Question: As shown in the figure, OA and OB are the perpendicular bisectors of segments MC and MD respectively, MD=5cm, MC=7cm, CD=10cm. A small ant starts from point M, crawls to any point E on the side OA, then crawls to any point F on the side OB, and finally crawls back to point M. The shortest path the ant can crawl is ()

Answer: 10cm

1. Let the intersection point of CD and OA be E, and the intersection point of CD and OB be F. According to the problem statement, draw auxiliary lines for geometric analysis.

2. Since OA is the perpendicular bisector of segment MC, according to the properties of the perpendicular bisector, we can deduce ME = CE.

3. Similarly, since OB is the perpendicular bisector of segment MD, according to the properties of the perpendicular bisector, we can deduce MF = DF.

4. Therefore, the path of the ant's crawl is ME + EF + FM = CE + EF + FD.

5. Given that the shortest segment between two points is a straight line, in order to minimize CE + EF + FD, C, E, F, and D must lie on the same straight line, thus the shortest path is CD.

6. Therefore, the shortest path for the ant's crawl is the length of CD, which is known to be 10cm.

7. Hence, the final answer is: the shortest path for the ant's crawl is 10cm.

垂直平分线; 线段; 等腰三角形; 反射

Definition of Perpendicular Bisector

In the figure of this problem, <note>line OA passes through the midpoint of segment MC</note>, and <note>line OA is perpendicular to segment MC</note>. Therefore, <note>line OA is the perpendicular bisector of segment MC</note>. Similarly, <note>line OB passes through the midpoint of segment MD</note>, and <note>line OB is perpendicular to segment MD</note>. Therefore, <note>line OB is the perpendicular bisector of segment MD</note>.

In the figure of this problem, <note>the perpendicular bisector of segment MC is line OA</note>, <note>point E is on line OA</note>. According to the properties of the perpendicular bisector, <note>the distances from point E to the endpoints M and C of segment MC are equal, i.e., ME = CE</note>. Similarly, <note>the perpendicular bisector of segment MD is line OB</note>, <note>point F is on line OB</note>. According to the properties of the perpendicular bisector, <note>the distances from point F to the endpoints M and D of segment MD are equal, i.e., MF = DF</note>.


Question: As shown in the figure, in order to green the environment, four sectors with a radius of 1 are drawn at the four corners of the rectangular vacant lot for greening. Then the total greening area is ()

Answer: π

1. Due to the greening of four sector areas at the four corners of the rectangular open space, each sector has a radius of 1.

2. Because the central angle of each sector is equal to the interior angle of the rectangle, each sector's central angle is 90 degrees.

3. Using the sector area formula, Area = Central Angle/360 × π × Radius^2.

4. For each sector, its area = 90/360 × π × 1^2 = π/4.

5. Since there are four such sectors, the total area is 4 × π/4.

6. After calculation, the total area of the four sectors is π.

矩形; 扇形; 圆心角; 圆

Definition of Sector

A sector is a figure formed by two radii of a circle and the arc enclosed between them.

<note>The radius of each sector is 1</note>, <note>the center of the circle is at the four vertices of the rectangle</note>, <note>the arc is the arc enclosed by these two radii</note>, so according to the definition of the sector, the figure formed by these two radii and the arc they enclose is a sector.

Definition of Central Angle

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

<note>In the figure of this problem, each sector's central angle is 90 degrees</note>, because <note>the sector's central angle is an interior angle of the rectangle</note>, and <note>each angle of the rectangle is 90 degrees</note>. Each sector's <note>central angle is formed by the lines connecting the center to the endpoints of the sector</note>.

Formula for the Area of a Sector

The area \( X \) of a sector can be calculated using the formula \( X = \frac{\theta}{360} \times \pi \times r^2 \), where \( \theta \) is the measure of the central angle in degrees, and \( r \) is the radius length.

<note>Each sector is located at the four corners of the rectangle</note>, each sector's <note>central angle is 90 degrees</note>, and the <note>radius length is 1</note>. According to the formula for the area of a sector, the area A of the sector can be calculated using the formula <note>A = ##(90/360) * π * 1^2##</note>, so <note>the area of each sector A = π/4</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 is a rectangle, with <note>all four interior angles being right angles (90 degrees)</note>, and <note>opposite sides are parallel and equal in length</note>.


Question: As shown in the figure, in the right triangle △ABC, ∠ACB=90°, AC=6, BC=8, AD is the angle bisector of ∠BAC. If P and Q are moving points on AD and AC respectively, then the minimum value of PC+PQ is ()

Answer: \frac{24}{5}

1. Given in the right triangle ABC, ∠ACB=90°, side AC=6, side BC=8, AD is the bisector of ∠BAC.

2. Through point C, draw a perpendicular line CM perpendicular to AB, intersecting AB at point M, and intersecting AD at point P'.

3. Through point P', draw a perpendicular line P'Q' perpendicular to AC, intersecting AC at point Q'.

4. According to the property of the bisector of a straight angle, P'Q' = P'M.

5. Therefore, when P'Q' = P'M, the minimum value of P'C + P'Q' is the length of CM.

6. Using the Pythagorean theorem, the length of AB can be obtained, AB = √(AC^2 + BC^2) = √(6^2 + 8^2) = 10.

7. Calculating the area of triangle ABC, we get S△ABC = 1/2 * AC * BC = 1/2 * 6 * 8 = 24.

8. The area of triangle ABC can also be expressed as: S△ABC = 1/2 * AB * CM.

9. Setting the two area formulas equal, we get the length of CM, 24 = 1/2 * 10 * CM, i.e., CM = 24/5.

10. Therefore, the minimum value of PC + PQ is 24/5.

直角三角形; 线段; 等腰三角形; 反射

Definition of Angle Bisector

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

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

Definition of Perpendicular Lines

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

<note>Line CM</note> and <note>Line AB</note> intersect to form an angle <note>∠CMB</note> of 90 degrees, so according to the definition of perpendicular lines, <note>Line CM</note> and <note>Line AB</note> are perpendicular to each other; similarly, <note>Line P'Q'</note> and <note>Line AC</note> intersect to form an angle <note>∠P'Q'A</note> of 90 degrees, so according to the definition of perpendicular lines, <note>Line P'Q'</note> and <note>Line AC</note> are perpendicular to each other.

Pythagorean Theorem

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

In the <note>right triangle ABC</note>, <note>∠ACB is a right angle (90 degrees)</note>, sides <note>AC and BC are the legs</note>, and side <note>AB is the hypotenuse</note>, so according to the <note>Pythagorean Theorem</note>, <note>AB^2 = AC^2 + BC^2</note>, that is, <note>10^2 = 6^2 + 8^2</note>.

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 BC is the height</note>. According to the area formula of a triangle, the area of triangle ABC is equal to <note>base AC multiplied by height BC</note> and then divided by 2, that is, <note>S△ABC = (AC * BC) / 2 = (6 * 8) / 2 = 24</note>. Another method is to use the base <note>AB and height CM</note> to find: <note>S△ABC = 1/2 * AB * CM</note>, thus obtaining <note>24 = 1/2 * 10 * CM</note>, and solving for <note>CM = 24/5</note>.

Property of Angle Bisector

Any point on the angle bisector is equidistant to the two sides of the angle.

In the figure of this problem, <note>angle BAC is bisected by the angle bisector AD</note>, <note>point P' is on the angle bisector AD</note>. According to the property of the angle bisector, <note>the distance from point P' to both sides of the angle AB and AC is equal</note>, <note>that is, P'Q' = P'M</note>.


Question: As shown in the figure, the perimeter of △ABC is 16. Point D is the midpoint of side AB, BD=2, through point D draw the perpendicular line l to AB, E is an arbitrary point on l, then the minimum value of the perimeter of △AEC is ()

Answer: 12

1. Given point D is the midpoint of side AB, BD=2, therefore AB=2BD=4.

2. Given the perimeter of △ABC is 16, then AC + BC = 16 - AB = 16 - 4 = 12.

3. Draw a perpendicular line l to AB through point D, intersecting at point E, connect BE.

4. Since point D is the midpoint of side AB and l⊥AB, l is the perpendicular bisector of AB, thus AE=BE.

5. Therefore, AE + CE = BE + CE.

6. According to the triangle inequality, ##BE + CE > BC##.

7. When points B, E, and C are collinear, the minimum value of BE + CE equals the length of BC.

8. Therefore, at this time, the minimum perimeter of △AEC is AC + BC.

9. Given AC + BC = 12, the minimum perimeter of △AEC is 12.

中点; 垂线; 直线; 普通三角形; 平移

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 diagram of this problem, the midpoint of line segment AB is point D</note>. According to the definition of the midpoint of a line segment, point D divides line segment AB into two equal parts, that is, <note>the lengths of line segments AD and DB are equal</note>. That is, AD = DB = 2.

Definition of Perpendicular Lines

The lines <note>l and AB</note> intersect to form <note>angles ∠ADE and ∠BDE that are 90 degrees</note>, so according to the <note>definition of perpendicular lines</note>, <note>lines l and AB are perpendicular to each other</note>.

Definition of Perpendicular Bisector

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

<note>Line l passes through the midpoint D of segment AB</note>, and <note>line l is perpendicular to segment AB</note>. Therefore, <note>line l is the perpendicular bisector of segment AB</note>.

Properties of the Perpendicular Bisector

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

<note>The perpendicular bisector of line segment AB is line l</note>, point E is on <note>line l</note>. According to <note>the properties of the perpendicular bisector</note>, the distance from point E to <note>the two endpoints A and B of line segment AB is equal, that is AE = BE</note>.

Theorem of Triangle Inequality

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

<note>Side BE, side CE, and side BC form a triangle</note>. According to the Theorem of Triangle Inequality, the sum of any two sides is greater than the third side, that is, <note>side BE + side CE > side BC</note>.


Question: Students have all played the seesaw game. The figure shows a schematic diagram of a seesaw, where the pillar OC is perpendicular to the ground, and OA = OB. When one end A of the seesaw touches the ground, ∠AOA′ = 50°, then when the other end B of the seesaw touches the ground, ∠COB′ equals ()

Answer: 65°

1. Given OA=OB, and OC is perpendicular to the ground, according to the problem statement, we can obtain ##∠ACO and ∠B′CO are right angles, i.e., ∠ACO=∠B′CO=90°##.

####

##2. Because OA=OB, and according to the principle of the seesaw, OB=OB′, so OA=OB′##.

##3. Because OA=OB′, in triangle AOB′, according to the theorem that the altitude, median, and angle bisector coincide in an isosceles triangle, OC is the angle bisector of ∠AOB′, so ∠COB′ is half of ∠AOB′##.

##4. Because ∠AOA′=50°, according to the definition of the full angle, ∠AOB′=180°-50°=130°##.

