Question: As shown in the figure, DC∥EF∥AB, if \frac{EG}{AB}=\frac{1}{2}, DC=6, then the length of GF is ()

Answer: 3

1. Given DC∥EF∥AB.

2. According to ∠DEG＝∠DAB, ∠DGE＝∠DBA, the similarity theorem of triangles (AA), triangle DEG is similar to triangle DAB, denoted as △DEG ∽ △DAB.

3. Because △DEG ∽ △DAB, according to the definition of similar triangles, we have DG/DB = EG/AB.

4. From the problem statement, we know EG/AB = 1/2, thus DG/DB = 1/2.

5. Because DG/DB = 1/2, it means point G is the midpoint of segment DB.

6. Given DC∥EF, and point G is the midpoint of DB, according to the parallel line axiom 2, ∠GFB＝∠DCB, and ∠GBF＝∠DBC, according to the similarity theorem of triangles (AA), triangle GFB is similar to triangle DCB, and according to the definition of similar triangles, GF = 1/2DC.

7. Therefore, GF = 1/2DC.

8. Given DC = 6, thus GF = (1/2)× 6 = 3.

Definition of Similar Triangles

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

<note>Triangles DEG and DAB are similar triangles</note>. According to the definition of similar triangles: <note>∠EDG = ∠ADB</note>, <note>∠DGE = ∠DBA</note>, <note>∠DEG = ∠DAB</note>; <note>DE/DA = DG/DB = EG/AB</note>. <note>Triangles GFB and DCB are similar triangles</note>. According to the definition of similar triangles: <note>∠GFB = ∠DCB</note>, <note>∠GBF = ∠DBC</note>, <note>∠CDB = ∠FGB</note>; <note>GB/DB = GF/DC = FB/CB</note>.

Similarity Theorem for Triangles (AA)

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

In <note>triangle DEG</note> and <note>triangle DAB</note>, if <note>angle DEG is equal to angle DAB</note>, and <note>angle DGE is equal to angle DBA</note>, then <note>triangle DEG is similar to triangle DAB</note>. In <note>triangle GFB</note> and <note>triangle DCB</note>, if <note>angle DCB is equal to angle GFB</note>, and <note>angle GBF is equal to angle DBC</note>, then <note>triangle GFB is similar to triangle DCB</note>.

Parallel Postulate 2 of Parallel Lines

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

In the diagram of this problem, two parallel lines <note>EF</note> and <note>DC</note> are intersected by a third line <note>CB</note>, forming the following geometric relationship:
1. Corresponding angles: <note>angle DCB and angle GFB are equal</note>.
Two parallel lines <note>EF</note> and <note>AB</note> are intersected by a third line <note>DA</note>, forming the following geometric relationship:
1. Corresponding angles: <note>angle DEF and angle DAB are equal</note>.
Two parallel lines <note>EF</note> and <note>AB</note> are intersected by a third line <note>DB</note>, forming the following geometric relationship:
1. Corresponding angles: <note>angle DGE and angle DBA are equal</note>.

Midpoint of a Line Segment

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

In the figure of this problem, <note>the midpoint of line segment DB is point G</note>. According to the definition of the midpoint of a line segment, <note>point G divides line segment DB into two equal parts</note>, that is, <note>the lengths of line segments DG and GB are equal</note>. That is, <note>DG = GB</note>.

Definition of Triangle

In the figure of this problem, <note>triangle DEG is a geometric figure composed of three non-collinear points D, E, G and their connecting line segments DE, DG, EG</note>. Points <note>D, E, G are the three vertices of the triangle</note>, and line segments <note>DE, DG, EG are the three sides of the triangle</note>. <note>Triangle DAB is a geometric figure composed of three non-collinear points D, A, B and their connecting line segments DA, DB, AB</note>. Points <note>D, A, B are the three vertices of the triangle</note>, and line segments <note>DA, DB, AB are the three sides of the triangle</note>. <note>Triangle GFB is a geometric figure composed of three non-collinear points G, F, B and their connecting line segments GF, GB, FB</note>. Points <note>G, F, B are the three vertices of the triangle</note>, and line segments <note>GF, GB, FB are the three sides of the triangle</note>. <note>Triangle DBC is a geometric figure composed of three non-collinear points D, B, C and their connecting line segments DB, BC, DC</note>. Points <

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.

Original text: <note>Two parallel lines GF and DC</note> are intersected by a <note>line CB</note>, where <note>angle GFB and angle DCB are on the same side of the intersecting line CB and on the same side of the intersected lines GF and DC</note>, thus <note>angle GFB and angle DCB are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle GFB equals angle DCB</note>. Two parallel lines <note>EF and AB</note> are intersected by a <note>line DA</note>, where <note>angle DEF and angle DAB are on the same side of the intersecting line DA and on the same side of the intersected lines EF and AB</note>, thus <note>angle DEF and angle DAB are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle DEF equals angle DAB</note>. Two parallel lines <note>EF and AB</note> are intersected by a <note>line DB</note>, where <note>angle DGE and angle DBA are on the same side of the intersecting line DB and on the same side of the intersected lines EF and AB</note>, thus <note>angle DGE and angle DBA are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle DGE


Question: As shown in the figure, a guy wire is fixed to a utility pole at a height of 5m above the ground. The guy wire forms a 60° angle with the ground. Find the length of the guy wire AC.

Answer: \frac{10√{3}}{3}m

1. According to the problem statement, it is known that the line at point C forms a 60° angle with the ground AB, and CD is perpendicular to AB, with CD having a length of 5 meters.

2. According to the ##sine function##, sin(##angle CAD##) = opposite side (i.e., CD) / hypotenuse (i.e., AC), where the opposite side is 5 meters, and ##angle CAD## = 60°.

3. Applying the ##sine function##, we get the length of AC: AC = CD / sin(##angle CAD##) = 5 / sin(60°).

4. Through calculation, sin(60°) = √3/2, thus substituting into the formula, we get: AC = 5 / (√3/2) = 5 * 2 / √3 = 10 / √3.

5. Rationalizing the result, we get AC = 10√3/3.

6. Through the above reasoning, the final answer is 10√3/3 meters.

直角三角形; 正弦; 线段

Definition of Sine Function

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

In the figure of this problem, in the right triangle ACD, <note>angle CAD is an acute angle</note>, <note>side CD is the opposite side of angle CAD</note>, <note>side AC is the hypotenuse</note>. According to the definition of the sine function, <note>the sine value of angle CAD is equal to the ratio of the opposite side CD to the hypotenuse AC</note>, that is, <note>sin(angle CAD) = CD / AC</note>.

Definition of Right Triangle

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

<note>Angle ∠CDA is a right angle (90 degrees)</note>, therefore triangle ACD is a right triangle. <note>Side CD and side AD are the legs</note>, <note>side AC is the hypotenuse</note>.


Question: As shown in the figure, in the right triangle △ABC, ∠C is a right angle, CD⊥AB at D, given AC=3, AB=5, then tan∠BCD equals ()

Answer: \frac{4}{3}

1. Given △ABC is a right triangle, ∠C is a right angle, and CD is perpendicular to AB at point D.

2. From △ABC, ∠C is a right angle, thus ∠A and ∠B are complementary, i.e., ∠A + ∠B = 90°.

3. Since CD is perpendicular to AB, it follows that ∠BCD + ∠B = 90°.

4. Based on the above conclusion, we get ∠BCD = ∠A.

5. Since ∠C is a right angle, and given AC = 3 and AB = 5, we can calculate the length of BC using the Pythagorean theorem.

6. According to the Pythagorean theorem, ##BC = √(AB^2 - AC^2) = √(5^2 - 3^2) = 4##.

7. In the right triangle ABC, tan∠A = BC / AC = 4 / 3.

8. Since ∠BCD = ∠A, therefore tan∠BCD = 4 / 3.

9. Through the above reasoning, the final answer is 4/3.

直角三角形; 垂线; 正切

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 figure of this problem, in <note>right triangle ABC</note>, <note>angle ∠A is an acute angle</note>, <note>side BC is the opposite side of angle ∠A</note>, <note>side AC is the adjacent side of angle ∠A</note>, so <note>the tangent value of angle ∠A is equal to the length of side BC divided by the length of side AC</note>, that is, <note>tan(∠A) = BC / AC = 4 / 3</note>. Similarly, since <note>∠BCD = ∠A</note>, therefore <note>tan∠BCD = 4 / 3</note>.

Pythagorean Theorem

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

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

Triangle Angle Sum Theorem

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

In the figure of this problem, since <note>CD⊥AB</note>, <note>both triangle ADC and triangle CDB are right triangles</note>. Given that <note>△ABC is a right triangle</note>, it can be seen from the figure that <note>∠ACD + ∠BCD = ∠ACB = 90°</note>. Also, because triangle ADC is a right triangle, according to the Triangle Angle Sum Theorem <note>∠CAD + ∠ACD = 90°</note>. Combining the reasoning, we get: <note>∠CAD = ∠BCD, that is, ∠BCD = ∠A</note>.

