Question: As shown in the figure, trees are planted on a slope. Given ∠A=30°, AC=3m, the slope distance between two adjacent trees AB is equal to ()

Answer: 2√{3}m

1. Given ∠BAC=30°, AC = 3m. Use the ##cosine function## in the right triangle.

2. According to the ##cosine function##, AC / cos(∠BAC) = AB.

3. Since ∠BAC=30°, cos(30°) = √3/2.

4. Substitute the cosine value of the angle into the formula, then AB = 3 / (√3 / 2) = 3 * 2 / √3 = 2√3.

5. Through the above reasoning, it is concluded that the slope distance AB between the two adjacent trees is 2√3 m.

Cosine Function

The cosine function is a trigonometric function that can be defined via the sides of a right triangle. For an angle α, its cosine value is equal to the ratio of the length of the adjacent side to the hypotenuse. In the unit circle, the cosine value is the x-coordinate of the point on the circle that makes an angle α with the positive x-axis.

In the <note>right triangle ABC</note>, <note>side AC is the adjacent side to angle BAC</note>, and <note>side AB is the hypotenuse</note>. Therefore, according to the definition of the cosine function, <note>the cosine value of angle BAC</note> is equal to <note>the ratio of the adjacent side AC to the hypotenuse AB</note>, i.e., <note>cos(∠BAC) = AC / AB</note>.

Definition of Right Triangle

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

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


Question: As shown in the figure, it is known that D, E, F are points on the sides BC, CA, and AB of the isosceles △ABC respectively. If AB=AC, ∠FDE=∠B, BD=2, CD=3, CE=4, AE=1, then the length of AF is ()

Answer: 3.5

1. Given AB = AC, according to the properties of an isosceles triangle, we get ∠ABD = ∠ACD.

2. Given ∠FDE = ∠C (equal to ∠B), so by the exterior angle theorem of the triangle ∠FBD + ∠BFD = ∠FDE + ∠EDC, we get ∠BFD = ∠EDC.

3. Because ∠B = ∠C, ∠BFD = ∠EDC, by the similarity theorem of triangles (AA), we know △BDF ∽ △CED.

4. According to the definition of similar triangles, we get BD:CE = BF:DC.

5. Given BD = 2, CD = 3, CE = 4, substituting these values into the ratio of sides gives: 2:4 = BF:3, solving for BF we get BF = 1.5.

6. Given AB = AC, we get AB = AC = AE + EC = 1 + 4 = 5.

7. From AF = AB - BF, we get AF = 5 - 1.5 = 3.5.

8. Through the above reasoning, the final answer is 3.5.

等腰三角形; 普通三角形; 点; 线段; 内错角

Definition of Isosceles Triangle

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

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

Definition of Similar Triangles

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

In the figure of this problem, <note>triangle BDF and triangle CED are similar triangles</note>. According to the definition of similar triangles: <note>∠BDF = ∠CED</note>, <note>∠BFD = ∠CDE</note>, <note>∠DBF = ∠DCE</note>; <note>BD/CE = BF/CD = DF/DE</note>. Based on the given sides <note>BD = 2</note>, <note>CE = 4</note>, <note>CD = 3</note>, the ratio is <note>2/4 = BF/3</note>, thus <note>BF = 1.5</note>.

Similarity Theorem for Triangles (AA)

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

In the figure of this problem, in triangles FBD and EDC, if <note>angle B is equal to angle C</note>, and <note>angle BFD is equal to angle EDC</note>, then <note>triangle BFD is similar to triangle EDC</note>.

Exterior Angle Theorem of Triangle

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

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

Properties of Isosceles Triangle

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

<note>In the isosceles triangle ABC, side AB and side AC are equal</note>. Therefore, according to the properties of isosceles triangles, <note>the angles opposite the equal sides are equal</note>, i.e., <note>angle ABC = angle ACB</note>.

Definition of Exterior Angle of a Polygon

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

The original text: <note>One interior angle of polygon BFD is ∠BDF</note>, <note>The angle ∠FDC formed by extending the adjacent sides BD and FD of this interior angle</note> is called the exterior angle of the interior angle <note>∠BDF</note>.


Question: In the mathematics practical exploration class, the teacher assigned the students to measure the height of the school flagpole. As shown in the figure, Xiaoming's study group measured the angle of elevation to the top of the flagpole to be 60° from a point 10 meters away from the base of the flagpole. Then the height of the flagpole is () meters.

Answer: 10√{3}

1. Let the triangle shown in the figure be triangle ABC, the top of the flagpole be point B, the bottom of the flagpole be point A, and the location of the study group be point C. According to the problem, the study group is 10 meters away from the bottom of the flagpole, and the angle of elevation measured by the theodolite from the study group to the top of the flagpole is 60°, i.e., AC=10, ∠ACB=60°. Since the flagpole is perpendicular to the ground, it is also perpendicular to the horizontal line AC, so AB⊥AC, then ∠CAB=90°. According to the definition of a right triangle, triangle ABC is a right triangle.

2. In the right triangle ABC, according to the definition of the tangent function, using the tangent function formula: tan(θ)=opposite/adjacent, we get tan(∠ACB)=AB/AC.