##5. Therefore, ##∠COB′=1/2×∠AOB′=1/2×130°=65°##.

##6##. Through the above reasoning, the final answer is 65°.

对顶角; 垂线; 旋转; 等腰三角形

Definition of Isosceles Triangle

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

The line segments <note>OA and OB</note> are equal, so <note>triangle AOB</note> is an isosceles triangle.

Definition of Full Angle

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

In the figure of this problem, <note>ray OB’ around endpoint O</note> rotates one full turn, forming <note>full angle A‘OB’</note>. According to the <note>definition of full angle</note>, <note>the degree of the full angle is 360 degrees</note>, that is, <note>full angle A‘OB’ = 360 degrees</note>.

Coincidence Theorem of Altitude, Median, and Angle Bisector in Isosceles Triangle

In an isosceles triangle, the angle bisector of the vertex angle not only bisects the vertex angle but also bisects the base and is perpendicular to the base.

In the diagram of this problem, in the isosceles triangle AOB', the vertex angle is <note>angle AOB'</note>, and the base is <note>side AB'</note>. The angle bisector OC of the vertex angle not only bisects <note>vertex angle AOB'</note>, but also bisects <note>base AB'</note>, making <note>AC = CB'</note>, and is also <note>perpendicular to the base AB'</note>, forming <note>right angles ACO and OCB' (90 degrees)</note>. Therefore, the segment <note>OC is both the angle bisector of the vertex angle, the median of the base, and the altitude</note>.


Question: As shown in the figure, ∠BAC=110°, if A and B are symmetric about line MP, and A and C are symmetric about line NQ, then the size of ∠PAQ is ()

Answer: 40°

1. Given ∠BAC = 110°.

2. Since A and B are symmetric about line MP and A and C are symmetric about line NQ, according to symmetry, line MP is the perpendicular bisector of segment AB, and line NQ is the perpendicular bisector of segment AC.

3. Because line MP is the perpendicular bisector of segment AB, ∠BAP = ∠ABP.

4. Because line NQ is the perpendicular bisector of segment AC, ∠QAC = ∠QCA.

5. By the triangle angle sum theorem, ∠ABC + ∠BCA = 180° - ∠BAC = 180° - 110° = 70°.

6. Also, because ∠BAP = ∠ABP and ∠QAC = ∠QCA, ∠BAP + ∠QAC = ∠ABP + ∠QCA = 70°.

7. From the above conclusion, ∠PAQ = ∠BAC - ∠BAP - ∠QAC = 110° - 70° = 40°.

8. Through the above reasoning, it is finally concluded that ∠PAQ = 40°.

对称; 邻补角; 三角形的外角

Symmetric Point

A symmetric point refers to a point that is symmetric with respect to a given line (or a given point) such that the two points are equidistant from the line (or point).

<note>Point A and point B are symmetric points with respect to line MP</note>, and <note>the distance from point A to line MP is equal to the distance from point B to line MP</note>. Similarly, <note>point A and point C are symmetric points with respect to line NQ</note>, and <note>the distance from point A to line NQ is equal to the distance from point C to line NQ</note>. According to the definition of symmetric points, <note>point A and point B are on opposite sides of line MP and satisfy the equal distance relationship, that is, distance AM = distance BM</note>. Likewise, <note>point A and point C are on opposite sides of line NQ and satisfy the equal distance relationship, that is, distance AN = distance CN</note>.

<note>The perpendicular bisector of segment AB is line MP</note>, <note>point P is on line MP</note>. According to the properties of the perpendicular bisector, <note>the distance from point P to both endpoints A and B of segment AB is equal, i.e., PA = PB</note>. Similarly, <note>the perpendicular bisector of segment AC is line NQ</note>, <note>point Q is on line NQ</note>, according to the properties of the perpendicular bisector, <note>the distance from point Q to both endpoints A and C of segment AC is equal, i.e., QA = QC</note>.

Triangle Angle Sum Theorem

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

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


Question: As shown in the figure, the sector OAB and the sector OCD, both with central angles of 90°, overlap. OA=3, OC=1, connect AC and BD respectively. Then the area of the shaded part in the figure is ()

Answer: 2π

1. According to the problem, the central angles of sector OAB and sector OCD are both 90°, and OA=3, OC=1. Connect AC and BD.

2. According to the description in the problem, △OAC can be completely overlapped with △ODB after rotating △OAC clockwise by 90°. Therefore, the area of △OAC is equal to the area of △ODB.

3. Calculate the area of sector OAB: The formula for calculating the area of a sector is πr^2×(θ/360°). When r=3 and θ=90°, the area of sector OAB is π×3^2×(90/360)=1/4×9π=2.25π.

4. Calculate the area of sector OCD: When r=1 and θ=90°, the area of sector OCD is π×1^2×(90/360)=1/4×1π=0.25π.

5. Calculate the area of the shaded part: The area of the shaded part is the area of sector OAB plus the area of △ODB minus the area of △OAC and sector OCD.

6. According to the previous calculations, △OAC's area is equal to △ODB's area. So S(△ODB)-S(△OAC)=0. The area of the shaded part is S(sector OAB) - S(sector OCD)=2.25π-0.25π=2π.

7. Finally, after the above reasoning, the area of the shaded part is 2π.

扇形; 圆心角; 线段; 圆

Definition of Sector

A sector is a figure formed by two radii of a circle and the arc enclosed between them.

Original: <note>In the figure of this problem, in sector OAB, radius OA and radius OB are two radii of the circle, and arc AB is the arc enclosed by these two radii</note>, so according to the definition of a sector, the figure formed by <note>these two radii and the arc AB they enclose is a sector</note>. Similarly, <note>in sector OCD, radius OC and radius OD are two radii of the circle, and arc CD is the arc enclosed by these two radii</note>, so according to the definition of a sector, the figure formed by <note>these two radii and the arc CD they enclose is a sector</note>.

Formula for the Area of a Sector

The area \( X \) of a sector can be calculated using the formula \( X = \frac{\theta}{360} \times \pi \times r^2 \), where \( \theta \) is the measure of the central angle in degrees, and \( r \) is the radius length.

Original text: <note>In sector OAB, the central angle ∠AOB is 90°</note>, <note>the length of radius OA is 3</note>. According to the formula for the area of a sector, <note>the area A of sector OAB can be calculated using the formula A = (90/360) * π * 3^2</note>, resulting in 2.25π. <note>In sector OCD, the central angle ∠COD is 90°</note>, <note>the length of radius OC is 1</note>. According to the formula for the area of a sector, <note>the area A of sector OCD can be calculated using the formula A = (90/360) * π * 1^2</note>, resulting in 0.25π.


Question: As shown in the figure, in a square grid with side length 1, connect grid points D, N and E, C. DN and EC intersect at point P. tan∠CPN is ()

Answer: 2

1. Given in a square grid with side length 1, connect grid points D, N, and E, C.

2. Let the grid point below point E be point M, the first grid point below point D be point O, the second grid point below point D be point A, and the grid point below point N be point B. Connect grid points MN and DM, thus defining an auxiliary line MN and a connecting line DM.

3. Since CN and EM are both sides of a square with side length 1, according to the definition of a square, CN and EM are equal and parallel. Therefore, according to the definition of a parallelogram, quadrilateral MNCE is a parallelogram.

4. Based on the given conditions, since all squares in the grid have a side length of 1, ∠DAM=∠NBM=90°, and AD=AM=OA+OD=1+1=2, NB=MB=1. According to the definition of an isosceles right triangle, △DAM and △MBN are both isosceles right triangles. Therefore, ∠DMA=∠NMB=45°. Using the Pythagorean theorem, DM=√(AD^2+AM^2)=√(2^2+2^2)=2√2 and MN=√(BN^2+BM^2)=√(1^2+1^2)=√2.

5. Since quadrilateral MNCE is a parallelogram, according to the properties of parallelograms, CE∥NM. By the parallel postulate 2 of parallel lines, ∠CPN=∠DNM (alternate interior angles are equal).

6. Then, through the definition of a straight angle, ∠DMN=180°-∠DMA-∠NMB=180°-45°-45°=90°.

7. According to the definition of the tangent function, tan∠DNM=opposite side/adjacent side=DM/MN=2√2/√2=2.

8. Through the above reasoning, it is finally concluded that tan∠CPN=2.

点; 线段; 正切; 三角形的外角; 正方形

Definition of Square

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

<note>Each small square in the square grid is a square</note>. Each small square's <note>four internal angles are right angles (90 degrees)</note>, and <note>opposite sides are parallel and equal in length</note>.

Definition of Parallelogram

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

<note>Quadrilateral MNCE is a parallelogram</note>, <note>side MN is parallel and equal to side EC</note>, <note>side MC is parallel and equal to side NE</note>.

Properties of Parallelogram Theorem

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

In the given problem diagram, in the parallelogram <note>MNCE</note>, the angles <note>ECN and EMN are equal</note>, the angles <note>CEM and CNM are equal</note>; the sides <note>CE and MN are equal</note>, the sides <note>CN and EM are equal</note>; the diagonals <note>CM and EN bisect each other</note>, that is, the intersection point divides the diagonal <note>CM into two equal segments</note>, and divides the diagonal <note>EN into two equal segments</note>.

Definition of Isosceles Right Triangle

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

In the figure of this problem, <note>triangle DAM and triangle NBM are isosceles right triangles</note>, in which <note>angles ∠DAM and ∠NBM are right angles (90 degrees)</note>, and <note>sides DA and MA as well as sides NB and BM are equal legs of the right angles</note>.

Definition of Tangent Function

In a right-angled triangle, the tangent value of an acute angle is equal to the length of the opposite side divided by the length of the adjacent side.

In the right triangle <note>DMN</note>, <note>∠DNM is an acute angle</note>, <note>side DM is the opposite leg to ∠DNM</note>, <note>side MN is the adjacent leg to ∠DNM</note>, so <note>the tangent of ∠DNM equals the length of side DM divided by the length of side MN</note>, that is, <note>tan∠DNM=DM/MN</note>.

Definition of Alternate Interior Angles

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

In the figure of this problem, two parallel lines CE and NM are intersected by a line PN, where angle CPN and angle DNM are located between the two parallel lines and on opposite sides of the intersecting line PN. Therefore, angle CPN and angle DNM are alternate interior angles. Alternate interior angles are equal, that is, angle CPN is equal to angle DNM.