Definition of Perpendicular Lines

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

The angle formed by the intersection of <note>line CD and line AB</note> is <note>∠CDB is 90 degrees</note>, so according to the definition of perpendicular lines, line CD and line AB are perpendicular to each other.


Question: As shown in the figure, a tangent to ⊙O is drawn through point A on ⊙O, intersecting the extension of diameter BC at point D. Connect AB. If ∠B = 25°, then the degree measure of ∠D is ()

Answer: 40°

1. Given that BA is a chord of ⊙O, BC is the diameter of ⊙O, AD is a tangent to ⊙O, and point O is connected to A. According to the problem, ∠B = 25°.

2. Connect OD. According to the inscribed angle theorem, ∠DOA is equal to 2 times ∠B, because ∠B is an inscribed angle and the corresponding central angle is ∠DOA. Therefore, ∠DOA = 2 * 25° = 50°.

3. Since AD is a tangent to ⊙O, according to the ##properties of the tangent to a circle##, ∠OAD = 90°.

4. Since △ODA is a triangle and the sum of its internal angles is 180°, we have ∠D + ∠OAD + ∠DOA = 180°.

5. Substituting the known angles, we get ∠D + 90° + 50° = 180°.

6. Solving for ∠D, we get ∠D = 180° - 90° - 50° = 40°.

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

圆; 圆周角; 切线; 直角三角形; 三角形的外角

Definition of Inscribed Angle

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

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

Definition of Tangent to a Circle

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

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

Definition of Radius

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

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

Definition of Diameter

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

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

Inscribed Angle Theorem

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

In circle O, points A, B, and C are on the circle, the arc AC corresponds to the central angle ∠DOA, and the inscribed angle is ∠DBA. According to the Inscribed Angle Theorem, ∠DBA is equal to half of the central angle ∠DOA corresponding to the arc AC, that is, ∠DBA = 1/2 ∠DOA.

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>Point A is the point of tangency between line AD and the circle</note>, segment <note>OA is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line AD is perpendicular to the radius OA at the point of tangency A</note>, i.e., <note>∠OAD=90 degrees</note>.

Triangle Angle Sum Theorem

<note>In triangle ODA, angles ∠ODA, ∠OAD, and ∠DOA are the three interior angles of triangle ODA</note>, according to the Triangle Angle Sum Theorem, <note>∠ODA + ∠OAD + ∠DOA = 180°</note>.

Definition of Central Angle

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

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


Question: The three views of a certain geometric body are shown in the figure, where both the front view and the left view are equilateral triangles with side length 2. Then the surface area of the geometric body is ()

Answer: 3π

1. According to the front view and left view, both being equilateral triangles with side length 2, the three views can determine that the geometric body is a cone, where the axial section of the cone is an equilateral triangle with side length 2.

2. The base area of the cone can be calculated using the formula ## S=πr^2 ##, where r is the radius of the cone's base. Given that the diameter of the cone's base is equal to the side length of the equilateral triangle, which is 2, the radius r = 1. Therefore, the base area of the cone S = ## π×1^2 ## = π.

3. The lateral area of the cone can be calculated using the formula S=πrl, where l is the slant height and r is the base radius. According to the definition of the slant height, we know that the slant height l = the side length of the equilateral triangle, which is 2.

4. Therefore, the lateral area of the cone is calculated as S=πr×l = ## π×1×2 = 2π ##.

5. Thus, the total surface area of the cone is the sum of the base area and the lateral area, i.e., S_total surface area of the cone = ## π + 2π ## = 3π.

6. Through the above reasoning, the final calculated result for the surface area is 3π.

等边三角形; 圆; 三视图; 圆锥

Cone

A cone is a geometric figure with a circular base and a single vertex. Its surface consists of a curved lateral surface extending from the base to the vertex.

<note>The front view and left view are both equilateral triangles</note>, determine the geometric body <note>as a cone</note>. The cone's <note>base is a circle</note>, with the circle's radius <note>being 1</note>, and the center <note>being O</note>. The cone's <note>vertex is V</note>, and the distance between the vertex <note>V and the center O is the height of the cone, denoted as h</note>. The cone's lateral surface is a curved surface, with the distance from <note>vertex V to any point on the circumference being the slant height, denoted as l</note>.

Definition of Equilateral Triangle

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

<note>The triangles in the front view and left view are equilateral triangles</note>. <note>The three sides of an equilateral triangle are all 2 units long</note>, and <note>each interior angle is 60°</note>.

Definition of Radius

<note>In the circular base of a cone, the center of the circle is the vertex of the cone</note>, any point on the circle is a point on the circumference of the base circle, <note>the length of the line segment from the center to any point on the circumference is 1</note>, therefore this line segment is the radius of the circle.

Definition of Diameter

The diameter of the base circle of the cone is equal to the side length of the equilateral triangle 2, connecting the <note>center O</note> and <note>two points on the circumference</note>, with a length of <note>2 times the radius</note>, i.e., <note>diameter = 2</note>.

Generatrix

The generatrix of a cone is the line segment that joins a point on the circumference of the base to the apex.

In the figure of this problem, in the cone, <note>point A on the circumference of the base</note>, <note>vertex B</note>, <note>the line segment AB connecting point A on the circumference of the base and vertex B</note> is the generatrix. The generatrix is the line segment from a point on the circumference of the base to the vertex in the cone.

Area Formula of a Circle

The area of a circle is given by the formula A = πr², where A represents the area, π (pi) is the mathematical constant approximately equal to 3.14159, and r is the radius of the circle.

<note>The base of the cone is a circle</note>, <note>The radius of the circle is 1</note>, according to the area formula of a circle, the area A of the circle is equal to the circumference π multiplied by <note>the square of radius 1</note>, that is, A = ##π×1^2## = π.

Development of a Cone

The development (or net) of a cone is a sector of a circle, where the radius of the sector is the slant height of the cone, and the arc length of the sector is equal to the circumference of the cone's base.

<note>In the diagram of this problem, the development of the cone is a sector</note>.<note>The radius of the sector is the slant height of the cone 2</note>.

Formula for the Surface Area of a Cone

The total surface area of a cone is equal to the sum of the base area and the lateral surface area.

Original text: <note>In the figure of this problem, the base of the cone is a circle with a radius of 1</note>, the base area is ##π×1^2##=π. <note>The lateral surface of the cone, when unfolded, is a sector, and the lateral area is equal to the area of the sector</note>, that is, the lateral surface area of the cone is calculated as S=πr×l = π×1×2 =2 π. <note>The total surface area of the cone is equal to the base area plus the lateral area</note>, so the total surface area is 3π.


Question: As shown in the figure, in △ABE, the perpendicular bisector MN of AE intersects BE at point C, ∠E=30°, and AB=CE. Then the degree of ∠BAE is ()

Answer: 90°

1. Given in △ABE, the perpendicular bisector MN of AE intersects BE at point C, ∠E=30°, and AB=CE.

2. Since MN is the perpendicular bisector of AE, ##according to the property of the perpendicular bisector:## CE=CA.

3. According to the given AB=CE, combined with step 2 where CE=CA, we get CA=AB.

4. In △ACE, CE=CA, ##according to the definition of an isosceles triangle: △ACB and △ACE## are isosceles triangles.

5. Because △ACE is an isosceles triangle, we can use the properties of an isosceles triangle, ##to get## ∠ACB=∠CEA##=30°##.

##6. Since ∠ACB is an exterior angle of △EAC, and the two interior angles ∠CEA and ∠EAC of △EAC are not adjacent to it,## ∠ACB=∠CEA+∠EAC (according to the exterior angle theorem of triangles), we get ∠ACB=30°##+30°=60°##.

##7. In the isosceles triangle ABC,## according to the properties of an isosceles triangle, ∠ACB=∠ABC.

##9##. In △BAC, ##according to the triangle interior angle sum theorem:## ∠BAC=180°-∠ACB-∠ABC=180°-##60°-60°=60°##.

##10##. Since ∠CAB## is part of ∠BAE##, ##and ∠CAB=60°,## ∠CAE=30°.

##11##. Since ##∠BAE##=∠CAB+∠CAE, therefore ∠BAE=60°+30°=90°.

##12##. Through the above reasoning, we finally get the degree of ∠BAE as 90°.

垂直平分线; 等腰三角形; 直角三角形

Definition of Perpendicular Bisector

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

In the figure of this problem, <note>line MN passes through the midpoint C of segment AE</note>, and <note>line MN is perpendicular to segment AE</note>. Therefore, <note>line MN is the perpendicular bisector of segment AE</note>.

Definition of Isosceles Triangle

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

In the figure of this problem, <note>in triangle ACE, side CE and side CA are equal</note>, therefore <note>triangle ACE is an isosceles triangle</note>. In triangle ACB, <note>side AB and side CA are equal</note>, therefore <note>triangle ACB is an isosceles triangle</note>.