3. Given: tan(∠ACB)=tan(60°)=√3, AC=10, therefore √3=AB/10.

5. Therefore, through the above calculations, the height of the flagpole is 10√3 meters.

仰角; 直角三角形; 正切

Definition of Right Triangle

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

Angle of Elevation

The angle formed between the horizontal line and the observer's line of sight when the observer looks upward towards an object is referred to as the angle of elevation.

<note>In the diagram of this problem, the observer is located at point C</note>, looking forward along the horizontal line to form segment AC. When the observer <note>looks up from point C to an object located at point B, the line of sight forms segment BC</note>. At this time, <note>the angle ∠ACB formed between the line of sight BC and the horizontal line AC is the angle of elevation</note>.

Definition of Tangent Function

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

<note>In the figure of this problem, in the right triangle ABC</note>, angle ACB is an acute angle, <note>side AB is the opposite side of angle ACB</note>, <note>side AC is the adjacent side of angle ACB</note>, so the tangent value of angle ACB is equal to the length of side AB divided by the length of side AC, that is <note>tan(∠ACB) = AB / AC</note>.


Question: As shown in the figure, in △ABC, AB=BC=2, ⊙O with AB as the diameter is tangent to BC at point B, then AC equals ()

Answer: 2√{2}

1. Given that circle O with diameter AB is tangent to BC at point B, according to the property of the tangent to a circle, the tangent at the point of tangency is perpendicular to the radius passing through the point of tangency, therefore ##∠ABC=90°##.

####

##2.## Using the Pythagorean theorem in triangle △ABC, we get ##AC?=AB?+BC?##.

##3.## Substituting the given values AB=2 and BC=2, we calculate ##AC=√(2?+2?)=√8=2√2##.

##4.## Through the above reasoning, we finally obtain the answer as 2√2.

等腰三角形; 圆; 切线; 直角三角形; 圆周角

Definition of Tangent to a Circle

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

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

Definition of Isosceles Right Triangle

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

In the figure of this problem, triangle ABC is an isosceles right triangle, where <note>angle ABC is a right angle (90 degrees)</note>, and <note>sides AB and BC are equal right-angle sides</note>.

Property of the Tangent Line to a Circle

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

Original text: <note>In the figure of this problem, in circle O, point B is the point of tangency between line CB and the circle</note>, <note>line segment OB is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>tangent line CB is perpendicular to the radius OB at the point of tangency B</note>, that is, <note>∠ABC=90 degrees</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, ∠ABC is a right angle (90 degrees)</note>, <note>sides AB and BC are the legs</note>, <note>side AC is the hypotenuse</note>, so according to the Pythagorean Theorem, ##AC?=AB?+BC?##.


Question: As shown in the figure, given that the radius of ⊙O is 6, M is a point outside ⊙O, and OM = 12. The line passing through M intersects ⊙O at A and B. The symmetric points of A and B with respect to OM are C and D, respectively. AD and BC intersect at point P. Find the length of OP.

Answer: 3

1. Given the radius of ⊙O is 6, OM=12, A and B are points symmetric to OM, respectively C and D. The line passing through M intersects ⊙O at A and B. Connect AC intersecting MO at point N, connect OA and OC.

2. According to symmetry, MN⊥AC and AN=CN, thus point P is on MN.

3. Since AP=CP, ∠APN=∠CPN, further obtaining ∠BPM=∠APO.

4. According to the perpendicular bisector theorem, the diameter perpendicular to the chord and bisecting the chord, ∠AON=∠CON=##1/2##∠AOC.

5. Also, according to the inscribed angle theorem, the inscribed angle of the same arc is half of the central angle, thus ∠ABC=##1/2##∠AOC, so ∠AON=∠ABC.

6. According to the triangle angle sum theorem, ∠AON=180°-(∠OAM+∠AMO), and ∠ABC=180°-(∠BPM+∠AMO), further obtaining ∠OAM=∠BPM, thus ∠OAM=∠APO.

7. Also, according to the similarity triangle criterion (AA), ∠AOP=∠MOA, thus △AOP∽△MOA.

8. According to the definition of similar triangles, ∠MOA##=##∠AOP, then ##OP/OA=OA/OM##.

9. Substituting the given OA=6, OM=12, obtaining ##OP/6=6/12##, solving: OP=3.

10. Through the above reasoning, the final answer is that the length of OP is 3.

圆; 对称; 线段; 直线; 弦

Definition of Radius

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

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

Definition of Chord

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

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

Symmetric Point

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

In the figure of this problem, <note>points A and C are symmetric points with respect to line OM</note>, and <note>the distance from point A to line OM is equal to the distance from point C to line OM</note>. Similarly, <note>points B and D are symmetric points with respect to line OM</note>, and <note>the distance from point B to line OM is equal to the distance from point D to line OM</note>. According to the definition of symmetric points, <note>points A and C are on opposite sides of line OM</note>, and satisfy the relationship of equal distances, i.e., <note>distance AM = distance CM</note>; <note>points B and D are on opposite sides of line OM</note>, and satisfy the relationship of equal distances, i.e., <note>distance BM = distance DM</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>Radius MO</note> is perpendicular to chord AC, then according to the Perpendicular Diameter Theorem, <note>Radius MO bisects chord AC</note>, that is <note>AM=MC</note>, and diameter MO bisects the two arcs subtended by chord AC, that is <note>arc AN=arc CN</note>.