Parallel Postulate 2 of Parallel Lines

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

<note>In the diagram of this problem, two parallel lines CE and NM are intersected by a line PN</note>, <note>where angles CPN and DNM are located between the two parallel lines and on opposite sides of the intersecting line PN</note>, so angles CPN and DNM are alternate interior angles. <note>Alternate interior angles are equal, i.e., angle CPN is equal to angle DNM</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 DAM, angle DAM is a right angle (90 degrees)</note>, sides DA and MA are the legs, and side DM is the hypotenuse, so according to the Pythagorean Theorem, DM? = DA? + MA? which means <note>DM=√(AD^2+AM^2)</note>=√(2^2+2^2)=2√2. <note>In the right triangle MBN, angle MBN is a right angle (90 degrees)</note>, sides NB and MB are the legs, and side MN is the hypotenuse, so according to the Pythagorean Theorem, MN? = NB? + MB? which means <note>MN=√(BN^2+BM^2)</note>=√(1^2+1^2)=√2.

Definition of Straight Angle

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

<note>Ray MA rotates around endpoint M to form a straight line with the initial side</note>, forming a <note>straight angle AMB</note>. According to the definition of a straight angle, the measure of a straight angle is <note>180 degrees</note>, i.e., <note>angle AMB = angle DMA + angle DMN + angle NMB = 180 degrees</note>.


Question: As shown in the figure, in △ABC, ∠ABC=110°, AM=AN, CN=CP, then ∠MNP=()

Answer: 35°

1. Given ∠ABC=110°, according to the triangle angle sum theorem, we get ∠BAC+∠BCA=180°-∠ABC=70°.

2. Given that segment AM=AN, so triangle AMN is an isosceles triangle, according to the properties of isosceles triangles, we get ∠ANM=∠NMA.

3. According to the triangle angle sum theorem, in △ANM, we get ∠ANM=∠NMA=(180°-∠BAC)/2.

4. Given that segment CN=CP, so triangle CNP is an isosceles triangle, according to the properties of isosceles triangles, we get ∠CNP=∠CPN.

5. According to the triangle angle sum theorem, in △CNP, we get ∠CNP=∠CPN=(180°-∠BCA)/2.

6. Since point N is on AC, ∠ANC is 180°, so ∠MNP=180°-∠ANM-∠CNP=180°-(180°-∠BAC)/2-(180°-∠BCA)/2.

7. Continuing to simplify, we get ∠MNP=1/2(∠BAC+∠BCA).

8. Since in the first step it was determined that ∠BAC+∠BCA=70°, so ∠MNP=1/2×70°=35°.

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

等腰三角形; 三角形的外角; 对称; 对顶角; 普通三角形

Definition of Isosceles Triangle

In triangle CNP, <note>side CN and side CP are equal</note>, therefore triangle CNP is an isosceles triangle. Similarly, in triangle AMN, <note>side AN and side AM are equal</note>, therefore triangle AMN is an isosceles triangle.

Properties of Isosceles Triangle

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

<note>In the isosceles triangle CNP, sides CN and CP are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle CNP = angle CPN</note>. Similarly, <note>in the isosceles triangle AMN, sides AM and AN are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle ANM = angle NMA</note>.

Triangle Angle Sum Theorem

<note>In triangle CNP, angle CNP, angle NPC, and angle PCN are the three interior angles of triangle CNP</note>, according to the Triangle Angle Sum Theorem, angle CNP + <note>angle NPC</note> + angle PCN = 180°. <note>In triangle AMN, angle AMN, angle ANM, and angle MAN are the three interior angles of triangle AMN</note>, according to the Triangle Angle Sum Theorem, angle AMN + angle ANM + angle MAN = 180°. <note>In triangle ABC, angle BAC, angle ABC, and angle ACB are the three interior angles of triangle ABC</note>, according to the Triangle Angle Sum Theorem, angle BAC + angle ABC + angle ACB = 180°.

Definition of Straight Angle

<note>Ray AN rotates around endpoint N to form a straight line with the initial side</note>, forming <note>straight angle ANC</note>. According to the <note>definition of straight angle</note>, the measure of a straight angle is 180 degrees, i.e., <note>angle ANC = 180 degrees</note>.


Question: At a certain time, there is a passenger ship at point P in the sea. It is measured that the lighthouse A is located 30° east of north from P, and the distance is 50 nautical miles. The passenger ship sails at a speed of 60 nautical miles per hour in the direction of 60° west of north for \frac{2}{3} hours to reach point B. Then tan∠BAP=()

Answer: \frac{4}{5}

1. Given that lighthouse A is located 30° north-east of point P and is 50 nautical miles away. Therefore, the length of segment AP is 50 nautical miles.

2. The passenger ship travels at a speed of 60 nautical miles per hour in the 60° north-west direction for ##2/3## hours to reach point B. According to the displacement formula ##s=vt##, we get BP = 60 × ##2/3## = 40 nautical miles.

3. At point P, consider the angle formed by the directions 30° north-east and 60° north-west. ##Therefore, ∠APB = 60°+30°= 90°, so the two directions are perpendicular.##

4. In the right triangle APB, given AP = 50 nautical miles and BP = 40 nautical miles. According to the definition of the ##tangent function##, tan∠BAP = opposite side BP / adjacent side AP.

5. Calculating, we get tan∠BAP = 40/50 = 4/5.

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

方向角; 普通三角形; 正切

Definition of Tangent Function

In the figure of this problem, in <note>right triangle APB</note>, <note>angle ∠BAP is an acute angle</note>, side <note>BP is the opposite leg of angle ∠BAP</note>, side <note>AP is the adjacent leg of angle ∠BAP</note>, so <note>the tangent value of angle ∠BAP is equal to the length of side BP divided by the length of side AP</note>, that is, <note>tan(∠BAP) = BP / AP = 40 / 50 = 4 / 5</note>.


Question: As shown in the figure, in quadrilateral ABCD, the angle bisector of ∠DAB intersects the external angle bisector of ∠ABC at point P, and ∠D + ∠C = 200°, then ∠P = ()

Answer: 10°

1. Given ∠D + ∠C = 200°, and the sum of the interior angles of a quadrilateral is 360°, we have ∠DAB + ∠ABC + ∠C + ∠D = 360°.

2. According to the above equation, we get ∠DAB + ∠ABC = 160°.

3. The angle bisector of ∠DAB intersects the exterior angle bisector of ∠ABC at point P.

4. Let ∠PAB be the bisected angle of ∠DAB, then ∠PAB = 0.5 * ∠DAB.

5. Let ∠PBE be the bisected angle of the exterior angle CBE, then ∠PBE = 0.5 * (180° - ∠ABC).

7. In △PAB, ∠PBE is an exterior angle, ∠P = ∠PBE - ∠PAB = 10°.

8. Through the above reasoning, we finally get ∠APB = 10°.

邻补角; 普通三角形; 三角形的外角

Polygon Interior Angle Sum Theorem

The sum of the interior angles of a polygon is equal to (n - 2) * 180°, where n represents the number of sides of the polygon.

In the original text: <note>Quadrilateral ABCD</note>, <note>ABCD is a polygon with 4 sides, where 4 represents the number of sides of the polygon</note>. According to the <note>Polygon Interior Angle Sum Theorem, the sum of the interior angles of this polygon is equal to (4-2) × 180° = 360°</note>.

Definition of Angle Bisector

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

In this problem diagram, <note>the vertex of angle DAB is point A</note>, a line AP is drawn from <note>point A</note>, <note>this line divides angle DAB into two equal angles</note>, namely <note>angle DAP and angle PAB are equal</note>. Therefore, <note>line AP is the angle bisector of angle DAB</note>.##<note>The vertex of angle CBE is point B</note>, a line BP is drawn from <note>point B</note>, <note>this line divides angle CBE into two equal angles</note>, namely <note>angle CBP and angle PBE are equal</note>. Therefore, <note>line BP is the angle bisector of angle CBE</note>.####

Exterior Angle Theorem of a Triangle

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

In <note>△PAB</note>, <note>∠PBE</note> is the exterior angle that is not adjacent to <note>∠PAB and ∠APB</note>. According to the Exterior Angle Theorem of a Triangle, <note>∠PAB + ∠APB = ∠PBE</note>.


Question: As shown in the figure, AB is a chord of ⊙O, OC ⊥ AB at point D, intersecting ⊙O at point C. If the radius is 5 and OD = 3, then the length of chord AB is ()

Answer: 8

1. Connect point A and point O to obtain segment AO. Because OC is perpendicular to AB and intersects AB at point D, according to the ##diameter perpendicular theorem##, point D is the midpoint of segment AB. Therefore, AD equals BD, and equals half of AB, i.e., AD=BD=1/2*AB.

2. In right triangle AOD, given that the length of OA is 5 and the length of OD is 3, according to the ##Pythagorean theorem##, the length of AD can be calculated as: AD=## √(OA^2 - OD^2 )= √(5^2 - 3^2 )= √( 25 - 9) = √16 ##=4.

3. Since AB equals 2 times AD, the length of AB is 2*AD=2*4=8.

圆; 弦; 垂线; 直角三角形; 线段

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 AOD, <note>∠ODA is a right angle (90 degrees)</note>, so triangle AOD is a right triangle. <note>Side OD and side AD are the legs</note>, <note>side OA is the hypotenuse</note>.

Pythagorean Theorem

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

Perpendicular Diameter Theorem

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

In <note>circle O</note>, if the <note>diameter is perpendicular to chord AB</note>, then according to the Perpendicular Diameter Theorem, the <note>diameter bisects chord AB, i.e., AD=DB</note>, and the <note>diameter bisects the arcs subtended by chord AB, i.e., arc AC = arc CB</note>.


Question: The side view of a certain staircase is shown in the figure. It is measured that the length of AB is 3 meters, and the slope ratio of the staircase BC is 1:2. Then the length of the staircase slope BC is ()

Answer: 3√{5}米

1. Given that the length of AB is 3 meters, the slope ratio of the staircase ramp BC is 1:2.

2. According to the slope ratio of 1:2, it can be derived that \rac{AB}{AC} = \rac{1}{2}.

3. Based on the ratio \rac{AB}{AC} = \rac{1}{2}, it can be calculated that AC = 2AB = 2 \	imes 3 = 6 meters.

4. Let BC be the unknown variable, according to the Pythagorean theorem BC = \\sqrt{AB^2 + AC^2}, calculate BC.

5. Substitute the values AB=3 meters, AC=6 meters, it can be obtained that BC = \\sqrt{3^2 + 6^2} = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5} meters.

6. Through the above reasoning, the final answer is 3\\sqrt{5} meters.

Definition of Slope

The slope is defined as the ratio of the vertical height to the horizontal distance of a line or a particular section of terrain.

<note>Let the stair slope be segment BC</note>, <note>the vertical height of segment BC is segment AB</note>, <note>the horizontal distance is segment AC</note>. According to the definition of slope, slope refers to the ratio of vertical height <note>AB to horizontal distance AC</note>, that is, <note>slope = AB/AC</note>.