Definition of Exterior Angle of a Polygon

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

In △EAC, one interior angle is ∠ACE, and the angle formed by extending the adjacent sides EC and AC of this interior angle is called the exterior angle of ∠ACE, i.e., ∠ACB.

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, in the isosceles triangle ACE, <note>side CE and side CA are equal</note>. Therefore, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, that is, <note>angle CAE = angle CEA</note>. <note>In the isosceles triangle ACB, side AB and side CA are equal</note>. Therefore, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, that is, <note>angle ACB = angle ABC</note>.

Properties of the Perpendicular Bisector

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

In the figure of this problem, the perpendicular bisector of the line segment <note>AE is the line MN</note>, and the point <note>C is on the line MN</note>. According to the properties of the perpendicular bisector, <note>the distance from point C to the endpoints A and E of the line segment AE is equal</note>, that is, <note>CA=CE</note>.

Triangle Angle Sum Theorem

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

Original: <note>In the figure of this problem, in triangle BAC</note>, <note>angle BAC, angle ACB, and angle ABC are the three interior angles of triangle BAC</note>, according to the Triangle Angle Sum Theorem, <note>angle BAC + angle ACB + angle ABC = 180°</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>Angle ACB is an exterior angle of the triangle</note>, <note>Angle CEA and Angle EAC are the two non-adjacent interior angles to the exterior angle ACB</note>, according to the Exterior Angle Theorem of Triangle, <note>the exterior angle ACB is equal to the sum of the two non-adjacent interior angles CEA and EAC</note>, that is, <note>Angle ACB = Angle CEA + Angle EAC</note>.


Question: As shown in the figure, the inclined angle ∠ABD of the 4m long staircase AB is 60°. To improve the safety performance of the staircase, it is planned to rebuild the staircase so that its inclined angle ∠ACD is 45°. Then the length of the adjusted staircase AC is ()

Answer: 2√{6}m

1. From the figure, ∠ADB is marked as a right angle. According to the definition of a right triangle, triangle ABD is a right triangle. In triangle ABD, AB is the hypotenuse, BD is the base, and AD is the height, i.e., AB=4m, ∠ABD=60°.

2. According to the definition of the sine function, in right triangle ABD, sin∠ABD = AD/AB.

3. Substituting the known conditions, sin∠ABD = sin60° = AD/4,

4. Given that the value of sin60° is √3/2, we get AD = 4 * (√3/2) = 2√3(m).

5. Next, we analyze the adjusted right triangle ACD. In right triangle ACD, AD is the height, AC is the hypotenuse, and the base is CD.

6. According to the definition of the sine function, in right triangle ACD, sin∠ACD = AD/AC.

7. Substituting the known conditions, sin∠ACD = sin45° = AD/AC.

8. Given that the value of sin45° is √2/2, we get (√2/2) = 2√3/AC.

9. Solving the above equation, we get AC = 2√3 / (√2/2) = (2√3 * 2)/√2 = 2√6(m).

10. Through the above reasoning, the final answer is 2√6(m).

直角三角形; 正弦; 余弦; 正切; 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 ABD, <note>angle ∠BDA is a right angle (90 degrees)</note>, therefore <note>triangle ABD is a right triangle</note>. <note>Side BD and side AD are the legs</note>, and side AB is the hypotenuse. In triangle ACD, <note>angle ∠CDA is a right angle (90 degrees)</note>, therefore <note>triangle ACD is a right triangle</note>. <note>Side CD and side AD are the legs</note>, and side AC is the hypotenuse.

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, in the right triangle ABD, <note>angle ∠ABD is an acute angle</note>, <note>side AD is the opposite side of angle ∠ABD</note>, <note>side AB is the hypotenuse</note>. According to the definition of the sine function, <note>the sine value of angle ∠ABD is equal to the ratio of the opposite side AD to the hypotenuse AB</note>, that is, <note>sin(∠ABD) = AD / AB</note>. In the right triangle ACD, <note>angle ∠ACD is an acute angle</note>, <note>side AD is the opposite side of angle ∠ACD</note>, <note>side AC is the hypotenuse</note>. According to the definition of the sine function, <note>the sine value of angle ∠ACD is equal to the ratio of the opposite side AD to the hypotenuse AC</note>, that is, <note>sin(∠ACD) = AD / AC</note>.


Question: As shown in the figure, in ⊙O, diameter AB ⊥ chord CD at point H, E is a point on ⊙O. If ∠BEC=25°, then the degree of ∠BAD is ()

Answer: 25°

2. ∵ Diameter AB ⊥ chord CD at point H, ∴ ##by the perpendicular bisector theorem, CH equals HD##

3. ##By the congruent triangles criterion (SAS), we know that triangle ACH is congruent to triangle ADH##

4. ##By the definition of congruent triangles, ∠CAH = ∠HAD##

5. ∵ ##By the corollary of the inscribed angle theorem 1, ∴ ∠BEC = ∠BAC## = 25°,

圆; 弦; 垂线; 圆周角; 直角三角形

Definition of Chord

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

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

Definition of Diameter

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

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

Definition of Foot of a Perpendicular

The intersection point of a perpendicular line with the segment it is perpendicular to is called the foot of the perpendicular.

In the figure of this problem, line AB intersects line CD at <note>point H</note>, and <note>line AB is perpendicular to line CD</note>, therefore <note>point H is the foot of the perpendicular from line AB to line CD</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.

Original text: <note>The vertex E of angle BEC is on the circumference</note>, and the two sides of angle BEC intersect circle O at points B and C, respectively. Therefore, <note>angle BEC is an inscribed angle</note>. In circle O, <note>the vertex A of angle BAC is on the circumference</note>, and the two sides of angle BAC intersect circle O at points B and C, respectively. Therefore, <note>angle BAC is an inscribed angle</note>.

Perpendicular Diameter Theorem

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

<note>In the figure of this problem, in circle O, diameter AB is perpendicular to chord CD at point H</note>, then according to the Perpendicular Diameter Theorem, <note>diameter AB bisects chord CD, that is, CH = DH</note>, and <note>diameter AB bisects the two arcs subtended by chord CD, that is, arc CB = arc BD, arc CA = arc AD</note>.

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

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

Triangular Congruence Theorem (SAS)

If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent.

In the original text: <note>Triangle ACH and Triangle AHD</note>, <note>side CH is equal to side HD</note>, <note>side AH is equal to side AH</note>, and <note>angle AHC is equal to angle AHD</note>, therefore, according to the Triangular Congruence Theorem (SAS), these two triangles are congruent.

Definition of Congruent Triangles

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

In the diagram of this problem, <note>triangle ACH and triangle ADH are congruent triangles</note>, the corresponding sides and angles of triangle ACH are equal to those of triangle ADH, namely:
side <note>CH = side HD</note>
side <note>AH = side AH</note>
side <note>AC = side AD</note>,
and the corresponding angles are also equal:
angle <note>AHC = angle AHD</note>
angle <note>ACH = angle ADH</note>
angle <note>HAC = angle HAD</note>.

Definition of Arc

An arc is a curve on a circle that is defined by two distinct points on the circle.

In the figure of this problem, there are two points C and B on circle O, <note>arc CB is a segment of the curve connecting these two points</note>. According to the definition of an arc, <note>arc CB is a segment of the curve between two points C and B on the circle</note>.


Question: As shown in the figure, lines AB and CD intersect at point O, OD bisects ∠AOE, ∠BOC=50°, then ∠EOB=()

Answer: 80°

1. Given that line AB and line CD intersect at point O, and OD bisects ∠AOE.

2. According to the problem, ∠BOC = 50°.

3. Since OD is the bisector of ∠AOE, ∠AOD = ∠DOE = 1/2∠AOE.

4. From the fact that line AB and line CD intersect at point O, we get ∠BOC = ∠AOD (vertical angles are equal).

5. Since ∠BOC = 50°, therefore ∠AOD = 50°.

6. From the previous step, we know ∠AOE = 2×∠AOD = 2×50° = 100°.

7. ∠EOB and ∠AOE are a pair of adjacent supplementary angles. According to the definition of adjacent supplementary angles, the sum of the two angles is 180°.

8. Therefore, ∠EOB = 180° - ∠AOE = 180° - 100° = 80°.

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

对顶角; 邻补角; 直线; 射线

Definition of Angle Bisector

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

In the diagram of this problem, <note>the vertex of angle AOE is point O</note>, <note>a line OD is drawn from point O</note>, this line <note>divides angle AOE into two equal angles</note>, namely <note>angle AOD and angle DOE are equal</note>. Therefore, <note>line OD is the angle bisector of angle AOE</note>.