Inscribed Angle Theorem

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

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

Triangle Angle Sum Theorem

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

In <note>triangle AOM</note>, <note>angle AOM</note>, <note>angle OAM</note>, and <note>angle AMO</note> are the three interior angles of <note>triangle AOM</note>. According to the Triangle Angle Sum Theorem, <note>angle AOM</note> + <note>angle OAM</note> + <note>angle AMO</note> = 180°. Similarly, in <note>triangle MPB</note>, <note>angle MPB</note>, angle BMP, and angle MPB are the three interior angles of <note>triangle MPB</note>. According to the Triangle Angle Sum Theorem, <note>angle MPB</note> + angle BMP + <note>angle MPB</note> = 180°.

Definition of Similar Triangles

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

<note>Triangle AOP and triangle MOA are similar triangles</note>. According to the definition of similar triangles: <note>∠AOP = ∠MOA</note>, <note>∠OAP = ∠OMA</note>, <note>∠OPA = ∠MAO</note>; <note>OP/OA = OA/OM</note>.

Symmetry

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

<note>In the diagram of this problem, the figure MOB has symmetry</note>. Specifically, <note>if the figure MOB is symmetric about the line OM</note>, then <note>the line OM is the axis of symmetry of the figure MOB</note>. Each part of the figure MOB has a symmetric corresponding part on <note>the other side of the axis of symmetry</note>. If <note>the figure MOB is symmetric about the point O</note>, then <note>the point O is the center of symmetry of the figure MOB</note>. Each part of the figure MOB has a symmetric corresponding part on <note>the other side of the center of symmetry</note>. If <note>the figure MOB remains unchanged after rotating θ degrees</note>, then <note>the figure MOB has rotational symmetry</note>, and θ degrees is its <note>angle of symmetry</note>. Through these symmetry operations, the figure MOB visually remains unchanged.

Definition of Central Angle

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

In the figure of this problem, <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 the lines OA and OC is called the central angle.

Definition of Inscribed Angle

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

In the figure of this problem, in circle O, <note>the vertex of angle ABC (point B) 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>.

Similarity Theorem for Triangles (AA)

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

<note>Angle AOP is equal to angle MOA</note>, and <note>angle OAM is equal to angle APO</note>, so <note>triangle AOP is similar to triangle MOA</note>.


Question: As shown in the figure, in parallelogram ABCD, diagonals AC and BD intersect at point O, and point E is the midpoint of side CD. Connect OE. If the perimeter of parallelogram ABCD is 24 and BD = 8, then the perimeter of △DOE is ()

Answer: 10

1. Given the perimeter of parallelogram ABCD is 24, according to the ##perimeter formula of parallelogram##, AB + BC + CD + DA = 24. Since the opposite sides of the parallelogram are equal, BC + CD = 12.

2. Since point O is the midpoint of diagonal BD, OD = 0.5 * BD = 0.5 * 8 = 4.

3. Given point E is the midpoint of side CD, so DE = 0.5 * CD.

4. Since BC + CD = 12, CD = 12 - BC.

5. According to the properties of parallelogram, BC = AD. Therefore, CD = 12 - AD.

6. Since point O is the midpoint of diagonal BD and point E is the midpoint of side CD, according to the midline theorem, OE = 0.5 * BC = 0.5 * AD.

7. In summary, the perimeter of triangle DOE is OE + DE + OD.

8. Adding the lengths of each side, OE + DE + OD = 0.5 * AD + 0.5 * (12 - AD) + 4 = 6 + 4.

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

平行四边形; 中点; 线段; 普通三角形; 对称

Definition of Parallelogram

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

<note>Quadrilateral ABCD is a parallelogram</note>, <note>sides AB and CD are parallel and equal</note>, <note>sides AD and BC are parallel and 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 CD is point E</note>. According to the definition of the midpoint of a line segment, <note>point E divides line segment CD into two equal parts</note>, that is, <note>the lengths of line segments CE and ED are equal</note>. That is, CE = ED.

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

Triangle Midline Theorem

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

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

Perimeter of a Parallelogram

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

In the figure of this problem, <note>parallelogram ABCD</note>, <note>sides BC and CD are adjacent sides</note>, <note>sides AB and AD are also adjacent sides</note>. According to the perimeter formula of a parallelogram, the perimeter of the parallelogram is equal to twice the sum of its two sets of adjacent sides, that is, <note>perimeter P = 2(BC + CD)</note>.

Perimeter Theorem of Triangle

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

<note>The three sides of triangle △OED are OE, OD, and ED</note>. According to the Perimeter Theorem of Triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, that is, <note>Perimeter P=OE+OD+ED</note>.


Question: As shown in the figure, in the right triangle △ABC, AD⊥BC at D, DE⊥AB at E, if AD=3 and DE=2, then AC=()

Answer: \frac{9}{2}

1. Given in the right triangle ABC, AD is perpendicular to BC and intersects at point D, DE is perpendicular to AB and intersects at point E.

