interior angles are supplementary.

<note>In the figure of this problem, two lines a and b are intersected by a third line c</note>, <note>angles 2 and 3 are on the same side of the intersecting line c</note>, and <note>within the intersected lines a and b</note>, so <note>angles 2 and 3 are consecutive interior angles</note>. <note>Consecutive interior angles 2 and 3 are supplementary</note>, that is, <note>angle 2 + angle 3 = 180 degrees</note>.


Question: From one corner of a cube with an edge length of 4, a smaller cube with an edge length of 2 is removed, resulting in a part as shown in the figure. The surface area of this part is ()

Answer: 96

1. Given that the surface area of the original cube is the same as the surface area of a cube with side length 4, according to the surface area formula for a cube, the surface area of the original cube is 4×4×6.

2. A small cube with side length 2 is removed from one corner of the large cube. Based on the decomposition of geometric shapes, removing the small cube will add three new faces from the six faces of the removed small cube to the original cube.

3. At the same time, removing the small cube will reduce three faces from the six faces of the removed small cube from the original cube.

4. Combining steps 2 and 3, it can be concluded that the overall surface area of the original cube will not change after removing the small cube.

5. After calculation, the surface area of the original cube is 4×4×6=96.

6. Therefore, through the above steps, the final surface area is 96, without any change.

立方体; 正方形; 点; 线段

Surface Area Formula for a Cube

The total surface area of a cube is equal to 6 times the square of the edge length of the cube.

<note>The total surface area of a cube is equal to 6 times the square of the cube's edge length</note>. In this problem's diagram, <note>the total surface area of the cube is calculated as 6×4²=96</note>. The total surface area of a cube is equal to 6 times the square of the cube's edge length. Given that the edge length of the cube is 4, therefore <note>the total surface area of the cube is 6×4²=96</note>.


Question: As shown in the figure, it is known that ∠1=60°, ∠A+∠B+∠C+∠D+∠E+∠F=(). Requirements: 1. Ensure all mathematical symbols are not translated 2. Do not translate figure labels (e.g., ABCD) 3. Output format: directly return the translated English problem without any additional content (no need to start with 'Question:')

Answer: 240°

1. Given that ∠1 is equal to 60°, according to the problem statement and the figure, ##let the intersection of BE and CF be G, the intersection of BE and AD be H, and the intersection of AD and FC be O. By the exterior angle theorem of a triangle, in triangle GEC, ∠FGE is the sum of ∠C and ∠E. In triangle HBD, ∠AHB is the sum of ∠B and ∠D. According to the definition of vertical angles, we know that ∠GOH = ∠1 = 60°. By the interior angle sum theorem of a triangle, we can determine that the total angle formed by ∠GOH and ∠B, ∠C, ∠D, and ∠E is equal to the interior angle sum of triangle OGH, which is 180°##.

2. Therefore: ##∠GOH## + ∠B + ∠C + ∠D + ∠E = 180°.

3. Substitute the known value of ##∠GOH## which is 60°, we get: 60° + ∠B + ∠C + ∠D + ∠E = 180°.

4. Therefore: ∠B + ∠C + ∠D + ∠E = 180° - 60° = 120°.

5. ##According to the interior angle sum theorem of a triangle, in triangle AFO##, we can determine that ∠1 along with ∠A and ∠F also form a 180° angle.

6. Therefore: ∠1 + ∠A + ∠F = 180°.

7. Substitute the known value of ∠1 which is 60°, we get: 60° + ∠A + ∠F = 180°.

8. Therefore: ∠A + ∠F = 180° - 60° = 120°.

9. Adding the two equations from steps 4 and 8, we get ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 120° + 120° = 240°.

10. Through the above reasoning, the final answer is 240°.

普通多边形; 五边形; 三角形的外角; 内错角; 邻补角

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>∠1 is a geometric figure formed by two rays AD and CF</note>, these <note>two rays have a common endpoint O</note>. This common <note>endpoint O is called the vertex of angle 1</note>, and the <note>rays AD and CF are called the sides of angle 1</note>.

Triangle Angle Sum Theorem

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

In the figure of this problem, <note>in triangle AF1, angle 1, angle A, and angle F are the three interior angles of triangle AF1</note>, according to the Triangle Angle Sum Theorem, <note>angle 1 + angle A + angle F = 180°</note>. Similarly, <note>in triangle OGH, angle GOH, angle GHO, and angle OGH are the three interior angles of triangle OGH</note>, according to the Triangle Angle Sum Theorem, <note>angle GOH + angle GHO + angle OGH = 180°</note>. In the figure of this problem, <note>in triangle GCE, angle FGE is an exterior angle of the triangle</note>, <note>angle C and angle E are the two non-adjacent interior angles to the exterior angle FGE</note>, according to the Exterior Angle Theorem, <note>the exterior angle FGE is equal to the sum of the two non-adjacent interior angles C and E, that is, angle FGE = angle C + angle E</note>. Similarly, <note>in triangle HBD, angle AHB is the exterior angle of the triangle</note>.

Definition of Vertical Angles

<note>Two intersecting lines AD and CF intersect at point O</note>, forming four angles: <note>angle AOF</note>, <note>angle GOH</note>, <note>angle AOC</note>, and <note>angle FOD</note>. According to the definition of vertical angles, <note>angle AOF and angle GOH are vertical angles</note>, <note>angle AOC and angle FOD are vertical angles</note>. Since the angles of vertical angles are equal, <note>angle AOF = angle GOH</note>, <note>angle AOC = angle FOD</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 AOF is a geometric figure composed of three non-collinear points A, O, F and their connecting line segments AO, OF, AF</note>. Points A, O, F are the three vertices of the triangle, and line segments AO, OF, AF are the three sides of the triangle. Similarly, <note>Triangle GOH is a geometric figure composed of three non-collinear points G, O, H and their connecting line segments GO, OH, GH</note>. Points G, O, H are the three vertices of the triangle, and line segments GO, OH, GH are the three sides of the triangle.