Definition of Linear Pair of Angles

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

<note>Angle AOE and angle EOB have a common side OE</note>, their <note>other sides OA and OB are extensions in opposite directions</note>, so <note>angle AOE and angle EOB are adjacent supplementary angles</note>.

Definition of Vertical Angles

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

In the figure of this problem, two intersecting lines <note>AB and CD intersect at point O</note>, forming four angles: ∠AOD, ∠BOC, <note>∠DOB</note>, and ∠AOC. According to the definition of vertical angles, <note>∠AOD and ∠BOC are vertical angles</note>, <note>∠DOB and ∠AOC are vertical angles</note>. Since the angles of vertical angles are equal, ∠AOD = ∠BOC, <note>∠DOB = ∠AOC</note>.


Question: As shown in the figure, in parallelogram ABCD, E is the midpoint of BC, and AE and BD intersect at point F. If the area of △BFE is 3, then the area of △ABF is ()

Answer: 6

1. Given that quadrilateral ABCD is a parallelogram, thus AD∥BC and AD=BC.

2. Since E is the midpoint of BC, it follows that BE=1/2BC=1/2AD.

3. Consider △AFD and △EFB, because AD∥BC and point E is the midpoint of BC, ##according to the parallel axiom 2 of parallel lines, ∠DAF=∠BEF and ∠ADF=∠FBE##. According to the similarity theorem (AA), △AFD∽△EFB.

4. According to the definition of similar triangles, the corresponding sides of similar triangles are proportional, i.e., AF/EF=AD/BE=AD/(1/2AD)=2.

5. Using the theorem that the area ratio of similar triangles equals the square of the similarity ratio, ##S△AFD/S△EFB##=(AF/EF)^2=2^2=4.

6. Because ##△ABD and △ABE have the same height (draw AH perpendicular to BC at point H, both heights are AH), and the ratio of the base is 2:1, the area ratio of △ABD and △ABE is 2:1##.

7. ##From △AFD＋△ABF＝2△ABF＋2△BFE##.

8. ##Substituting numbers, after the above calculations,## the final answer is 6.

平行四边形; 中点; 普通三角形

Definition of Parallelogram

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

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

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

Similarity Theorem for Triangles (AA)

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

<note>Triangles AFD and EFB are similar triangles</note>. According to the definition of similar triangles: <note>∠DAF = ∠EFB</note>, <note>∠ADF = ∠BEF</note>; <note>AF/EF = AD/BE = AD/(1/2AD) = 2</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>Triangles AFD and EFB are similar triangles</note>. According to the <note>definition of similar triangles</note>, we have: <note>∠DAF = ∠EFB</note>, <note>∠ADF = ∠BEF</note>; <note>AF/EF = AD/BE = AD/(1/2AD) = 2</note>.

Properties of Parallelogram Theorem

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

In the figure of this problem, in the parallelogram <note>ABCD</note>, the opposite angles <note>∠DAB and ∠BCD</note> are equal, and the opposite angles <note>∠ABC and ∠CDA</note> are equal; the sides <note>AB and CD</note> are equal, and the sides <note>AD and BC</note> are equal; the diagonals <note>AC and BD</note> bisect each other, that is, the intersection point <note>F</note> divides the diagonal <note>AC into two equal segments AF and FC</note>, and divides the diagonal <note>BD into two equal segments BF and FD</note>.

Theorem on the Area Ratio of Similar Triangles

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

<note>Triangle AFD and triangle EFB are similar triangles</note>, the ratio of side <note>AD to side BE is 2, which means the similarity ratio is 2</note>. Therefore, <note>the ratio of the area of triangle AFD to the area of triangle EFB is equal to 4</note>.

Parallel Postulate 2 of Parallel Lines

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

<note>Two parallel lines AD and BC are intersected by a third line BD</note>, forming the following geometric relationship: <note>alternate interior angles: angle ADF and angle EBF are equal</note>. In the diagram of this problem, <note>two parallel lines AD and BC are intersected by a third line AE</note>, forming the following geometric relationship: <note>alternate interior angles: angle DAF and angle BEF are equal</note>.

Definition of Triangle

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

<note>Triangle ADF is a geometric figure composed of three non-collinear points A, D, F and their connecting line segments AD, AF, DF</note>. <note>Points A, D, F</note> are the three vertices of the triangle, and <note>line segments AD, AF, DF</note> are the three sides of the triangle. <note>Triangle BEF is a geometric figure composed of three non-collinear points B, E, F and their connecting line segments BE, EF, BF</note>. <note>Points B, E, F</note> are the three vertices of the triangle, and <note>line segments BE, EF, BF</note> are the three sides of the triangle. <note>Triangle ABD is a geometric figure composed of three non-collinear points A, B, D and their connecting line segments AD, AB, BD</note>. <note>Points A, B, D</note> are the three vertices of the triangle, and <note>line segments AD, AB, BD</note> are the three sides of the triangle. <note>Triangle ABE is a geometric figure composed of three non-collinear points A, B, E and their connecting line segments AE, AB, BE</note>. <note>Points A, B,

Area Formula of a Triangle

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

<note>In triangle ABD, side AD is the base, and segment AH is the height</note>. According to the area formula of a triangle, <note>the area of triangle ABD is equal to the base AD multiplied by the height AH and then divided by 2</note>, i.e., Area = (AD * AH) / 2. <note>In triangle ABE, side BE is the base, and segment AH is the height</note>. According to the area formula of a triangle, <note>the area of triangle ABE is equal to the base BE multiplied by the height AH and then divided by 2</note>, i.e., Area = (BE * AH) / 2.


Question: As shown in the figure, A, B, C are three points on ⊙O, ∠ABC=25°, then the degree of ∠AOC is ()

Answer: 50°

1. Given that points A, B, and C are on ⊙O, according to the problem, ∠ABC=25°.

2. In the circle, the central angle is twice the inscribed angle subtended by the same chord. According to the inscribed angle theorem, we get ∠AOC=2∠ABC.

3. Substitute ∠ABC=25°, we get ∠AOC=2×25°.

4. After calculation, we get ∠AOC=50°.

5. The final answer is ∠AOC=50°.

Definition of Inscribed Angle

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

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

Definition of Central Angle

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

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

Inscribed Angle Theorem

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

The central angle <note>∠AOC</note> corresponding to the arc <note>∠ABC 所对的弧 AC</note> is <note>∠AOC</note>. According to the Inscribed Angle Theorem, the central angle <note>∠AOC 是圆周角 ∠ABC 的两倍</note>, that is, <note>∠AOC = 2 × ∠ABC</note>.


Question: As shown in the figure, in △ABC, points D and E are on sides AB and AC respectively, and DE ∥ BC. If AD = 5, BD = 10, DE = 3, then the length of BC is ()

Answer: 9

1. Given AD=5, BD=10, point D is on side AB, so AB=AD+BD=5+10=15.

2. Given DE∥BC, ##according to the parallel axiom 2 of parallel lines, ∠ADE = ∠ABC (corresponding angles are equal), ∠DEA = ∠BCA (corresponding angles are equal), so △ADE∽△ABC (according to the AA criterion for similar triangles)##.

3. ##According to the definition of similar triangles##, we get: \( \frac{AD}{AB}=\frac{DE}{BC} \).

4. Substitute the given values, we get: \( \frac{5}{15}=\frac{3}{BC} \).

5. By solving the proportion, we can find the length of BC: ##BC=9##.

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

平行线; 普通三角形; 位似

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>In the figure of this problem, two parallel lines DE and BC are intersected by a line AB, where angle ADE and angle ABC are on the same side of the intersecting line AB and on the same side of the two intersected lines DE and BC</note>, therefore <note>angle ADE and angle ABC are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle ADE is equal to angle ABC</note>. Similarly, <note>two parallel lines DE and BC are intersected by a line AC, where angle DEA and angle BCA are on the same side of the intersecting line AC and on the same side of the two intersected lines DE and BC</note>, therefore <note>angle DEA and angle BCA are corresponding angles</note>. Corresponding angles are equal, that is, <note>angle DEA is equal to angle BCA</note>.

Definition of Similar Triangles

In the problem diagram, <note>△ADE∽△ABC</note>. According to the definition of similar triangles, <note>∠ADE = ∠ABC</note>, <note>∠DEA = ∠BCA</note>, <note>∠EAD = ∠CAB</note>; <note>\( \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC} \)</note>. Substituting the known values, we get: <note>\( \frac{5}{15}=\frac{3}{BC} \)</note>. Solving the proportion, we get <note>BC=9</note>.

Parallel Postulate 2 of Parallel Lines

<note>Two parallel lines DE and BC</note> are intersected by <note>a third line AB</note>, forming the following geometric relationships:
1. Corresponding angles: <note>Angle ADE and angle ABC are equal, angle DEA = angle BCA</note>.
2. Alternate interior angles: None.
3. Consecutive interior angles: None.
These relationships indicate that when two parallel lines are intersected by a third line, corresponding angles are equal, alternate interior angles are equal, and consecutive interior angles are supplementary.