2. Because AD is perpendicular to BC, angle ADC is equal to 90 degrees. Since DE is perpendicular to AB, angle DEA is equal to 90 degrees. Moreover, ABC is a right triangle, so angle BAC is also equal to 90 degrees.

3. Therefore, it can be concluded that angle EAD plus angle DAC equals 90 degrees.

4. According to the triangle angle sum theorem and angle ADC equals 90 degrees. So angle C plus angle CAD equals 90 degrees, it can be concluded that angle C equals angle EAD.

5. According to the definition of the sine function, sin(angle EAD) equals sin(angle C), which equals DE divided by AD, and also equals AD divided by AC.

6. Given AD equals 3, DE equals 2, so 2 divided by 3 equals 3 divided by AC.

7. Solving this equation gives: AC equals 9 divided by 2.

8. Through the above reasoning, the final answer is 4.5.

直角三角形; 垂线; 垂直平分线; 线段

Definition of Right Triangle

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

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

Definition of Perpendicular Lines

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

<note>The angle ∠ADC formed by the intersection of line AD and line BC is 90 degrees</note>, therefore, according to the definition of perpendicular lines, line AD and line BC are perpendicular to each other; <note>The angle ∠DEA formed by the intersection of line DE and line AB is 90 degrees</note>, therefore, according to the definition of perpendicular lines, line DE and line AB are perpendicular to each other.

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.

<note>In right triangle ADC, 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 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>.

Triangle Angle Sum Theorem

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


Question: As shown in the figure, the 'V-shaped' steel frame has inclined beams AB = AC, the top angle ∠BAC = 120°, and the span BC = 10m. AD is the pillar (i.e., the median of the base BC). Two support frames DE ⊥ AB and DF ⊥ AC. Then DE + DF equals ()

Answer: 5m

1. Given △ABC, AB=AC, and ∠BAC=120°. According to the properties of an isosceles triangle, ∠ABC=∠ACB.

2. Since the sum of the interior angles of a triangle is 180°, ∠ABC + ∠ACB + ∠BAC = 180°.

3. Because ∠BAC=120°, then ∠ABC + ∠ACB = 60°.

4. Since ∠ABC = ∠ACB, 2∠ABC = 60°, thus ∠ABC=30°, ∠ACB=30°.

5. Given AD is the median of the triangle, and DE ⊥ AB, DF ⊥ AC, E and F are the feet of the perpendiculars respectively.

7. Since AD is the median of △ABC, D is the midpoint of BC, and BC=10 meters, thus BD=DC=5 meters.

8. Since DE ⊥ AB, DF ⊥ AC, and E and F are the feet of the perpendiculars, according to the properties of a 30°-60°-90° triangle, DE is half of BD, and DF is half of DC.

8. ∴DE = 1/2 * BD and DF = 1/2 * DC. Since BD = DC = 5 meters, DE = 1/2 * 5 = 2.5 meters, DF = 1/2 * 5 = 2.5 meters.

9. Therefore, DE + DF = 2.5 meters + 2.5 meters = 5 meters.

10. Through the above reasoning, the final answer is 5 meters.

等腰三角形; 垂线; 垂直平分线; 直角三角形

Definition of Isosceles Triangle

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

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

Definition of Median of a Triangle

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

In the figure of this problem, in triangle ABC, <note>vertex A is a vertex of the triangle</note>, <note>the opposite side BC is the side opposite to vertex A</note>. <note>Point D is the midpoint of side BC</note>, <note>segment AD is the segment from vertex A to midpoint D of opposite side BC</note>, therefore <note>AD is a median of triangle ABC</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.

<note>DE intersects AB at point E</note>, and <note>DE is perpendicular to AB</note>, therefore <note>point E is the foot of the perpendicular from DE to AB</note>; similarly, <note>DF intersects AC at point F</note>, and <note>DF is perpendicular to AC</note>, therefore <note>point F is the foot of the perpendicular from DF to AC</note>.

Properties of Isosceles Triangle

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

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

Triangle Angle Sum Theorem

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

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

Properties of 30°-60°-90° Triangle

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

In the problem diagram, <note>30°-60°-90° triangles BDE and CDF</note>, <note>angle B and angle C are 30 degrees</note>, <note>angle BDE and angle CDF are 60 degrees</note>, <note>angle BED and angle CEF are 90 degrees</note>. <note>Sides BD and CD are the hypotenuses</note>, <note>sides DE and DF are the sides opposite the 30-degree angles</note>, <note>sides BE and CF are the sides opposite the 60-degree angles</note>. According to the properties of 30°-60°-90° triangles, <note>side DE is equal to half of side BD</note>, <note>side DF is equal to half of side CD</note>. That is: <note>BE = 1/2 * BD</note>, <note>DF = 1/2 * CD</note>.


Question: As shown in the figure, in parallelogram ABCD, E is the midpoint of DC, the area of △DEF is 2, then the area of △ABF is ()

Answer: 8

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

2. Since E is the midpoint of DC, DE = EC = 1/2 DC.

3. ##According to the definition of a parallelogram##, AB = CD, thus DE = 1/2 AB.