Pythagorean Theorem

<note>∠CAB is a right angle (90 degrees)</note>, sides <note>CA and BA are the legs</note>, side <note>BC is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>BC^2 = AB^2 + AC^2</note>.


Question: As shown in the figure, in the equilateral △ABC, BD bisects ∠ABC and intersects AC at point D. From point D, draw DE⊥BC at point E, and CE=1.5. Then the length of AB is ()

Answer: 6

1. Given △ABC is an equilateral triangle, according to the definition of an equilateral triangle, we have AB=BC=CA, and ∠ABC=∠BCA=∠CAB=60°.

2. Since DE⊥BC, we get ∠CED=90°.

3. Because ∠CED=90° and ∠DCE=60°, according to the triangle angle sum theorem, we get ∠EDC=30°.

4. In △CDE, since EC=1.5, according to the properties of a 30°-60°-90° triangle, we get CD=2*EC=3.

5. Because BD bisects ∠ABC and △ABC is an equilateral triangle, according to the theorem that the altitude, median, and angle bisector coincide in an isosceles triangle, we get AD=CD=3.

6. Since △ABC is an equilateral triangle, we have AB=AC.

7. Because AB=AC=AD+CD, according to the previous calculations, we get AD=3 and CD=3, so AB=3+3=6.

8. Through the above reasoning, we finally get the length of AB as 6.

Definition of Equilateral Triangle

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

<note>Triangle ABC is an equilateral triangle</note>.<note>The lengths of sides AB, BC, and CA are equal</note>, and <note>the measures of angles ABC, BCA, and CAB are equal, each being 60°</note>.

Triangle Angle Sum Theorem

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

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

In an isosceles triangle, the angle bisector of the vertex angle not only bisects the vertex angle but also bisects the base and is perpendicular to the base.

In the figure of this problem, in <note>isosceles triangle ABC</note>, the vertex angle is <note>angle ABC</note>, and the base is <note>side AC</note>. The <note>angle bisector BD</note> of the vertex angle not only bisects <note>vertex angle ABC</note>, but also bisects the base <note>AC, making DA = DC</note>, and is <note>perpendicular to the base AC</note>, forming a <note>right angle BDC (90 degrees)</note>. Therefore, <note>segment BD is both the angle bisector of the vertex angle, the median of the base, and the altitude</note>.

Properties of 30°-60°-90° Triangle

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

<note>△CDE is a 30°-60°-90° triangle</note>, where <note>∠CED=90°</note>, <note>∠DCE=60°</note>, <note>∠CDE=30°</note>. <note>Side CE is opposite the 30-degree angle</note>, <note>side CD is opposite the 60-degree angle</note>, <note>side DE is the hypotenuse</note>. According to the properties of a 30°-60°-90° triangle, <note>side CE is half the length of side CD</note>, <note>side DE is √3 times the length of side CE</note>. That is: <note>CE = 1/2 * CD, DE = CE * √3</note>.


Question: As shown in the figure, in △ABC, BD and CE are angle bisectors, AM⊥BD at point M, AN⊥CE at point N. The perimeter of △ABC is 30, BC=12. Then the length of MN is ()

Answer: 3

1. From the given conditions, we know that the perimeter of △ABC is 30, and BC=12.

2. According to the ##formula for the perimeter of a triangle##, we can derive that AB+AC=30-BC=18.

3. Extend AN and AM to intersect BC at points F and G respectively.

4. Since ##BD## is the angle bisector of ∠ABC, according to the ##definition## of an angle bisector, we have ##∠CBD=∠ABD##.

5. Since ##BD## is perpendicular to AG, therefore ##according to the triangle's interior angle sum theorem: ∠ABD + ∠BAG=180°-∠AMB=180°-90°=90°##, and ##∠AGB + ∠CBD##=180°-∠BMG=180°-90°=90°.

6. Thus, we can derive that ##∠BAG##=∠AGB, hence AB=BG.

7. According to ##step 6, we can deduce that triangle ABG is an isosceles triangle, and BM is its perpendicular line. According to the theorem that the height, median, and angle bisector coincide in an isosceles triangle, we get AM=MG##. Similarly, using the same method, we get AC=CF, ##AN=NF##.

8. From the definition of the midline, we know that MN is the midline of △AFG, hence MN=## 1/2 ##GF.

9. GF=BG+CF-BC, according to the previous conclusion BG=##AB##, CF=##AC##, thus GF=AB+AC-BC.

10. Finally, MN=## 1/2 ##(AB+AC-BC)=## 1/2 ##(18-12)=3.

11. Through the above reasoning, we finally derive that the length of MN is 3.

普通三角形; 垂线; 线段

Definition of Angle Bisector

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

Definition of Perpendicular Lines

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

<note>Line AM and line BD</note> intersect to form <note>angle ∠AMB is 90 degrees</note>, so according to the definition of perpendicular lines, <note>line AM and line BD</note> are perpendicular to each other. Similarly, <note>line AN and line CE</note> intersect to form <note>angle ∠ANC is 90 degrees</note>, so according to the definition of perpendicular lines, <note>line AN and line CE</note> are perpendicular to each other.

Triangle Midline Theorem

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

In this problem, in triangle AFG, <note>point N is the midpoint of side AF</note>, <note>point M is the midpoint of side AG</note>, <note>line segment MN connects these two midpoints</note>. According to the Triangle Midline Theorem, <note>line segment MN is parallel to the third side FG and equal to half of the third side FG</note>, that is, <note>MN || FG</note>, and <note>MN = 1/2 * FG</note>.

Formula for the Perimeter of a Triangle

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

<note>The three sides of triangle ABC are AB, AC, and BC</note>. According to the formula for the perimeter of a triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, <note>Perimeter P = AB + AC + BC</note>.

Definition of Right Triangle

In the figure of this problem, in triangle AMB, <note>angle ∠AMB is a right angle (90 degrees)</note>, so triangle AMB is a right triangle. <note>Side AM and side BM are the legs</note>, <note>side AB is the hypotenuse</note>. Similarly, in triangle ANC, <note>angle ∠ANC is a right angle (90 degrees)</note>, so triangle ANC is a right triangle. <note>Side AN and side NC are the legs</note>, <note>side AC is the hypotenuse</note>.

Triangle Angle Sum Theorem

<note>In triangle AMB, angle ABD, angle BAG, and angle AMB are the three interior angles of triangle AMB</note>, according to the Triangle Angle Sum Theorem, <note>angle ABD + angle BAG + angle AMB = 180°</note>. <note>In triangle BMG, angle AGB, angle CBD, and angle BMG are the three interior angles of triangle BMG</note>, according to the Triangle Angle Sum Theorem, <note>angle AGB + angle CBD + angle BMG = 180°</note>.

Midpoint of a Line Segment

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

<note>The midpoint of line segment BD is point M</note>, <note>The midpoint of line segment CE is point N</note>. According to the definition of the midpoint of a line segment, <note>point M divides line segment BD into two equal parts, that is, BM = MD</note>; <note>point N divides line segment CE into two equal parts, that is, CN = NE</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 ABG, side AB and side BG are equal</note>, therefore <note>triangle ABG is an isosceles triangle</note>. <note>In triangle ACF, side AC and side CF are equal</note>, therefore <note>triangle ACF is an isosceles triangle</note>.

In the figure of this problem, in the isosceles triangle ABG, <note>the vertex angle is angle WBG</note>, <note>the base is side AG</note>. <note>The angle bisector AM of the vertex angle not only bisects the vertex angle ABG</note>, but also bisects the base AG, making <note>AM = GM and perpendicular to the base AG, forming a right angle AMB (90 degrees)</note>. Therefore, segment BM is both the angle bisector of the vertex angle, the median of the base, and the altitude. In the isosceles triangle ACF, <note>the vertex angle is angle ACF</note>, <note>the base is side AF</note>. <note>The angle bisector CN of the vertex angle not only bisects the vertex angle ACF</note>, but also bisects the base AF, making <note>AN = FN and perpendicular to the base AF, forming a right angle ANC (90 degrees)</note>. Therefore, segment CN is both the angle bisector of the vertex angle, the median of the base, and the altitude.


Question: As shown in the figure, in △ABC, ∠A=90°, AB=AC=3. Now △ABC is rotated counterclockwise around point B by a certain angle, and point C′ falls exactly on the line where the height from side BC is located. Then the area swept by side BC during the rotation is ()

Answer: 3π

1. Given ∠A = 90°, AB = AC = 3, draw the altitude AD, then point C' is located on the extension line of AD in the opposite direction.

2. Because ∠A = 90°, AB = AC = 3, ##definition of isosceles right triangle##, △ABC is an isosceles right triangle, according to the Pythagorean theorem, BC = ##√(AB?+AC?) = √(3?+3?) = 3√2##.

3. ##According to the theorem that the altitude, median, and angle bisector coincide in an isosceles triangle##, BD = DC.

4. Since △ABC rotates counterclockwise around point B by a certain angle, point C' happens to fall on the line where the altitude of side BC is located. ##Because BC' = BC = 3√2, BD = 1/2 BC, so BD = 1/2 BC', and because in the right triangle BC'D, according to the properties of a 30°-60°-90° triangle, ∠BC'D = 30°, ∠DBC' = 60°##.

5. According to the area formula of a sector ##(60/360) * π * (3√2)^2 = 3π, so## the area swept by side BC during the rotation is 3π.

直角三角形; 旋转; 垂线; 垂直平分线

Definition of Right Triangle

<note>In triangle ABC, angle BAC is a right angle (90 degrees)</note>, therefore triangle ABC is a right triangle. <note>Sides AB and AC are the legs, and side BC is the hypotenuse</note>. In <note>triangle BC'D, angle BDC' is a right angle (90 degrees)</note>, therefore triangle BC'D is a right triangle. <note>Sides BD and C'D are the legs, and side BC' is the hypotenuse</note>.

Definition of Isosceles Triangle

<note>Side AB is equal to side AC</note>, therefore <note>triangle ABC is an isosceles triangle</note>.

Pythagorean Theorem

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

In the figure of this problem, in the right triangle <note>ABC</note>, <note>angle BAC is a right angle (90 degrees)</note>, <note>sides AB and AC are the legs</note>, <note>side BC is the hypotenuse</note>, so according to the Pythagorean Theorem, ##<note>BC</note>? = <note>AB</note>? + <note>AC</note>?##, that is, <note>BC = √(AB? + AC?) = √(3? + 3?) = 3√2</note>.

Definition of Isosceles Right Triangle

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

<note>Triangle ABC is an isosceles right triangle</note>, in which <note>angle BAC is a right angle (90 degrees)</note>, and <note>sides AB and AC are equal right-angle sides</note>.

Formula for the Area of a Sector

The area \( X \) of a sector can be calculated using the formula \( X = \frac{\theta}{360} \times \pi \times r^2 \), where \( \theta \) is the measure of the central angle in degrees, and \( r \) is the radius length.