Similarity Theorem for Triangles (AA)

<note>Angle ADE</note> is equal to <note>Angle ABC</note>, and <note>Angle DEA</note> is equal to <note>Angle BCA</note>, so triangle <note>ADE</note> is similar to triangle <note>ABC</note>.


Question: As shown in the figure, in △ABC, D and E are the midpoints of AB and AC respectively. Given that the area of △ADE is 1, then the area of △ABC is ()

Answer: 4

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

2. From the Midline Theorem, we know that DE∥BC and the length of DE is half the length of BC.

3. According to the Similar Triangles Criterion (SSS), △ADE and △ABC are similar because their corresponding sides are proportional.

4. Since DE is half of BC, the ratio coefficient is 1/2. According to the theorem that the area ratio of similar triangles equals the square of the similarity ratio, we have ##S△ADE/S△ABC=(1/2)^2=1/4##.

5. Given that the area of △ADE is 1, we can deduce that the area of △ABC is 4 times the area of △ADE, that is, S△ABC=4.

6. Through the above reasoning, we finally conclude that the area of △ABC is 4.

中点; 普通三角形; 平行线

Midpoint of a Line Segment

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

Triangle Midline Theorem

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

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

<note>Triangle △ADE and triangle △ABC are similar triangles</note>, the ratio of side <note>DE to side BC is 1/2</note>, that is, <note>the similarity ratio is 1/2</note>. Therefore, <note>the area ratio of triangle △ADE to triangle △ABC is equal to ##(1/2)^2##=1/4</note>.

Similarity Theorem for Triangles (SSS)

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

In <note>triangles ADE and ABC</note>, <note>side AD corresponds to side AB</note>, <note>side DE corresponds to side BC</note>, <note>side AE corresponds to side AC</note>, and <note>side AD/side AB = side DE/side BC = side AE/side AC</note>. Therefore, according to the Side-Side-Side Similarity Theorem (<note>SSS Similarity Theorem</note>), <note>triangle ADE is similar to triangle ABC</note>.


Question: As shown in the figure, in △ABC, DE∥BC, intersecting AB and AC at points D and E respectively. If AE=3 and EC=6, then the value of \frac{DE}{BC} is ()

Answer: \frac{1}{3}

1. Given that in △ABC, DE∥BC, intersecting AB and AC at points D and E respectively.

2. According to the parallel line axiom 2, corresponding angles are equal, i.e., ∠ADE = ∠B. Since ∠A is a common angle, △ADE∽△ABC. According to the definition of similar triangles, rac{DE}{BC} = rac{AE}{AC}.

3. Based on the given conditions, AE = 3, EC = 6.

4. Therefore, AC = AE + EC = 3 + 6 = 9.

5. Substituting the ratio, we get rac{DE}{BC} = rac{AE}{AC} = rac{3}{9} = rac{1}{3}.

6. Through the above reasoning, the final answer is 1/3.

Definition of Similar Triangles

In the figure of this problem, since <note>DE∥BC</note>, according to the definition of similar triangles, <note>△ADE ∽ △ABC</note>. Based on the definition of similar triangles, we have <note>∠ADE = ∠ABC</note>, <note>∠DEA = ∠BCA</note>, <note>∠DAE = ∠BAC</note>; <note>\(\frac{DE}{BC} = \frac{AE}{AC} = \frac{AD}{AB}\)</note>.

Similarity Theorem for Triangles (AA)

Original: <note>Angle ADE is equal to angle ABC</note>, and <note>angle DAE is equal to angle BAC</note>, so <note>triangle ADE is similar to triangle ABC</note>.

Definition of Corresponding Angles

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

Parallel Postulate 2 of Parallel Lines

<note>In the diagram of this problem, the two parallel lines DE and BC are intersected by a third line AB</note>, forming the following geometric relationship: <note>Corresponding angles: angle ADE and angle ABC are equal</note>.


Question: As shown in the figure, in △ABC, if DE∥BC, rac{AD}{AB}=rac{1}{3}, DE=4cm, then the length of BC is ()

Answer: 12cm

1. Given DE∥BC, according to the theorem of proportional segments in parallel lines, △ADE is similar to △ABC.

2. According to the proportionality of corresponding sides in similar triangles: AD/AB = DE/BC.

3. Given AD/AB = 1/3 and DE = 4 cm, substitute these values into the previous formula to get 1/3 = 4/BC.

4. Solve the equation 1/3 = 4/BC to get BC = 12 cm.

5. Through the above reasoning, the final answer is 12 cm.

平行线; 普通三角形; 线段; 位似

Definition of Similar Triangles

In the figure of this problem, <note>triangle ADE and triangle ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠ADE = ∠ABC</note>, <note>∠DEA = ∠BCA</note>, <note>∠EAD = ∠CAB</note>; <note>AD/AB = DE/BC = AE/AC</note>.

Proportional Segments Theorem

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

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

Definition of Parallel Lines

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

In the figure of this problem, <note>line DE and line BC lie in the same plane</note>, and <note>they do not intersect</note>, so according to the <note>definition of parallel lines</note>, <note>line DE and line BC are parallel lines</note>.

SAS Criterion for Similar Triangles

If two triangles have two pairs of corresponding sides in proportion and the included angle between those sides is equal, then the two triangles are similar.

In the figure of this problem, in triangles ADE and ABC, <note>side AD corresponds to side AB</note>, <note>side DE corresponds to side BC</note>, and <note>side AD/side AB = side DE/side BC</note>, and <note>angle DAE = angle BAC</note>. Therefore, according to the Side-Angle-Side (SAS) Criterion for Similar Triangles, <note>triangle ADE is similar to triangle ABC</note>.


Question: As shown in the figure, place the right-angle vertex of the triangle containing a 30° angle (∠A=30°) on one of the two parallel lines. If ∠1=38°, then the degree of ∠2 is ()

Answer: 22°

1. According to the problem statement and the definition of a right triangle, it is known that ∠A=30°, let the right angle vertex be C, and the other angle be B, so ∠ACB=90°. Using the triangle angle sum theorem, we can deduce ∠ABC=180° - 30° - 90°=60°.

2. Let the two parallel lines be GH and EF, with the upper line being GH and the lower line being EF. Extend AB to intersect the line EF at point E. According to the problem statement, it is known that ∠1=38°.

3. Using the exterior angle theorem of triangles, ∠BEC=∠ABC-∠1=60°-38°=22°.

4. Since line GH is parallel to line EF, according to the parallel postulate 2, alternate interior angles are equal, it can be deduced that ∠2=∠BEC=22°.

5. Finally, the degree of ∠2 is 22°.

平行线; 同位角; 内错角; 直角三角形; 三角形的外角

Triangle Angle Sum Theorem

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

<note>Angle CAB</note>, <note>Angle ABC</note>, and <note>Angle ACB</note> are the three interior angles of triangle <note>ABC</note>. According to the Triangle Angle Sum Theorem, <note>Angle CAB + Angle ABC + Angle ACB = 180°</note>. Specifically, given the conditions of this problem, it is known that <note>Angle CAB = 30°</note> and <note>Angle ACB = 90°</note>. Finally, we can find that <note>Angle ABC = 180° - 30° - 90° = 60°</note>.

Definition of Right Triangle

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

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

Exterior Angle Theorem of Triangle

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

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

Parallel Postulate 2 of Parallel Lines

In the figure of this problem, line GH is parallel to line EF, therefore according to Parallel Postulate 2 of Parallel Lines, the following geometric relationships can be derived:
1. Corresponding angles: ##no corresponding relationship##.
2. Alternate interior angles: ##<note>∠2 is equal to ∠BEC</note>##.
3. Same-side interior angles: no corresponding relationship.
Therefore, <note>∠2 = ∠BEC = 22°</note>.

Definition of Alternate Interior Angles

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

<note>Two parallel lines GH and EF</note> are intersected by a line <note>AE</note>, where <note>angle 2</note> and <note>angle BEC</note> are located between the two parallel lines and on <note>opposite sides of the intersecting line AE</note>, therefore <note>angle 2 and angle BEC are alternate interior angles</note>. <note>Alternate interior angles are equal, that is, angle 2 is equal to angle BEC</note>.

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

In the figure of this problem, an interior angle of the polygon <note>BEC</note> is <note>∠EBC</note>. The angle formed by extending the adjacent sides <note>EB</note> and <note>BC</note> of this interior angle is called the exterior angle of the interior angle <note>∠EBC</note>.