4. ##According to the parallel axiom 2 of parallel lines, ∠BAE=∠DEA, ∠EDB=∠DBA, and according to the similarity theorem (AA) of triangles, △DEF and △BAF form similar triangles##.

5. The ratio of corresponding sides DE and AB of the two similar triangles is 1:2, thus the similarity ratio is 1:2, i.e., DE/AB = 1/2.

6. According to the theorem that the ratio of areas is equal to the square of the similarity ratio (the area ratio of similar triangles is equal to the square of the ratio of corresponding side lengths), ##S△DEF / S△BAF## = (DE/AB)^2 = (1/2)^2 = 1/4.

7. Given that the area of △DEF is 2, then ##S△BAF = S△DEF## / (1/4) = 2 / (1/4) = 2 * 4 = 8.

8. Through the above reasoning, the final area of △ABF is 8.

平行四边形; 中点; 普通三角形

Definition of Parallelogram

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

Midpoint of a Line Segment

In the figure of this problem, <note>the midpoint of line segment DC is point E</note>. According to the definition of the midpoint of a line segment, <note>point E divides line segment DC into two equal parts</note>, that is, <note>the lengths of line segments DE and EC are equal</note>. That is, <note>DE = EC = 1/2 DC</note>.

Definition of Alternate Interior Angles

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

<note>Two parallel lines AB and CD</note> are intersected by <note>lines AE and DB</note>, where <note>∠BAE and ∠DEA</note>, <note>∠EDB and ∠DBA</note> are located between the two parallel lines and on opposite sides of the intersecting lines <note>AE and DB</note>. Therefore, <note>∠BAE and ∠DEA are alternate interior angles</note>, <note>∠EDB and ∠DBA are alternate interior angles</note>. <note>Alternate interior angles are equal</note>, that is, <note>∠BAE=∠DEA</note>, <note>∠EDB=∠DBA</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 DEF and triangle BAF are similar triangles</note>, <note>the ratio of side DE to side AB is 1:2, which means the similarity ratio is 1:2</note>. Therefore, <note>the area ratio of triangle DEF to triangle BAF is equal to (1/2)^2 = 1/4</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>In the diagram of this problem, two parallel lines AB and CD are intersected by lines AE and DB</note>, forming the following geometric relationships: <note>Alternate interior angles: ∠BAE=∠DEA, ∠EDB=∠DBA</note>.

Similarity Theorem for Triangles (AA)

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

In <note>△DEF and △BAF</note>, <note>∠BAE=∠DEA</note>, <note>∠EDB=∠DBA</note>, so <note>△DEF and △BAF are similar</note>.


Question: As shown in the figure, in △ABC, E and F are the midpoints of AB and AC, respectively. If the area of △AEF is 1, then the area of quadrilateral EBCF is ()

Answer: 3

1. Given that point E and point F are the midpoints of segment AB and segment AC respectively.

2. Since point E and point F are the midpoints of segment AB and segment AC respectively, according to the midline theorem, segment EF is parallel to segment BC and EF = 1/2 BC, thus ∠AEF = ∠ABC (corresponding angles are equal).

3. According to the criteria for similar triangles, △AEF is similar to △ABC because ∠A = ∠A and ∠AEF = ∠ABC.

4. According to the area ratio of similar triangles, the area ratio is equal to the square of the similarity ratio, thus (area△AEF) / (area△ABC) = (EF/BC)² = (1/2)² = 1/4.

5. Given that the area of △AEF = 1, through the proportional relationship, the area of △ABC is 4.

6. The area of quadrilateral EBCF is equal to the area of △ABC minus the area of △AEF, thus the area of quadrilateral EBCF is 4 - 1 = 3.

7. Through the above reasoning, the final answer is 3.

中点; 平行线; 普通三角形; 普通四边形; 平行四边形

Midpoint of a Line Segment

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

In the figure of this problem, the midpoint of line segment AB is point <note>E</note>, and the midpoint of line segment AC is point <note>F</note>. According to the definition of the midpoint of a line segment, point <note>E divides line segment AB into two equal parts, that is, AE = EB</note>, and point <note>F divides line segment AC into two equal parts, that is, AF = FC</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 E is the midpoint of side AB</note>, <note>point F is the midpoint of side AC</note>, <note>segment EF connects these two midpoints</note>. According to the Triangle Midline Theorem, <note>segment EF is parallel to the third side BC and is equal to half of the third side BC</note>, that is, <note>EF || BC</note>, and <note>EF = 1/2 * BC</note>.

Parallel Postulate 2 of Parallel Lines

<note>Two parallel lines EF and BC are intersected by a third line AB</note>, forming the following geometric relationships:
<note>Corresponding angles: angle AEF and angle ABC are equal</note>;
<note>Angle AFE 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.

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

Definition of Similar Triangles

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

<note>Triangle AEF and triangle ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠AEF=∠ABC</note>, <note>∠A = ∠A</note>, <note>∠AFE = ∠ACB</note>; <note>AE/AB = AF/AC = EF/BC</note>.