In this problem, the sector formed after rotation is defined by <note>∠DBC = 60°</note> and has a radius of <note>BC = 3√2</note>. According to the formula for the area of a sector, the area A of the sector can be calculated using the formula <note>A = (θ/360) * π * r²</note>, where <note>θ is the central angle in degrees, and r is the radius length</note>. Therefore, the area A of the sector is <note>(60/360) * π * (3√2)²</note> = 3π.

Properties of 30°-60°-90° Triangle

<note>In the 30°-60°-90° triangle BC'D</note>, <note>angle BC'D is 30 degrees</note>, <note>angle C'BD is 60 degrees</note>, <note>angle C'DB is 90 degrees</note>. <note>Side BC' is the hypotenuse</note>, <note>side BD is opposite the 30-degree angle</note>, <note>side C'D is opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle, <note>side BD is equal to half of side BC'</note>, <note>side C'D is equal to √3 times side BC'</note>. That is: <note>BD = 1/2 * BC'</note>, <note>C'D = BC' * √3</note>.

In <note>isosceles triangle ABC</note>, <note>the vertex angle is angle A</note>, and <note>the base is side BC</note>. <note>The angle bisector of the vertex angle AD</note> not only bisects <note>the vertex angle A</note>, but also bisects <note>the base BC</note>, making <note>BD = DC</note>, and is perpendicular to <note>the base BC</note>, thus forming <note>right angle ADB</note> and <note>right angle ADC (90 degrees)</note>. Therefore, <note>line segment AD</note> is both <note>the angle bisector of the vertex angle</note> and <note>the median and altitude of the base</note>.

Definition of Altitude

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

<note>The line segment AD from vertex A perpendicular to the opposite side BC is the altitude of vertex A</note>。<note>The line segment AD forms a right angle (90 degrees) with side BC</note>，which indicates that <note>the line segment AD is the perpendicular distance from vertex A to the opposite side BC</note>。


Question: In order to measure the width of the river AB, which is parallel on both sides, it is measured that ∠ACB=30°, ∠ADB=60°, CD=60m. What is the width of the river AB?

Answer: 30√{3}m

1. Given ∠ACB = 30°, ∠ADB = 60°, CD = 60 meters.

2. According to the definition of adjacent supplementary angles, ∠CDA is the adjacent supplementary angle of ∠ADB, ∠CDA = 180° - ∠ADB = 180° - 60° = 120°.

3. According to the triangle angle sum theorem, in triangle ACD, ∠CAD = 180° - ∠ACD - ∠ADC = 180° - 30° - 120° = 30°. Similarly, in right triangle ABD, ∠DAB = 180° - ∠ABD - ∠ADB = 180° - 90° - 60° = 30°.

4. According to the properties of an isosceles triangle, ∠CAD = ∠ACD = 30°, so AD = CD = 60 meters.

5. According to the properties of a 30°-60°-90° triangle, in right triangle ABD, BD : AB : AD = 1 : √3 : 2, so AB = √3/2 AD = √3/2 * 60 = 30√3.

7. Through the above reasoning, the final answer is 30√3 meters.

普通三角形; 直角三角形; 正弦; 余弦; 正切

Properties of Isosceles Triangle

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

In the <note>isosceles triangle ACD</note>, <note>angle DCA and angle CAD are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the sides opposite the equal angles are equal, i.e., side CD = side AD</note>.

Triangle Angle Sum Theorem

<note>In triangle ADC, angle ADC, angle DCA, and angle CAD are the three interior angles of triangle ADC</note>, according to the Triangle Angle Sum Theorem, <note>angle ADC + angle DCA + angle CAD = 180°</note>. Similarly, <note>In triangle ADB, angle ADB, angle DAB, and angle DBA are the three interior angles of triangle ADB</note>, according to the Triangle Angle Sum Theorem, <note>angle ADB + angle DAB + angle DBA = 180°</note>.

Definition of Linear Pair of Angles

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

The angles <note>∠CDA and ∠ADB have a common side AD, and their other sides CD and DB are extensions in opposite directions</note>, so <note>∠CDA and ∠ADB are adjacent supplementary angles</note>.

Properties of 30°-60°-90° Triangle

<note>In the figure of this problem, in the 30°-60°-90° triangle ADB, angle DAB is 30 degrees</note>, <note>angle ADB is 60 degrees</note>, <note>angle ABD is 90 degrees</note>. <note>Side AD is the hypotenuse</note>, <note>side DB is the side opposite the 30-degree angle</note>, <note>side AB is the side opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle, <note>side DB is equal to half of side AD</note>, <note>side AB is equal to side DB times √3</note>. That is: <note>DB = 1/2 * AD</note>, <note>AB = DB * √3</note>.


Question: As shown in the figure, it is known that point D is the midpoint of side AB, AF∥BC, CG:GA=3:1, BC=8, then AF equals ()

Answer: 4

1. Given AF∥BC, ##it can be known that ∠AFD=∠DEB (Parallel lines axiom 2, alternate interior angles are equal), and because ∠ADF=∠BDE (definition of vertical angles), then according to the similarity theorem (AA)##, we get △AFD∽△BED.

2. According to the proportionality of corresponding sides of similar triangles, we get AF:BE=AD:BD.

3. Since D is the midpoint of AB, i.e., AD:BD = 1:1, therefore AF:BE = 1:1, i.e., AF = BE.

4. Given AF∥BC, ##it can be known that ∠AFG=∠CEG (Parallel lines axiom 2, alternate interior angles are equal), and because ∠AGF=∠CGE (definition of vertical angles), then according to the similarity theorem (AA)##, we get ##△AGF##∽△CGE.

5. According to the proportionality of corresponding sides of similar triangles, we get CE:AF=CG:GA.

6. Given CG:GA=3:1, therefore CE:AF=3:1, i.e., CE=3AF.

7. Given BC=8, and BC=##CE-BE##, according to ##AF = BE and CE=3AF##, we get 8 =##3AF - AF##, i.e., BC=2AF.

8. Solving the above equation, we get 2AF = 8, i.e., AF = 4.

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

中点; 平行线; 平行四边形; 线段; 普通三角形

Midpoint of a Line Segment

Original: <note>The midpoint of line segment AB is point D</note>. According to the definition of the midpoint of a line segment, <note>point D divides line segment AB into two equal parts</note>, that is, <note>the lengths of line segments AD and DB are equal</note>. That is, <note>AD = DB</note>.

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 segments AF and BC are located in the same plane</note>, and <note>they do not intersect</note>, so according to the definition of parallel lines, <note>line segments AF and BC are parallel lines, i.e., AF∥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.

<note>In the figure of this problem, triangle AFD and triangle BED are similar triangles.</note> According to the definition of similar triangles: <note>∠AFD = ∠BED</note>, <note>∠ADF = ∠BDE</note>, <note>∠DFA = ∠DEB</note>; <note>AF/BE = AD/BD = DF/DE</note>. <note>In the similar triangles AFG and CEG</note>, according to the definition of similar triangles: <note>∠AFG = ∠CEG</note>, ∠FAG = ∠GCE, <note>∠GAF = ∠GCE</note>; <note>AF/CE = AG/CG = FG/EG</note>.

Similarity Theorem for Triangles (AA)

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

<note>In the figure of this problem, in triangle ADF and triangle BDE, angle ADF is equal to angle BDE</note>, and <note>angle AFD is equal to angle DEB</note>, so <note>triangle ADF is similar to triangle BDE</note>. Similarly, <note>in triangle AGF and triangle CGE, angle AGF is equal to angle CGE</note>, and <note>angle AFG is equal to angle GEC</note>, so <note>triangle AGF is similar to triangle CGE</note>.

Definition of Vertical Angles

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

<note>In the figure of this problem, two intersecting lines AB and EF intersect at point D</note>, forming four angles: <note>angle ADF, angle BDE, angle ADE, and angle BDF</note>. According to the definition of vertical angles, <note>angle ADF and angle BDE are vertical angles</note>, <note>angle ADE and angle BDF are vertical angles</note>. Since vertical angles are equal, <note>angle ADF = angle BDE</note>, <note>angle ADE = angle BDF</note>. Similarly, <note>two intersecting lines AC and EF intersect at point G</note>, forming four angles: <note>angle AGF, angle CGE, angle AGE, and angle CGF</note>. According to the definition of vertical angles, <note>angle AGF and angle CGE are vertical angles</note>, <note>angle AGE and angle CGF are vertical angles</note>. Since vertical angles are equal, <note>angle AGF = angle CGE</note>, <note>angle AGE = angle CGF</note>.

Definition of Alternate Interior Angles

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

<note>In the diagram of this problem, two parallel lines AF and CE are intersected by line AB</note>, where <note>angle AFD and angle DEB are located between the two parallel lines and on opposite sides of the intersecting line AB</note>, therefore <note>angle AFD and angle DEB are alternate interior angles</note>. <note>Alternate interior angles are equal, that is, angle AFD is equal to angle DEB</note>. Similarly, <note>two parallel lines AF and CE are intersected by line AC</note>, where <note>angle AFG and angle CEG are located between the two parallel lines and on opposite sides of the intersecting line AC</note>, therefore <note>angle AFG and angle CEG are alternate interior angles</note>. <note>Alternate interior angles are equal, that is, angle AFG is equal to angle CEG</note>.

Parallel Postulate 2 of Parallel Lines

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

<note>In the diagram of this problem, the two parallel lines AF and CE</note> are intersected by the third line <note>AC and AB</note>, forming the following geometric relationships: alternate interior angles: <note>∠AFD=∠DEB, ∠AFG=∠CEG</note>. These relationships indicate that when two parallel lines are intersected by a third line, the alternate interior angles are equal.


Question: As shown in the figure, DE∥BC, BD and CE intersect at O, rac{EO}{OC}=rac{1}{3}, AE=3, then EB=()

Answer: 6

1. Given DE∥BC, according to the parallel line axiom 2, alternate interior angles are equal, we know ∠EDO=∠CBO, ∠DEO=∠BCO. Then, according to the similarity theorem (AA), we get △EOD∽△COB.

2. From △EOD∽△COB, according to the definition of similar triangles, we have: DE/BC=EO/OC=1/3.

3. Since DE∥BC, according to the parallel line axiom 2, corresponding angles are equal, we know ∠AED=∠ABC, ∠ADE=∠ACB. Then, according to the similarity theorem (AA), we can conclude △AED∽△ABC.