Question: As shown in the figure, ⊙O is the circumcircle of △ABC, ∠BCO=40°, then the degree of ∠A is equal to ()

Answer: 50°

1. Given ⊙O is the circumcircle of △ABC, OC=OB, and ∠BCO=40°.

2. Since OC=OB, ##according to the definition of an isosceles triangle, △OBC is an isosceles triangle, thus according to the properties of isosceles triangles we get ∠OBC=∠BCO=40°.##

3. According to the triangle angle sum theorem, we get ∠BOC=180° - ∠OBC - ∠BCO = 180° - 40° - 40° = 100°.

4. According to the inscribed angle theorem, in the same circle, ##the inscribed angle is equal to half of the central angle corresponding to its intercepted arc, thus the inscribed angle ∠BAC is half of ∠BOC##, i.e., ∠BAC=100° × 1/2 = 50°.

5. To sum up, the measure of ∠BAC is 50°.

圆; 圆周角; 三角形的外角; 圆内接四边形

Definition of Isosceles Triangle

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

<note>In triangle OBC, side OC and side OB are equal</note>, therefore triangle OBC 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 OBC, sides OC and OB are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle OBC = angle BCO = 40°</note>.

Triangle Angle Sum Theorem

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

Inscribed Angle Theorem

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

Definition of Central Angle

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

Definition of Inscribed Angle

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


Question: At 9:00 AM, a ship departs from point A and sails due east at a speed of 40 nautical miles per hour. It reaches point B at 9:30 AM, as shown in the figure. From points A and B, the island M is observed at a bearing of 45° north of east and 15° north of east, respectively. What is the distance between point B and island M?

Answer: 20√{2}海里

1. Given that from 9:00 AM to 9:30 AM, the ship sails at a speed of 40 nautical miles per hour, then AB = 40 nautical miles × 0.5 hours = 20 nautical miles.

2. According to North 45° East and North 15° East, we can obtain ##∠ABM = 90° + 15° = 105°##.

3. Draw BN perpendicular to AM through point B, ##with the foot of the perpendicular at point N, so triangle ABN is a right triangle. That is, ∠BNA = 90°, so## angle ∠ABN = 45°.

4. In the right triangle ABN, calculate the length of BN using the sine function: BN = AB sin(45°) = 20 nautical miles × √2/2 = 20√2/2 = 10√2 nautical miles.

5. By solving the right triangle BNM, ##given ∠ABM = 105°, so ∠MBN = ∠ABM - ∠ABN = 105° - 45° = 60°, and since ∠MAB = 45°, according to the triangle sum theorem, in △BMA, ∠BMA = 30°##.

6. Using the sine function definition in ##the right triangle NBM## to find the length of BM: according to the sine formula, BM = BN / sin(30°) = 10√2 nautical miles × 2 = 20√2 nautical miles.

7. Through the above reasoning, the final answer is 20√2 nautical miles.

等腰三角形; 方向角; 直角三角形

Definition of an Angle

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

In the figure of this problem, <note>angle ∠ABM is a geometric figure formed by two rays BA and BM</note>, these <note>two rays have a common endpoint B</note>. This <note>common endpoint B is called the vertex of angle ∠ABM</note>, and <note>rays BA and BM are called the sides of angle ∠ABM</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 right triangle <note>△ABN</note>, <note>∠NAB is an acute angle</note>, <note>side BN is the opposite side of angle ∠NAB</note>, and <note>side AB is the hypotenuse</note>. According to the definition of the sine function, <note>the sine of angle ∠NAB is the ratio of the opposite side BN to the hypotenuse AB</note>, that is, <note>sin(45°) = BN / AB = √2/2</note>. In the right triangle <note>△MBN</note>, according to the definition of the sine function, <note>the sine of angle ∠NMB is the ratio of the opposite side BN to the hypotenuse MB</note>, that is, <note>sin(30°) = BN / MB = 1/2</note>. Therefore, <note>MB = 2BN = √2AB</note>.

Definition of Right Triangle

<note>Point B constructs BN perpendicular to AM</note>, so <note>∠ANB and ∠MNB are right angles (90 degrees)</note>, therefore <note>△ABN and △MBN are right triangles</note>. Hence, <note>sides AN and BN are the legs of △ABN, and side AB is the hypotenuse</note>. <note>Sides MN and BN are the legs of △MBN, and side MB is the hypotenuse</note>.

Triangle Angle Sum Theorem

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


Question: As shown in the figure, points A, B, and C are on ⊙O. If ∠ABC=35°, then the degree of ∠AOC is ()

Answer: 70°

1. Given that points A, B, and C are on circle O, ##according to the definition of the inscribed angle and the central angle, ∠ABC is the inscribed angle, and ∠AOC is the central angle, it is known that## ∠ABC=35°.

2. According to the inscribed angle theorem, the inscribed angle subtended by the same arc is half of the central angle subtended by that arc. Therefore, we have ∠AOC=2∠ABC.

3. By calculation, we get ∠AOC=2×35°=70°.

4. Through the above reasoning, the final answer is that the degree of ∠AOC is 70°.

圆; 圆周角; 圆心角; 弧; 弦

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

<note>In the figure of this problem, in circle O</note>, point A and point C are two points on the circle, <note>the center of the circle is point O</note>. The angle <note>∠AOC formed by the lines OA and OC 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 figure of this problem, points A, B, and C are on circle O, the central angle corresponding to arc AC and arc CA is <note>∠AOC</note>, and the inscribed angle is <note>∠ABC</note>. According to the Inscribed Angle Theorem, <note>∠ABC equals half of the central angle ∠AOC corresponding to arc AC</note>, i.e., <note>∠ABC = 1/2 ∠AOC</note>.


Question: As shown in the figure, in △ABC, DE∥BC, rac{AD}{DB}=rac{1}{2}, DE=4, find the length of BC.

Answer: 12

1. Given DE∥BC, according to ##Parallel Lines Axiom 2##, we have ∠ADE = ∠ABC and ∠DEA = ∠BCA. Therefore, according to ##Similarity Triangle Criterion (AA)##, we get △ADE ∽ △ABC.

2. According to ##Definition of Similar Triangles##, we can obtain: DE / BC = AD / AB.

3. Given AD / DB = 1 / 2, then AD / (AD + DB) = AD / AB = 1 / 3.

4. Substituting the above conclusion into the similarity ratio DE / BC = 1 / 3, and given DE = 4, we get 4 / BC = 1 / 3, thus calculating BC = 12.

5. Through the above reasoning, the final answer is 12.

平行线; 普通三角形; 线段

Definition of Similar Triangles

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

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

Similarity Theorem for Triangles (AA)

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

<note>DE ∥ BC</note>, according to <note>Parallel Line Axiom 2</note>, we get <note>∠ADE = ∠ABC</note>, <note>∠DEA = ∠ACB</note>, thus satisfying the conditions of the <note>Similarity Theorem for Triangles (AA)</note>, resulting in <note>△ADE ∽ △ABC</note>.

Parallel Postulate 2 of Parallel Lines

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

<note>Two parallel lines DE and BC are intersected by a third line AB</note>, forming the following geometric relationships:
1. Corresponding angles: <note>Angle ADE and angle ABC are equal</note>.
####<note>Two parallel lines DE and BC are intersected by a third line AC</note>, forming the following geometric relationships:
1. Corresponding angles: <note>Angle AED and angle ACB are equal</note>.

Definition of Corresponding Angles

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

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


Question: As shown in the figure, A, B, C are points on ⊙O, ∠ACB=32°, then ∠AOB equals ()

Answer: 64°

1. As shown in the figure, it is known that points A, B, and C are on ⊙O, and ∠ACB = 32°.

2. According to the inscribed angle theorem, the arc corresponding to the inscribed angle ∠ACB is AB, and the corresponding central angle is ∠AOB.

3. The inscribed angle theorem states that in a circle, the central angle corresponding to a given arc is twice the inscribed angle corresponding to that arc.

4. Therefore, ∠AOB = 2 * ∠ACB.

5. Substituting the known condition, ∠AOB = 2 * 32° = 64°.

6. Through the above reasoning, the final answer is 64°.

Definition of Circle

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

The original text: <note>The circle is ⊙O</note>, <note>point O is the center of the circle</note>, <note>the radius is OA, OB, OC</note>. All <note>points that are at a distance equal to OA, OB, OC from point O</note> are on <note>circle ⊙O</note>.

Definition of Inscribed Angle

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

Definition of Central Angle

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

Inscribed Angle Theorem

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


Question: As shown in the figure, points A, B, and C are all on ⊙O. If ∠C = 34°, then ∠AOB is ()

Answer: 68°

1. Given that point A, point B, and point C are all on circle O, ∠ACB=34°.

2. According to the inscribed angle theorem, the inscribed angle subtended by the same arc is half of the central angle. Specifically, if the inscribed angle subtended by arc AB is ∠ACB, then the central angle ∠AOB is twice the inscribed angle ∠ACB.