Similarity Theorem for Triangles (AA)

<note>Triangle AEF and triangle ABC have two pairs of corresponding angles that are equal</note>, namely <note>∠EAF = ∠BAC</note> and <note>∠AEF = ∠ABC</note>, therefore <note>triangle AEF is similar to triangle ABC</note>.

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


Question: As shown in the figure, in the square ABCD with side length 9, F is a point on AB. Connect CF. Through point F, draw FE perpendicular to CF, intersecting AD at point E. If AF = 3, then AE equals ()

Answer: 2

1. Given that quadrilateral ABCD is a square, according to the properties of a square, we have AD=AB=BC=9, and ∠DAB=∠ABC=90°.

2. According to the problem statement, FE⊥CF, thus ∠EFC=90°.

5. Because ∠EFC=90°, ∠AFE + ∠CFB = 90°, combining the conclusions from (3) and (4), we get ∠AEF = ∠CFB.

6. Since △AEF and △BFC have ∠AEF = ∠CFB, and ∠A=∠B=90°, according to the AA similarity criterion, we get △AEF ∽ △BFC.

7. According to the properties of similar triangles, corresponding sides are proportional, thus AE/BF = AF/BC.

8. Since AF=3 and BC=9, BF=BC-AF=9-3=6, substituting these values into the proportional relationship from the previous step, we get AE/6 = 3/9.

9. By solving the proportion equation, we get AE = 2.

10. Through the above reasoning, we finally get the length of AE as 2.

正方形; 垂线; 直角三角形; 线段

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°).

In the quadrilateral <note>ABCD</note>, <note>sides AB, BC, CD, and DA are equal</note>, and <note>angles DAB, ABC, BCD, and CDA are all right angles (90 degrees)</note>, so <note>ABCD is a square</note>.

Definition of Perpendicular Lines

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

The original text: <note>The angle ∠EFC formed by the intersection of line segment FE and line segment CF is 90 degrees</note>, therefore, according to the definition of perpendicular lines, <note>line segment FE and line segment CF are perpendicular to each other</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 A</note> is a right angle (90 degrees), so triangle AEF is a right triangle. Side AF and <note>side AE</note> are the legs, <note>side FE</note> is the hypotenuse. In triangle CFB, <note>angle B</note> is a right angle (90 degrees), so triangle CFB is a right triangle. <note>Side CB</note> and side FB are the legs, <note>side CF</note> is the hypotenuse.

Similarity Theorem for Triangles (AA)

<note>In the figure of this problem, in triangles AEF and CFB ##∠A=∠B=90°##</note>, and <note>∠AEF = ∠CFB</note>, so <note>triangle AEF is similar to triangle CFB</note>.

Definition of Similar Triangles

In the figure of this problem, <note>△AEF</note> and <note>△BFC</note> are similar triangles. According to the definition of similar triangles: <note>∠FEA</note> = <note>∠CFB</note>, <note>∠AFE</note> = <note>∠BCF</note>, <note>∠A</note> = <note>∠B</note>; <note>\( \frac{AE}{BF} = \frac{AF}{BC} \)</note>. AF=3, BC=9, BF=BC-AF=6, therefore <note>\( \frac{AE}{6} = \frac{3}{9} \)</note>. By solving the proportion equation, we get <note>AE = 2</note>.


Question: Let BF intersect AC at point P, AE intersect DF at point Q. If ∠APB=126°, ∠AQF=100°, then ∠A-∠F=()

Answer: 46°

1. According to the problem statement, it is known that ∠APB = 126°, ∠AQF=100°. Let the intersection of AE and BF be G.

3. In triangle APG, according to the exterior angle theorem, we get ∠APB = ∠A + ∠AGP.

4. In triangle QFG, according to the triangle angle sum theorem, we get ∠F + ∠FGQ + ∠GQF=180°.

5. Substitute ∠APB = 126°, ∠GQF=100°.

6. We get: 126° = ∠A + ∠AGP, 80° = ∠F + ∠QGF.

7. Since ∠AGP and ∠QGF are vertical angles, ∠AGP= ∠QGF.

8. Solve the equations, we get ∠A - ∠F = 46°.

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

对顶角; 普通三角形; 邻补角

Triangle Angle Sum Theorem

In the original problem diagram, in triangle QGF, angle F, angle GQF, and angle QGF are the three interior angles of triangle APB, according to the Triangle Angle Sum Theorem, <note>angle F + angle GQF + angle QGF = 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.

In the diagram of this problem, in triangle APG, <note>∠APB is an exterior angle of the triangle</note>, <note>∠A and ∠AGP are the two non-adjacent interior angles</note>. According to the Exterior Angle Theorem of Triangle, <note>the exterior angle ∠APB is equal to the sum of the two non-adjacent interior angles ∠A and ∠AGP, i.e., ∠APB = ∠A + ∠AGP</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, <note>line segment BF intersects AE at point G</note>, forming four angles: ∠BGE, ∠AGF, <note>∠AGP and ∠FGQ</note>. According to the definition of vertical angles, <note>∠AGP and ∠FGQ are vertical angles</note>. Since the angles of vertical angles are equal, <note>∠AGP=∠FGQ</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 polygon APG is <note>∠APG</note>. The angle <note>∠APB</note> formed by extending the adjacent sides <note>GP</note> and <note>AP</note> of this interior angle is called the exterior angle of the interior angle <note>∠APG</note>.