4. From △AED∽△ABC, according to the definition of similar triangles, we get: DE/BC=AE/AB.

5. Combining step 2 and step 4, we get: AE/AB=1/3.

6. Given AE=3, therefore, according to the proportion, we can solve AB=3*3=9.

7. Finally, since AB=AE+EB, we can conclude EB=AB-AE=9-3=6.

8. After the above reasoning, the final answer is 6.

平行线; 普通三角形; 平移; 线段; 位似

Definition of Similar Triangles

In the figure of this problem, <note>triangle EOD and triangle COB are similar triangles</note>. According to the definition of similar triangles: <note>∠EOD = ∠COB</note>, <note>∠ODE = ∠OBC</note>, <note>∠DEO = ∠BCO</note>; <note>DE/BC = EO/OC = DO/BO</note>. Similarly, <note>triangle AED and triangle ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠AED = ∠ABC</note>, <note>∠ADE = ∠ACB</note>, <note>∠DAE = ∠BAC</note>; <note>DE/BC = AE/AB = AD/AC</note>.

Parallel Postulate 2 of Parallel Lines

In the diagram of this problem, <note>two parallel lines DE and BC</note> are intersected by a third line DB, forming the following geometric relationship: alternate interior angles: <note>angle EDO and angle CBO are equal</note>. Similarly, <note>two parallel lines DE and BC</note> are intersected by a third line EC, forming the following geometric relationship: alternate interior angles: <note>angle DEO and angle BCO are equal</note>. Similarly, <note>two parallel lines DE and BC</note> are intersected by a third line AB, forming the following geometric relationship: corresponding angles: <note>angle AED and angle ABC are equal</note>. Similarly, <note>two parallel lines DE and BC</note> are intersected by a third line AC, forming the following geometric relationship: corresponding angles: <note>angle ADE and angle ACB are equal</note>. These relationships illustrate that when <note>two parallel lines are intersected by a third line</note>, corresponding angles are equal, and alternate interior angles are equal.

Definition of Alternate Interior Angles

In the diagram of this problem, <note>two parallel lines DE and BC</note> are intersected by a <note>line BD</note>, where <note>angle EDO and angle CBO are located between the two parallel lines</note> and <note>on opposite sides of the intersecting line BD</note>. Therefore, <note>angle EDO and angle CBO are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle EDO is equal to angle CBO</note>. Similarly, <note>two parallel lines DE and BC</note> are intersected by a <note>line EC</note>, where <note>angle DEO and angle BCO are located between the two parallel lines</note> and <note>on opposite sides of the intersecting line EC</note>. Therefore, <note>angle DEO and angle BCO are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle DEO is equal to angle BCO</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 a <note>line AB</note>, where <note>angle AED and angle ABC are on the same side of the transversal AB</note>, on the same side of the intersected <note>lines DE and BC</note>, therefore, <note>angle AED and angle ABC are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle AED is equal to angle ABC</note>. Similarly, two <note>parallel lines DE and BC</note> are intersected by a <note>line AC</note>, where <note>angle ADE and angle ACB are on the same side of the transversal AC</note>, on the same side of the intersected <note>lines DE and BC</note>, therefore, <note>angle ADE and angle ACB are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle ADE is equal to angle ACB</note>.

Similarity Theorem for Triangles (AA)

<note>Angle EDO is equal to angle CBO</note>, <note>angle DEO is equal to angle BCO</note>, so triangle EDO is similar to triangle CBO. Similarly, in triangles AED and ABC, <note>angle AED is equal to angle ABC</note>, <note>angle ADE is equal to angle ACB</note>, so triangle AED is similar to triangle ABC.


Question: As shown in the figure, in the isosceles △ABC, one of the equal sides AB is 4 cm long. Through any point D on the base BC, draw lines parallel to the two equal sides, intersecting the two equal sides at E and F respectively. Then the perimeter of quadrilateral AEDF is ()

Answer: 8厘米

1. According to the problem statement, draw auxiliary lines DE∥AC and DF∥AB, intersecting the two legs AB and AC at points E and F respectively.

2. Based on ##Parallel Line Axiom 2, corresponding angles are equal##, we obtain ∠EDB=∠ACB and ∠FDC=∠ABC.

3. In the isosceles triangle ABC, it is known that AB=AC, therefore according to the properties of isosceles triangles, we have ∠ABC=∠ACB.

4. From the previous two steps, we conclude that ∠EDB=∠ABC and ∠FDC=∠ACB.

5. ##Since ∠EDB=∠ABC and ∠FDC=∠ACB, based on the properties and definition of isosceles triangles##, we obtain DE=BE and DF=FC.

6. The perimeter of quadrilateral AEDF is AE + ED + DF + AF.

7. Among these, AE + ED = AB and DF + AF = AC.

8. Since AB=AC=4 cm, therefore AE + ED + DF + AF = AB + AC = 2AB = 2×4 = 8 cm.

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

等腰三角形; 平行线; 平行四边形; 线段

Definition of Isosceles Triangle

<note>In the diagram of this problem, in triangle ABC, side AB and side AC are equal</note>, therefore triangle ABC is an isosceles triangle. <note>Similarly, in triangle EBD, side EB and side ED are equal</note>, therefore triangle EBD is an isosceles triangle. Similarly, <note>in triangle FDC, side FD and side FC are equal</note>, therefore triangle FDC is an isosceles triangle.

Definition of Parallel Lines

In the diagram of this problem, <note>line DE and line AC</note> lie in the same plane and do not intersect, so according to the definition of parallel lines, <note>line DE and line AC are parallel lines</note>. Similarly, <note>line DF and line AB</note> lie in the same plane and do not intersect, so according to the definition of parallel lines, <note>line DF and line AB are parallel lines</note>.

Parallel Postulate 2 of Parallel Lines

<note>In the diagram of this problem, two parallel lines DE and AC are intersected by a third line DC</note>, forming the following geometric relationship: <note>corresponding angles ∠EDB and ∠ACB are equal</note>. <note>Similarly, two parallel lines DF and AB are intersected by a third line BD</note>, forming the following geometric relationship: <note>corresponding angles ∠FDC and ∠ABC are equal</note>. These relationships illustrate that when two parallel lines are intersected by a third line, the corresponding angles are equal.

Properties of Isosceles Triangle

In the diagram of this problem, <note>in isosceles triangle ABC, sides AB and AC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠ABC = ∠ACB</note>. Similarly, <note>in isosceles triangle EBD, sides EB and ED are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠EDB = ∠ABC</note>. Similarly, <note>in isosceles triangle FDC, sides FD and FC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>∠FDC = ∠ACB</note>.

Definition of Corresponding Angles

<note>Two parallel lines ED and AC are intersected by a line DC</note>, where <note>angle EDB and angle ACB are on the same side of the intersecting line DC and on the same side of the two intersected lines ED and AC</note>, therefore, <note>angle EDB and angle ACB are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle EDB is equal to angle ACB</note>. Similarly, <note>two parallel lines FD and AB are intersected by a line BD</note>, where <note>angle ABC and angle FDC are on the same side of the intersecting line BD and on the same side of the two intersected lines AB and FD</note>, therefore, <note>angle ABC and angle FDC are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle ABC is equal to angle FDC</note>.


Question: As shown in the figure, in △ABC, AB=AC, ∠A=40°, DE is the perpendicular bisector of AC, then the measure of ∠BCD is equal to ()

Answer: 30°

1. Given AB=AC and ∠A=40°, according to the properties of an isosceles triangle, we get ∠ABC=∠ACB.

2. By the triangle angle sum theorem, we get ∠ABC=∠ACB= (180° - ∠A) / 2 = 70°.

3. According to the properties of the perpendicular bisector, AD=DC.

4. According to the properties of an isosceles triangle, ∠ACD=∠A=40°.

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

等腰三角形; 垂直平分线; 垂线; 三角形的外角

Definition of Isosceles Triangle

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

Properties of Isosceles Triangle

In the figure of this problem, in <note>isosceles triangle △ABC</note>, <note>sides AB and AC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle ABC = angle ACB</note>. In <note>isosceles triangle △ADC</note>, <note>sides AD and DC are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle A = angle DCA</note>.

Triangle Angle Sum Theorem

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

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

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

<note>Line AC is the perpendicular bisector of segment DE</note>, <note>point E is on line AC</note>. According to the properties of the perpendicular bisector, <note>the distances from point D to the endpoints A and C of segment AC are equal, i.e., AD = CD</note>.


Question: As shown in the figure, in △ABC, D is a point on side AC. If ∠DBC=∠A, BC=3, AC=6, then the length of CD is ()

Answer: \frac{3}{2}

1. Given ∠DBC = ∠BAC, and BC=3, AC=6.

2. Since ∠DBC = ∠BAC, and ∠BCA = ∠BCA (common angle), according to the triangle similarity criterion (AA), we get △BCD ∽ △ACB.

3. According to the properties of similar triangles, corresponding sides are proportional, we get: CD/BC = BC/AC.

4. Substituting the given conditions, CD/3 = 3/6.

5. Solving the proportion, we get: CD = 3/2.

6. Through the above reasoning, we finally get the answer CD = 3/2.

普通三角形; 等腰三角形; 线段

Definition of an Angle

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

<note>∠BAC is a geometric figure composed of rays AB and AC</note>, these two rays have a <note>common endpoint A</note>. This <note>common endpoint A is called the vertex of angle BAC</note>, and <note>rays AB and AC are called the sides of angle BAC</note>. Similarly, <note>∠DBC is a geometric figure composed of rays DB and BC</note>, these two rays have a <note>common endpoint D</note>. This <note>common endpoint D is called the vertex of angle DBC</note>, and <note>rays DB and BC are called the sides of angle DBC</note>. <note>∠BCA is a geometric figure composed of rays BC and CA</note>, these two rays have a <note>common endpoint C</note>. This <note>common endpoint C is called the vertex of angle BCA</note>, and <note>rays BC and CA are called the sides of angle BCA</note>.

Similarity Theorem for Triangles (AA)

<note>In triangles BCD and ACB</note>, <note>∠DBC is equal to ∠BAC</note>, and <note>∠BCA is equal to ∠BCA (common angle)</note>, so <note>triangle BCD is similar to triangle ACB</note>.

Definition of Similar Triangles

<note>Triangles BCD and ACB are similar triangles</note>. According to the definition of similar triangles: <note>∠DBC = ∠BAC</note>, ∠BCA = ∠BCA; <note>CD/BC = BC/AC</note>.


Question: As shown in the figure, it is known that PA and PB are tangents to ⊙O, A and B are points of tangency, AC is the diameter of ⊙O, ∠P=40°, then the degree of ∠BAC is ()

Answer: 20°

1. Given that PA and PB are tangents to ⊙O, A and B are points of tangency, and AC is the diameter, ##connecting BC and OB##, according to (Corollary 2 of the Inscribed Angle Theorem) the inscribed angle subtended by the diameter is a right angle, ##we obtain ∠ABC=90°##.