3. Given that ∠ACB=34°, then ∠AOB=2 * 34°=68°.

4. Through the above reasoning, the final answer is ∠AOB=68°.

圆; 圆周角; 圆心角; 弦

Definition of Inscribed Angle

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

Inscribed Angle Theorem

Original: <note>在本题图中，圆 O 中</note>，<note>点 A、点 B 和点 C 都在圆上</note>，<note>弧 AB 对应的圆心角为 ∠AOB</note>，<note>圆周角为 ∠ACB</note>。根据<note>圆周角定理</note>，<note>∠ACB 等于它所对的弧 AB 所对应的圆心角 ∠AOB 的一半</note>，即 <note>∠ACB = 1/2 ∠AOB</note>。

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

Definition of Central Angle

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


Question: As shown in the figure, in △ABC, D and E are points on AB and AC respectively, and DE ∥ BC. If AD = 5, DB = 3, DE = 4, then BC equals ()

Answer: \frac{32}{5}

1. Given DE∥BC, according to the parallel axiom 2 of parallel lines, the corresponding angles between the two segments are equal, so ∠ADE=∠ABC.

####

2. When ∠A is the common angle, then ∠DAE=∠BAC, and when ∠ADE=∠ABC, according to the similarity theorem (AA) of triangles, △ADE is similar to △ABC. Since △ADE∽△ABC, AD/AB = DE/BC.

3. Given AD=5, DB=3, DE=4, according to the segment addition, AB=AD + DB = 5 + 3 = 8.

4. Substituting the known data into the proportion equation, we get: 5/8= 4/BC.

5. Solving the above equation by cross-multiplication, we get: 5BC = 32.

6. Dividing both sides by 5, we get: BC = 32/5.

7. Through the above reasoning, the final answer is: BC = 32/5.

平行线; 普通三角形; 线段; 位似

Parallel Postulate 2 of Parallel Lines

Original: <note>Two parallel lines DE and BC</note> are intersected by <note>a third line AB</note>, forming the following geometric relationship: <note>corresponding angles: angle ADE and angle ABC are equal</note>.

Definition of Corresponding Angles

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

Similarity Theorem for Triangles (AA)

In the original text: <note>Triangles ADE and ABC</note>, if <note>angle ADE is equal to angle ABC</note>, and <note>angle DAE is equal to angle BAC</note>, then <note>triangle ADE is similar to triangle ABC</note>.

Definition of Similar Triangles

In the figure of this problem, <note>triangle ADE and triangle ABC are similar triangles</note>. According to the <note>definition of similar triangles</note>, we have: <note>∠ADE = ∠ABC</note>, <note>∠DEA = ∠BCA</note>, <note>∠DAE = ∠BAC</note>; <note>AD/AB = DE/BC = AE/AC</note>.


Question: As shown in the figure, AB and CD are two chords of ⊙O. Connect AD and BC. If ∠BCD=70°, then the degree of ∠BAD is ()

Answer: 70°

1. Given that the measure of angle BCD is 70 degrees.

2. According to Corollary 1 of the Inscribed Angle Theorem, an inscribed angle is equal to the inscribed angle subtended by the same arc.

3. Since angle BCD and angle BAD both subtend arc BD, according to Corollary 1 of the Inscribed Angle Theorem, angle BCD is equal to angle BAD.

4. Therefore, angle BAD is 70 degrees.

圆; 圆周角; 弦; 对顶角; 圆内接四边形

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

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

Definition of Inscribed Angle

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


Question: As shown in the figure, in ⊙O, ∠ABC=130°, then ∠AOC equals ()

Answer: 100°

1. Given that in circle O, ∠ABC=130°. ##According to the definition of central angle and the definition of inscribed angle, for the major arc AC, the opposite angle of ∠AOC is the central angle, and ∠ABC is the inscribed angle.##

2. According to the inscribed angle theorem, the inscribed angle is equal to half of the central angle it subtends, so the opposite angle of ∠AOC##=2∠ABC.

3. By the inscribed angle theorem, we can deduce that the opposite angle of ∠AOC##=2×130°=260°.

####

##4##. ##Since the actual angle we are looking for is the central angle corresponding to the minor arc AC, we need to calculate the difference,## so ∠AOC = 360° -##the opposite angle of ∠AOC## =##360° - 260°=100°##.

##5##. After the above reasoning, we finally obtain the answer as 100°.

圆; 圆周角; 圆心角; 弧; 三角形的外角

Definition of Circle

In the figure of this problem, <note>in circle O, point O is the center of the circle</note>. <note>All points in the figure that are at a fixed distance from point O</note> are on <note>circle O</note>.

Definition of Central Angle

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

Definition of Inscribed Angle

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

Inscribed Angle Theorem

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


Question: AB is the diameter of ⊙O, point C is on ⊙O, if ∠C = 15°, then ∠BOC = ()

Answer: 30°

1. Given AB is the diameter of ⊙O, point C is on ⊙O, ∠C = 15°.

2. Because OA = OC, △OAC is an isosceles triangle (according to the definition of an isosceles triangle), ∴∠OAC = ∠C = 15° (according to the properties of an isosceles triangle).

3. According to the inscribed angle theorem: In a circle, the inscribed angle is equal to half of the central angle that subtends the same arc, we get ∠BOC = 2∠OAC.

4. Substitute the value of ∠OAC, we get ∠BOC = 2×15° = 30°.

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

Definition of Isosceles Triangle

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

Inscribed Angle Theorem

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

Properties of Isosceles Triangle

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

Definition of Inscribed Angle

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

Definition of Central Angle

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


Question: As shown in the figure, there is a square DEFG in △ABC, where D is on AC, E and F are on AB, and the line AG intersects DE and BC at points M and N respectively. If ∠B=90°, AB=8, BC=6, EF=2, then the length of BN is ()

Answer: \frac{24}{7}

1. Given that quadrilateral DEFG is a square, therefore ∠AED=∠AFG=90°.

2. ∵ ∠B=90°, ####∠DAE=∠CAB, ∠GAF=∠NAB.

3. Based on the above conditions, according to the similarity of triangles with two equal angles (AA similarity), we get △DAE∽△CAB and △GAF∽△NAB.

4. Because AB=8, BC=6, EF=2, and DE=EF=2. According to the definition of similar triangles, we get ##AE/AB=DE/BC, i.e., AE/8=2/6, thus AE=8/3##.

5. After obtaining AE, we find AF = AE + EF = ##8/3## + 2 =##14/3##.

6. From △GAF∽△NAB, according to the definition of similar triangles, we get ##BN/GF=AB/AF, i.e., BN/2=8/14/3##.

7. Solving the equation, we get BN=##24/7##.

8. Through the above reasoning, the final answer is ##24/7##.

直角三角形; 正方形; 线段; 垂线

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>Side DE</note>, <note>Side EF</note>, <note>Side FG</note>, and <note>Side GD are equal</note>, and <note>Angle ∠DEF</note>, <note>Angle ∠EFG</note>, <note>Angle ∠FGD</note>, and <note>Angle ∠GDE are all right angles (90 degrees)</note>, so <note>DEFG is a square</note>.

Similarity Theorem for Triangles (AA)

In the diagram of this problem, in triangles DAE and CAB, if <note>∠DAE=∠CAB</note> and <note>∠AED=∠ABC</note> (both are 90°), then triangle DAE is similar to triangle CAB. Similarly, in triangles GAF and NAB, if <note>∠GAF=∠NAB</note> and <note>∠AFG=∠ABN</note> (both are 90°), then triangle GAF is similar to triangle NAB.

Definition of Similar Triangles

In the figure of this problem, <note>triangle DAE and triangle CAB are similar triangles</note>. According to the definition of similar triangles: <note>∠DAE = ∠CAB</note>, <note>∠ADE = ∠ACB</note>, <note>∠DEA = ∠CBA</note>; <note>AE/AB = DE/BC = AD/AC</note>. Similarly, <note>triangle GAF and triangle NAB are similar triangles</note>. According to the definition of similar triangles: <note>∠GAF = ∠NAB</note>, <note>∠AGF = ∠ANB</note>, <note>∠AFG = ∠ABN</note>; <note>BN/GF = AB/AF = AN/AG</note>.


Question: As shown in the figure, in quadrilateral ABCD, AB=AC=AD, ∠CBD=23°, then ∠CAD is ()

Answer: 46°

1. Given in quadrilateral ABCD, AB=AC=AD, ##so points B, C, and D are on the circumference of the circle with center A and radius AD##.

2. According to the ##inscribed angle theorem##: the central angle is twice the inscribed angle.

3. From the given condition ∠CBD=23°, and AD is the radius of the circle, so in this circle, ∠CAD and ∠CBD have the relationship of ##central angle## and inscribed angle.

4. Therefore, according to this theorem, in this problem we can get: ∠CAD=2∠CBD=2×23°=46°.

5. Through the above reasoning, the final answer is 46°.

等腰三角形; 对顶角; 普通三角形

Definition of Inscribed Angle

<note>In the figure of this problem, with point A as the center and AD as the radius of the circle</note>, <note>the vertex B of angle CBD is on the circumference</note>, <note>the two sides of angle CBD intersect the circle at points C and D respectively</note>. Therefore, <note>angle CBD is an inscribed angle</note>.