Question: As shown in the figure, the perimeter of rhombus ABCD is 16, ∠A=60°, then the length of diagonal BD is ()

Answer: 4

1. Given that the perimeter of rhombus ABCD is 16, according to the formula for the perimeter of a rhombus, the perimeter of a rhombus is equal to four times its side length. Let the side length of the rhombus be a, thus 4*a=16, we get AB=AD=CD=BC=a=4.

2. Since AB=AD, triangle ABD is an isosceles triangle. According to the properties of an isosceles triangle, ∠ABD=∠ADB.

3. In △ABD, given ∠A=60°, according to the triangle angle sum theorem, we get ∠ABD=∠ADB=(180°-∠A)/2=60°. Using the properties of an equilateral triangle, we conclude that △ABD is an equilateral triangle, that is BD=AB=AD=4.

4. Therefore, the length of diagonal BD is 4.

菱形; 等边三角形; 正弦

Definition of Rhombus

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

In the figure of this problem, all sides of the quadrilateral <note>ABCD</note> are equal, specifically <note>AB, BC, CD, AD</note>. Therefore, the quadrilateral <note>ABCD</note> is a rhombus.

Definition of Equilateral Triangle

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

In the figure of this problem, <note>triangle ABD is an equilateral triangle</note>. <note>The lengths of sides AB, AD, and BD are equal, all being 4</note>, and <note>the measures of angles BAD, ABD, and ADB are equal, all being 60°</note>.

Properties of Equilateral Triangle

In an equilateral triangle, each interior angle measures 60°.

In the figure of this problem, in the <note>equilateral triangle ABD</note>, the sides <note>AB, AD, and BD are equal</note>, thus according to the properties of an equilateral triangle, <note>each interior angle of triangle ABD is 60°</note>. That is to say, <note>angle ABD, angle ADB, and angle BAD are all 60°</note>.

Triangle Angle Sum Theorem

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

<note>In triangle ABD, angles BAD, ABD, and ADB are the three interior angles of triangle ABD</note>, according to the Triangle Angle Sum Theorem, <note>angle BAD + angle ABD + angle ADB = 180°</note>. In the figure of this problem, <note>in triangle xxx, sides xx and xx are equal</note>, therefore <note>triangle xxx is an isosceles triangle</note>.

Definition of Isosceles Triangle

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

In triangle ABD, side AB and side AD are equal, therefore triangle ABD 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 figure of this problem, in the isosceles triangle ABD, sides AB and AD are equal</note>. Therefore, according to the properties of isosceles triangles, the angles opposite the equal sides are equal, that is, <note>angle ABD = angle ADB</note>.


Question: As shown in the figure, AB∥CD, AC and BD intersect at O, BO=6, DO=3, AC=12, then the length of AO is ()

Answer: 8

1. Given AB∥CD, ##according to the parallel axiom 2 of parallel lines, alternate interior angles are equal##, we can obtain ∠DCA=∠CAB (##alternate interior angles are equal##) and ∠BDC=∠DBA (##alternate interior angles are equal##).

####

##2##. According to the ##AA criterion of similar triangles##, we have: ∠DCA = ∠CAB and ∠BDC = ∠DBA, satisfying the condition that corresponding angles are equal, the similarity theorem of triangles holds, that is, △DOC and △BOA are similar.

##3##. Therefore, according to the ##definition of similar triangles##, DO/BO = OC/AO.

##4##. Substitute the given conditions into the proportional relationship, given BO=6, DO=3, AC=12, let AO=x, we get: 3/6=(12-x)/x.

####

##5##. Solve this proportion, we calculate: 1/2 = (12 - x) / x, x = 8.

##6##. Through the above reasoning steps, we finally obtain the length of AO as 8.

平行线; 内错角; 线段; 普通三角形

Definition of Similar Triangles

<note>In the diagram of this problem, triangle DOC and triangle BOA are similar triangles</note>. According to the definition of similar triangles: <note>∠DCO = ∠BAO</note>, <note>∠DOC = ∠BOA</note>, <note>∠ODC = ##∠OBA##</note>; <note>DO/BO = OC/AO</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.

Original: <note>Two parallel lines AB and CD</note> are intersected by <note>a line BD</note>, where <note>angle BDC and angle DBA are located between the two parallel lines and on opposite sides of the intersecting line BD</note>, thus <note>angle BDC and angle DBA are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle BDC is equal to angle DBA</note>. Similarly, <note>two parallel lines AB and CD are intersected by a line AC</note>, where <note>angle DCA and angle CAB are located between the two parallel lines and on opposite sides of the intersecting line AC</note>, thus <note>angle DCA and angle CAB are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle DCA is equal to angle CAB</note>.