2. Given that PA and PB are tangents to ⊙O, A and B are points of tangency, according to ##the property of the tangent to a circle##, we obtain ∠OAP=90° and ∠OBP=90°.

3. Since ##the sum of the interior angles of a quadrilateral is 360°##, we obtain ∠AOB=360°-∠OAP-∠OBP-∠P=360°-90°-90°-40°=140°.

4. According to the Inscribed Angle Theorem, ##the inscribed angle ACB is equal to half of the central angle AOB subtended by arc AB##, therefore we obtain ∠ACB=∠AOB / 2=70°.

5. In triangle ABC, since ∠ABC=90° and ∠ACB=70°, by the Triangle Sum Theorem we obtain ∠BAC=180°-90°-∠ACB=20°.

6. Through the above reasoning, we finally obtain the answer as 20°.

圆; 切线; 等腰三角形; 圆周角; 三角形的外角

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>AC is the diameter</note>, connecting <note>the center O</note> and <note>points A and C on the circumference</note>, with a length of <note>2 times the radius</note>, that is, <note>AC = 2 * OA</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 O and the lines PA and PB have only one common point A and B respectively, these common points are called <note>points of tangency</note>. Therefore, <note>the lines PA and PB are tangents to the circle O</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.

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

Definition of Central Angle

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

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

Inscribed Angle Theorem

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

In the diagram of this problem, in circle O, points A, B, and C are on the circle, the central angle corresponding to <note>arc AB</note> is <note>∠AOB</note>, and the inscribed angle is <note>∠ACB</note>. According to the Inscribed Angle Theorem, <note>∠ACB is equal to half of the central angle ∠AOB corresponding to the 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, points A and B are the points of tangency where lines PA and PB touch the circle</note>, <note>line segments OA and OB are the radii of the circle</note>. According to the property of the tangent line to a circle, <note>tangent lines PA and PB are perpendicular to the radii OA and OB at points of tangency A and B</note>, that is, <note>∠OAP=90° and ∠OBP=90°</note>.

Triangle Angle Sum Theorem

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

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

In the figure of this problem, the quadrilateral <note>AOBP</note> has four interior angles: <note>角AOB</note>, <note>角OAP</note>, <note>角OBP</note>, and <note>角P</note>. According to the Sum of Interior Angles of a Quadrilateral Theorem, the sum of these four interior angles is 360°, that is, <note>角AOB + 角OAP + 角OBP + 角P = 360°</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 angle subtended by the diameter AC is a right angle (90 degrees).


Question: As shown in the figure, it is known that a fisherman on a fishing boat at point A sees the lighthouse M in the direction of 60° north-east. The fishing boat sails eastward at a speed of 28 nautical miles per hour and reaches point B half an hour later. At point B, the fisherman sees the lighthouse M in the direction of 15° north-east. At this time, the distance between the lighthouse M and the fishing boat is ()

Answer: 7√{2}海里

1. Given that the fishing boat travels eastward at a speed of 28 nautical miles per hour and reaches point B after half an hour, then AB = ##1/2## × 28 = 14 nautical miles.

##2. Draw BN perpendicular to AM through point B and intersect at point N.##

3. From the angles in the figure, it is known that ∠MAB = ##30°，∠NBA equals 60°，obviously ∠NBM equals 90°-60°＋15°##

4. In the right triangle △ABN, ∠BAN = 30 degrees, according to the definition of a right triangle, BN = AB × sin(∠BAN) = 14 × sin(30 degrees) = 14 × ##1/2## = 7 nautical miles.

5. Because in the right triangle △BNM, ∠MBN = 45 degrees, according to the properties of a right triangle, this triangle is an isosceles right triangle, so BN = MN.

6. Given BN = 7 nautical miles, then MN = 7 nautical miles.

7. Through the properties of a right triangle, BM = ##√（BN^2 + MN^2） =√（7^2 + 7^2） = √（49 + 49） = 7√2## nautical miles.

8. Through the above reasoning, it is finally concluded that BM = 7√2 nautical miles.

方向角; 普通三角形; 直角三角形

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 ABN, angle ∠ABN is a right angle (90 degrees)</note>, therefore triangle ABN is a right triangle. Side AB and side BN are the legs, and side AN is the hypotenuse. <note>In triangle BNM, angle ∠BNM is a right angle (90 degrees)</note>, therefore triangle BNM is a right triangle. Side BN and side NM are the legs, and side BM is the hypotenuse.

Triangle Angle Sum Theorem

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

Definition of Sine Function

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

In the figure of this problem, <note>in the right triangle △ABN</note>, <note>angle ∠BAN is an acute angle</note>, <note>side BN is the opposite side of angle ∠BAN</note>, <note>side AB is the hypotenuse</note>. According to the definition of the sine function, <note>the sine value of angle ∠BAN is equal to the ratio of the opposite side BN to the hypotenuse AB</note>, that is, <note>sin(∠BAN) = BN / AB</note>.

Definition of Isosceles Right Triangle

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

<note>Triangle BNM is an isosceles right triangle</note>, in which <note>angle ∠BNM is a right angle (90 degrees)</note>, <note>sides BN and MN are equal right-angle sides</note>.

Pythagorean Theorem

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

In the figure of this problem, in the right triangle BNM, <note>angle BNM is a right angle (90 degrees)</note>, <note>sides BN and MN are the legs</note>, <note>side BM is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>##BM^2 = BN^2 + MN^2##</note>.


Question: As shown in the figure, in △ABC, ∠ACB=90°, D is a point on AB, connect CD, ∠ACD=∠B, if BC=13cm, CD=5cm, then BD=()

Answer: 12cm

####

##1##. Since ∠ACD=∠B, and ∠A=∠A, therefore △ADC∽△ACB (the condition for similar triangles is that two corresponding angles of the two triangles are equal respectively).

##2. According to the definition of similar triangles##, corresponding angles are equal, corresponding sides are proportional, thus ∠ADC=∠ACB.

##3##. Since ∠ACB=90°, therefore ∠ADC=90°, i.e., ##triangle BCD is a right triangle##, and ∠BDC=90°.

##4##. By ##Pythagorean theorem##, we have BD=√(##BC^2 - CD^2##),

##5##. Substituting the given conditions BC=13cm and CD=5cm, we get BD=√(##13^2 - 5^2##)=√(169 - 25)=√144=12cm.

##6##. Through the above reasoning, the final answer is BD=12cm.

直角三角形; 等腰三角形; 线段; 三角形的外角; 垂线

Similarity Theorem for Triangles (AA)

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

<note>In the given diagram, in triangles ACD and ABC</note>, if <note>angle A is equal to angle A</note> and <note>angle ACD is equal to angle B</note>, then <note>triangle ACD is similar to triangle ABC</note>.

Definition of Right Triangle

In the figure of this problem, in triangle ABC, <note>angle ACB is a right angle (90 degrees)</note>, so triangle ABC is a right triangle. <note>Side AC and side BC are the legs</note>, <note>side AB is the hypotenuse</note>. Similarly, in triangle BDC, <note>angle BDC is a right angle (90 degrees)</note>, so triangle BDC is a right triangle. <note>Side CD and side BD are the legs</note>, <note>side BC 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 diagram of this problem, <note>triangle ADC and triangle ACB are similar triangles</note>. According to the definition of similar triangles: <note>∠DAC = ∠BAC</note>, <note>∠ACD = ∠ABC</note>, <note>∠ADC = ∠ACB</note>; <note>AD/AC = DC/BC = AC/AB</note> (common side AC).

Pythagorean Theorem

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


Question: As shown in the figure, the slope of the conveyor belt and the ground is 1:2. It transports an object from point A on the ground to point B, which is 2 meters above the ground. Then the distance the object travels from A to B is ()

Answer: 2√{5}

1. Given that the slope between the conveyor belt and the ground is 1:2, and the object is transported from point A to point B which is 2 meters above the ground.

2. ##According to the definition of a right triangle,## in the right triangle ABC, AB is the hypotenuse, BC is the vertical height raised by the conveyor belt, and AC is the horizontal projection of the conveyor belt on the ground.

3. Let the length of BC be 2 meters, then according to the slope ratio of 1:2, we get BC:AC=1:2.

4. According to the ##slope## relationship, BC=2 meters, so AC=2×2=4 meters.

5. ##According to## the Pythagorean theorem, we have ##AB^2 = AC^2 + BC^2##, that is ##AB^2 = 4^2 + 2^2##=16+4=20.

6. Therefore, AB=√20 = 2√5.

7. Through the above reasoning, it is concluded that the distance the object travels from A to B is 2√5 meters.

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

Slope

The slope refers to the ratio of the vertical height to the horizontal distance of a particular line or segment.

In the figure of this problem, the slope formed by the conveyor belt and the ground has a <note>slope of 1:2, that is, the ratio of the vertical height BC to the horizontal distance AC is 1:2</note>. From the problem, it is known that <note>BC=2 meters</note>, so we can deduce that <note>AC=4 meters</note>.

Definition of Right Triangle

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


Question: As shown in the figure, a beam of light starts from point A(-3,3), passes through point C on the y-axis, and then reflects to pass through point B(-1,0). The length of the path of the light from point A to point B is ()

Answer: 5

1. Given the coordinates of point A are (-3, 3) and the coordinates of point B are (-1, 0), a ray from point A passes through point C on the y-axis and reflects to pass through point B.

2. Construct the symmetric point B′ of point B with respect to the y-axis, then the coordinates of B′ are (1, 0), and CB = CB′.

3. Construct AD perpendicular to the x-axis intersecting at point D, then the coordinates of point D are (-3, 0).

4. According to the properties of a right triangle, AD = 3, DB′ = 1 + 3 = 4.

5. Use the Pythagorean theorem to calculate the length of AB′. In the right triangle ADB′, AB′ = ##$	ext{sqrt}(AD^2 + DB′^2)$ = $	ext{sqrt}(3^2 + 4^2)$ = $	ext{sqrt}(9 + 16)$ = $	ext{sqrt}(25)$ = 5##.

6. Since the ray reflects at point C, the path includes AC and CB, and the path length equals AB′.

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

反射; 线段; 直线; 等腰三角形; 对称

Symmetric Point

A symmetric point refers to a point that is symmetric with respect to a given line (or a given point) such that the two points are equidistant from the line (or point).

<note>Point B(-1, 0) is symmetric to the y-axis at point B′(1, 0)</note>, because <note>the distance from point B to the y-axis is equal to the distance from point B′ to the y-axis</note>, and <note>the two points are on opposite sides of the y-axis</note>.

Definition of Right Triangle

<note>In triangle ADB′, angle ADB′ is a right angle (90 degrees)</note>, therefore triangle ADB′ is a right triangle. <note>Sides AD and DB′ are the legs</note>, <note>side AB′ is the hypotenuse</note>.