Definition of Central Angle

<note>The circle has center A</note>, <note>AD is the radius</note>. <note>Point C and point D are two points on the circle</note>, <note>the center of the circle is point A</note>. <note>The angle ∠CAD formed by the lines AC and AD is called the central angle</note>.

Inscribed Angle Theorem

<note>Points B, C, and D are on the circle centered at A</note>, <note>the central angle corresponding to arc BD and arc BC is ∠CAD</note>, <note>the inscribed angle is ∠CBD</note>. According to the Inscribed Angle Theorem, <note>∠CAD is equal to half of the central angle ∠CAD corresponding to arc BD</note>, that is, <note>∠CAD = 2∠CBD</note>.


Question: As shown in the figure, place the vertex of the 45° angle of the right triangle at the center of the circle O, with the hypotenuse and one of the right-angle sides intersecting ⊙O at points A and B respectively. C is any point on the major arc AB (not coinciding with A or B). Then the measure of ∠ACB is ()

Answer: 22.5°

1. Given that the vertex of the 45° angle of a right triangle is placed at the center of the circle O, the hypotenuse and one of the right-angle sides intersect the circle O at points A and B respectively.

2. According to the definition of central angles and inscribed angles, ∠AOB is a central angle and ∠ACB is an inscribed angle. At the same time, according to the inscribed angle theorem, the angle on the circumference corresponding to the central angle ∠AOB is half of the central angle.

4. According to the inscribed angle theorem, ∠ACB=1/2∠AOB.

5. Calculating, we get ∠ACB=1/2×45°=22.5°.

6. Through the above reasoning, the final answer is 22.5°.

圆; 圆周角; 直角三角形; 弧

Definition of Central Angle

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

Inscribed Angle Theorem

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

Definition of Inscribed Angle

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


Question: As shown in the figure, A, B, and C are on ⊙O. If ∠BAC=24°, then the degree of ∠BOC is ()

Answer: 48°

1. Given that points A, B, and C are on circle O, ∠BAC=24°.

2. By observation, ∠BOC is subtended by chord BC.

3. According to the inscribed angle theorem, the central angle subtended by a chord in a circle is twice the inscribed angle subtended by the same chord, i.e., in this problem, ∠BOC is twice ∠BAC.

4. Therefore, ∠BOC = 2 × ∠BAC.

5. Substituting the given condition, we get ∠BOC = 2 × 24° = 48°.

6. Hence, the measure of ∠BOC is 48°.

圆; 圆心角; 圆周角; 弧

Definition of Central Angle

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

Definition of Inscribed Angle

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

Inscribed Angle Theorem

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


Question: The diameter of the protractor coincides with the hypotenuse AB of the right triangle ABC, where the endpoint N of the protractor O's scale line coincides with point A. The ray CP starts from CA and rotates clockwise at a speed of 3 degrees per second. CP intersects the semicircular arc of the protractor at point E. At the 20th second, the reading corresponding to point E on the protractor is ()

Answer: 120°

1. Given that the diameter of the protractor coincides with the hypotenuse AB of the right triangle ABC, and the endpoint N of the protractor's O scale line coincides with point A. Ray CP starts from CA and rotates clockwise at a speed of 3 degrees per second, so at the 20th second, ∠PCA = 3°×20 = 60°.

2. Since ∠ACB = 90°, ##AB is the diameter of circle O, so according to (corollary 2 of the inscribed angle theorem) the inscribed angle subtended by the diameter is a right angle, it can be concluded that ∠ACB is an inscribed angle, ##i.e., point C is on circle O.

3. According to the inscribed angle theorem, in a circle, ##the central angle is twice the inscribed angle, ##since C is on circle O, so according to the definition of the inscribed angle, ∠ECA is the inscribed angle of arc EA, ∠EOA is the central angle of arc EA##, ∠EOA = 2∠ECA = 2×60° = 120°.

4. Through the above reasoning, it is finally concluded that the reading corresponding to point E on the protractor is 120°.

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

Definition of Inscribed Angle

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

Inscribed Angle Theorem

<note>∠EOA is the central angle, ∠ECA is the corresponding inscribed angle</note>, according to the Inscribed Angle Theorem, <note>the central angle ∠EOA = 2 × ∠ECA</note>, therefore <note>∠EOA = 2 × 60° = 120°</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 diameter AB subtends the angle ∠ACB at the circumference, which is a right angle (90 degrees).

Definition of Central Angle

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


Question: As shown in the figure, in ⊙O, the diameter AB=5cm, the chord AC=4cm, then the distance from point O to the line AC is ()

Answer: 1.5cm

1. As shown in the figure, ##through point O, draw OD perpendicular to the diameter AC at point D##, and connect BC. ##According to the perpendicular bisector theorem, segment OD bisects chord AC##, AD is equal to CD, which means point D is the midpoint of chord AC.

2. Since AB is the diameter of ⊙O, according to the theorem of the angle subtended by a diameter, we get ∠ACB=90°.

3. By the Pythagorean theorem, in the right triangle △ACB, the length of BC can be calculated, which is BC equal to ##√(AB? - AC?)##.

4. Given AB=5cm, chord AC=4cm, substituting into the calculation gives BC=##√(5? - 4?)##=3 cm.

5. Since point O is the midpoint of diameter AB, ##according to the midline theorem of a triangle, OD: BC=1/2, the length of OD is half of BC, given BC=3, thus OD=BC/2=3/2=1.5cm##.

6. Through the above reasoning, the final distance from point O to line AC is 1.5cm.

圆; 垂线; 弦; 直角三角形; 线段

Definition of Right Triangle

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

Definition of Chord

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

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

Triangle Midline Theorem

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

In <note>triangle ABC</note>, <note>point D is the midpoint of side AC</note>, <note>point O is the midpoint of side AB</note>, <note>segment OD connects these two midpoints</note>. According to the Triangle Midline Theorem, <note>segment OD is parallel to the third side BC</note> and <note>equals half of the third side BC</note>, that is, <note>OD || BC</note> and <note>BC = 1/2 * OD</note>.

Pythagorean Theorem

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

<note>In the diagram of this problem, in the right triangle △ACB, ∠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>AB² = AC² + BC²</note>.

Perpendicular Diameter Theorem

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

In the figure of this problem, in circle O, <note>segment OD</note> is perpendicular to chord AC. Then, according to the Perpendicular Diameter Theorem, <note>segment OD</note> bisects chord AC, that is, AD=DC, and <note>segment OD</note> bisects the arcs subtended by chord AC, that is, <note>arc AD=arc CD</note>.


Question: As shown in the figure, points A, B, and C are on circle O. Connect AB and AC. If ∠BOC=100°, then the sum of ∠B and ∠C is () degrees.

Answer: 50°

1. Given ∠BOC=100°, according to the inscribed angle theorem, ∠A=1/2∠BOC=50°. Connect AO in the figure.

2. Since OA=OB=OC are radii, triangles OAB and OAC are isosceles triangles, so ∠BAO=∠B and ∠CAO=∠C.

3. Therefore, ∠B+∠C=∠BAO+∠CAO=∠A=50°

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

圆; 圆心角; 圆周角; 圆内接四边形

Definition of Radius

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

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

Inscribed Angle Theorem

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

Definition of Isosceles Triangle

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

<note>Side OA and side OB are equal</note>, therefore triangle OAB is an isosceles triangle; similarly, <note>side OA and side OC are equal</note>, therefore triangle OAC 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>in the isosceles triangle OAB, sides OA and OB are equal</note>. Therefore, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, that is, <note>∠BAO = ∠OBA</note>. <note>In the isosceles triangle OAC, sides OA and OC are equal</note>. Therefore, according to the properties of an isosceles triangle, the angles opposite the equal sides are equal, that is, <note>∠CAO = ∠OCA</note>.

Definition of Central Angle

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

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

Definition of Inscribed Angle

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


Question: As shown in the figure, AB is the diameter of ⊙O, ∠ABC=25°, then the degree of ∠D is ()

Answer: 65°

1. Given AB is the diameter of ⊙O, ##according to the theorem of the inscribed angle, the inscribed angle subtended by the diameter is a right angle##, thus ∠ACB is a right angle, which is 90°.

2. According to the given condition ∠ABC = 25°, using the triangle angle sum theorem, we get ∠CAB = ##∠ACB## - ∠ABC = 90° - 25° = 65°.

3. ##According to the theorem of the inscribed angle, the inscribed angles subtended by arc BC are equal, thus ∠CAB = ∠CDB = 65°##.

4. Based on the previous step, ##∠CDB = 65°##, we can conclude that ∠D = 65°.

5. Finally, we can conclude that the measure of ∠D is 65°.

圆; 圆周角; 直角三角形; 圆内接四边形

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

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

Triangle Angle Sum Theorem

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

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