Pythagorean Theorem

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

Law of Reflection

When a ray of light reflects, the angle of incidence is equal to the angle of reflection, and the reflected ray lies on the same plane as the incident ray and the normal.

<note>The light ray reflects off point C on the y-axis from point A and then passes through point B</note>. According to the law of reflection, <note>the angle of incidence is equal to the angle of reflection</note>, and the reflected ray lies on the same side of the plane as the incident ray and the normal. Since <note>point B's symmetric point with respect to the y-axis is B′</note>, the <note>path lengths of AC and CB equal AB′</note>, which means <note>the path length is 5</note>.


Question: As shown in the figure, in △ABC, ∠BAC=90°, AD⊥BC at D, if AB=3 and BC=5, then the length of DC is ()

Answer: \frac{16}{5}

1. Given ∠BAC=90°, AD⊥BC at D, and AB=3, BC=5. In △ABC and △ADB, ∠ABC=∠ABD.

2. Because AD⊥BC, ∠ADB=90°. At the same time, ∠CAB=90°, according to the similarity theorem (AA), △ADB∽△ABC.

####

3. According to the definition of similar triangles, we get AB/BC=BD/BA.

4. Based on the given conditions AB=3, BC=5, we can write: 3/5 = BD/3.

5. Solving this proportion equation, we get BD = (3*3)/5 = 9/5.

6. Since BD and DC are parts of line segment BC, DC=BC-BD.

7. Given BC=5, BD=9/5, we get: DC = 5 - (9/5) = 25/5 - 9/5 = 16/5.

8. Through the above reasoning, the final answer is 16/5.

直角三角形; 垂线; 线段

Definition of Right Triangle

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

Definition of Perpendicular Lines

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

In the diagram of this problem, <note>line segment AD and line segment BC</note> intersect to form <note>angle ∠ADB is 90 degrees</note>, therefore according to the <note>definition of perpendicular lines</note>, line segment AD and line segment BC are perpendicular to each other.

Definition of Similar Triangles

<note>Triangle ABD and triangle ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠BAD = ∠BCA</note>, <note>∠ADB = ∠CAB</note>, <note>∠ABD = ∠ABC</note>; <note>AB/BC = BD/AB = AD/AC</note>.

Definition of Line Segment

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

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

Similarity Theorem for Triangles (AA)

<note>In the given problem diagram, in triangles ABD and ABC, ##if angle ABD is equal to angle ABC, and angle ADB is equal to angle CAB##, then triangle ABD is similar to triangle ABC.####</note>


Question: As shown in the figure, in the right triangle △ABC, ∠BAC=90°, AD⊥BC at D, DE⊥AB at E, AD=3, DE=2, then the length of CD is ()

Answer: \frac{3√{5}}{2}

1. Given AD is perpendicular to BC, DE is perpendicular to AB, therefore ∠ADC=∠AED=90°.

2. Given ∠BAC=90°, ##so ∠DAE+∠DAC=90°##, according to the triangle angle sum theorem and ∠ADC=90°, in triangle ADC ∠C+∠DAC=180°-90°=90°, therefore ∠DAE=∠C.

3. ##Since ∠ADC=∠AED=90°, ∠DAE=∠C, according to the similarity criterion for triangles (AA)## Rt△ACD is similar to Rt△DAE.

4. ##According to the definition of similar triangles##, the similarity ratio of △ACD and △DAE is AD/AC=DE/AD.

5. Given AD=3 and DE=2, therefore substituting the values into the similarity ratio, we get 3/AC=2/3.

7. In Rt△ACD, applying the Pythagorean theorem: ##CD^2+AD^2=AC^2##.

8. Substituting the values into the Pythagorean theorem, we get ##CD^2+3^2=(9/2)^2##.

11. Through the above reasoning, the final answer is 3√5/2.

直角三角形; 垂线; 平行线

Definition of Right Triangle

In <note>triangle ADC and triangle AED</note>, <note>angle ADC and angle AED are right angles (90 degrees)</note>, therefore <note>triangle ADC and triangle AED are right triangles</note>. <note>Sides AD and CD are the legs of triangle ADC</note>, <note>side AC is the hypotenuse</note>. <note>Sides AE and DE are the legs of triangle AED</note>, <note>side AD is the hypotenuse</note>.

Definition of Similar Triangles

<note>Triangles ACD and DAE are similar triangles</note>. According to the definition of similar triangles: <note>∠ACD = ∠DAE</note>, <note>∠ADC = ∠AED</note>, <note>∠CAD = ∠ADE</note>; <note>AD/AC = DE/AD</note>.

Pythagorean Theorem

<note>In the right triangle ACD, angle ∠ADC is a right angle (90 degrees)</note>, <note>sides AD and CD are the legs</note>, <note>side AC is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AC^2 = AD^2 + CD^2</note>.

Triangle Angle Sum Theorem

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

<note>Angles DEA, DAE, and ADE are the three interior angles of triangle AED</note>, according to the Triangle Angle Sum Theorem, <note>angle DEA + angle DAE + angle ADE = 180°</note>. <note>Angles ACD, ADC, and DAC are the three interior angles of triangle ACD</note>, according to the Triangle Angle Sum Theorem, <note>angle ACD + angle ADC + angle DAC = 180°</note>.

Definition of Perpendicular Lines

Original: <note>AD⊥BC于D</note>, <note>DE⊥AB于E</note>, therefore according to the definition of perpendicular lines, <note>∠ADC=∠AED=90°</note>.

Similarity Theorem for Triangles (AA)

In the original text: <note>△ACD and △DAE</note>, <note>∠ADC=∠AED=90°</note>, and <note>∠DAE=∠C</note>, so <note>triangles △ACD and △DAE are similar</note>.


Question: As shown in the figure, in quadrilateral ABCD, ∠BAD=130°, ∠B=∠D=90°, points E and F are moving points on segments BC and DC respectively. When the perimeter of △AEF is minimized, the degree of ∠EAF is ()

Answer: 80°

1. Construct points A' and A" as the symmetric points of A with respect to segments BC and CD respectively. Connect A'A" and intersect BC at point E, and intersect CD at point F.

2. Since A' and A" are the symmetric points of A with respect to BC and CD respectively, according to symmetry ####, A'A" is the minimum perimeter of triangle AEF.

3. Extend line DA and take point H such that AH is on the extended line DA.

4. ## Connect AA' and AA", since ∠BAD = 130°, by the triangle angle sum theorem, we know ∠AA'A" + ∠AA"A' = 180° - ∠BAD = 50° ##

5. ## Observing triangle AEF, by symmetry we know AE = EA', AF = FA", by the definition of an isosceles triangle, triangles AEA' and AFA" are both isosceles triangles. By the properties of isosceles triangles, we know ∠AA'E = ∠A'AE, ∠FAA" = ∠FA"A ##

6. ## Therefore, by the exterior angle theorem of triangles, we know ∠AEF + ∠AFE = 2(∠AA'A" + ∠AA"A') = 100° ##

7. ## Therefore, according to the triangle angle sum theorem, we know ∠EAF = 180° - 100° = 80°. ##

普通四边形; 直角三角形; 垂线; 垂直平分线; 内错角

Symmetric Point

<note>A' is the symmetric point of A with respect to BC</note>, <note>A" is the symmetric point of A with respect to CD</note>, and <note>A' and A" are located on BC and CD respectively</note>, such that <note>A'A is perpendicular and bisected by BC</note>, <note>A"A is perpendicular and bisected by CD</note>.

Symmetry

Symmetry refers to a geometric figure or pattern remaining invariant under certain operations such as rotation, reflection, or translation.

In the problem diagram, <note>the shape AEA' has symmetry</note>. Specifically, if <note>the shape AEA' is symmetric about the line BC</note>, then <note>the line BC is the axis of symmetry of the shape AEA'</note>. Each part of the shape AEA' has a symmetric corresponding part on the other side of the axis of symmetry. If <note>the shape AEA' is symmetric about the point O</note>, then <note>the point O is the center of symmetry of the shape AEA'</note>. Each part of the shape AEA' has a symmetric corresponding part on the other side of the center of symmetry. If <note>the shape AEA' remains unchanged after rotating by θ degrees</note>, then <note>the shape AEA' has rotational symmetry</note>, and <note>θ degrees is its angle of symmetry</note>. Through these symmetrical operations, the shape AEA' visually remains unchanged. In the problem diagram, <note>the shape AFA" has symmetry</note>. Specifically, if <note>the shape AFA" is symmetric about the line CD</note>, then <note>the line CD is the axis of symmetry of the shape AFA"</note>. Each part of the shape AFA" has a symmetric corresponding part on the other side of the axis of symmetry. If

Triangle Angle Sum Theorem

In the figure of this problem, in triangle AEA', <note>sides AE and EA' are equal</note>, therefore triangle AFA" is an <note>isosceles triangle</note>. In triangle AFA", <note>sides AF and FA' are equal</note>, therefore triangle AFA" is an isosceles triangle.

Properties of Isosceles Triangle

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

In the figure of this problem, <note>the sides AE and EA' of the isosceles triangle AEA are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle AA'E = angle A'AE</note>. <note>The sides AF and FA" of the isosceles triangle AFA" are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle FA"A = angle FAA"</note>.

Exterior Angle Theorem of Triangle

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

<note>In the figure of this problem, in triangle AEA', angle AEF is an exterior angle of the triangle</note>, <note>angle AA'E and angle A'AE are the two non-adjacent interior angles to the exterior angle AEF</note>, according to the exterior angle theorem of a triangle, <note>the exterior angle AEF is equal to the sum of the two non-adjacent interior angles AA'E and A'AE</note>, that is, <note>angle AEF = angle AA'E + angle A'AE</note>. <note>In triangle AFA", angle AFE is an exterior angle of the triangle</note>, <note>angle FAA'' and angle FA''A are the two non-adjacent interior angles to the exterior angle AFE</note>, according to the exterior angle theorem of a triangle, <note>the exterior angle AFE is equal to the sum of the two non-adjacent interior angles FAA'' and FA''A</note>, that is, <note>angle AFE = angle FAA'' + angle FA''A</note>.

Definition of Exterior Angle of a Polygon

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

The interior angle of <note>polygon AEA</note> is <note>∠AEA'</note>. The angle <note>∠AEF</note> formed by extending the adjacent sides <note>AE</note> and <note>A'E</note> of this interior angle is called the exterior angle of the interior angle <note>∠AEA'</note>. The interior angle of <note>polygon AFA"</note> is <note>∠AFA"</note>. The angle <note>∠AFE</note> formed by extending the adjacent sides <note>AF</note> and <note>A''F</note> of this interior angle is called the exterior angle of the interior angle <note>∠AFA"</note>.