Parallel Postulate 2 of Parallel Lines

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

In the diagram of this problem, <note>two parallel lines AB and DC</note> are intersected by <note>a third line AC</note>, forming the following geometric relationship: <note>alternate interior angles: angle DCA and angle CAB are equal</note>. Similarly, <note>two parallel lines AB and DC</note> are intersected by <note>a third line BD</note>, forming the following geometric relationship: <note>alternate interior angles: angle BDC and angle DBA are equal</note>. These relationships illustrate that when two parallel lines are intersected by a third line, the alternate interior angles are equal.

Similarity Theorem for Triangles (AA)

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

In the given diagram, in triangles DOC and BOA, <note>angle DCA is equal to angle CAB</note>, and <note>angle BDC is equal to angle DBA</note>, so <note>triangle DOC is similar to triangle BOA</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.

In the problem diagram, <note>triangle DOC is a geometric figure composed of three non-collinear points D, O, C and their connecting line segments DO, OC, CD</note>. <note>Points D, O, C are the three vertices of the triangle</note>, <note>line segments DO, OC, CD are the three sides of the triangle</note>. Similarly, <note>triangle BOA is a geometric figure composed of three non-collinear points B, O, A and their connecting line segments BO, OA, AB</note>. <note>Points B, O, A are the three vertices of the triangle</note>, <note>line segments BO, OA, AB are the three sides of the triangle</note>.


Question: As shown in the figure, in rhombus ABCD, ∠BAD=120°, BC=10, then the length of diagonal AC is equal to ()

Answer: 10

1. Given quadrilateral ABCD is a rhombus, according to the properties of a rhombus, AB=BC=CD=DA.

3. Since the diagonals of a rhombus are perpendicular bisectors of each other, let the diagonals AC and BD intersect at point O, then BO=OD, ∠AOB=∠AOD=90°, AO=AO. According to the congruence theorem (SAS), △ABO≌△ADO. Therefore, according to the definition of congruent triangles, ∠BAC=∠DAC=1/2∠BAD=60°.

4. In △ABC, since AB=BC and ∠BAC=60°, according to the theorem of equilateral triangles (60-degree angle of an isosceles triangle), △ABC is an equilateral triangle.

5. According to the definition of an equilateral triangle, the sides of an equilateral triangle are equal, therefore AC=BC.

6. It is known that BC=10, therefore AC=10.

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

Definition of Rhombus

<note>In quadrilateral ABCD, all sides AB, BC, CD, and DA are equal</note>, so quadrilateral ABCD is a rhombus. Additionally, <note>the diagonals AC and BD of quadrilateral ABCD are perpendicular bisectors of each other</note>, that is, <note>the diagonals AC and BD intersect at point O</note>, and <note>angle AOB is a right angle (90 degrees)</note>, and <note>AO=OC and BO=OD</note>.

Definition of Equilateral Triangle

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

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

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

In the given problem diagram, it is known that △ABC is an isosceles triangle, <note>sides AB and BC are equal</note>, and there is a <note>60-degree interior angle, i.e., ∠BAC=60°</note>. According to the Equilateral Triangle Identification Theorem, if an isosceles triangle has <note>a 60-degree interior angle</note>, then the triangle's <note>three sides are equal in length</note>, and <note>all three interior angles are 60 degrees</note>. Therefore, it can be determined that △ABC is an <note>equilateral triangle</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.

<note>Side BO is equal to side OD</note>, <note>side AO is equal to side AO</note>, and <note>angle AOB is equal to angle AOD</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.

<note>Triangle AOB</note> and <note>Triangle AOD</note> are congruent triangles, <note>the corresponding sides and corresponding angles of Triangle AOB are equal to those of Triangle AOD</note>, namely: <note>side AB = side AD</note> <note>side AO = side AO</note> <note>side BO = side OD</note>, and the corresponding angles are also equal: <note>angle BAO = angle DAO</note> <note>angle AOB = angle AOD</note> <note>angle ABO = angle ADO</note>


Question: As shown in the figure, the cross-section of a small reservoir dam is a right trapezoid. The width of the dam top BC is 6m, the height of the dam is 14m, and the slope of CD is i=1:2. Then the length of the dam bottom AD is ()

Answer: 34m

1. Given BC=6 meters, the height of the dam is 14 meters. Draw an auxiliary line CE perpendicular to AD through point C, and ##intersect AD at point E##.

2. Since CE is perpendicular to AD, the quadrilateral AECB is a rectangle. Among them, AD is parallel and equal to BC, so AE=6 meters.

3. CE is the height of the dam, so CE=14 meters.

4. Given the slope of CD is i=1:2, which means CE:DE=1:2.

5. CE=14 meters, according to the proportional relationship CE:DE=1:2, we get DE=2*14=28 meters.

6. Since AD is composed of AE and DE, AD=AE+DE=6 meters+28 meters=34 meters.

7. Through the above reasoning, the final answer is 34 meters.

梯形; 直角三角形; 正切; 线段

Right Angle Trapezoid

A trapezoid is classified as a right angle trapezoid if and only if its two bases are parallel (∥), and one of its non-parallel sides (leg) is perpendicular (⊥) to the bases.

<note>In the figure of this problem, in the right angle trapezoid ##ADCB##</note>, <note>the base AE and the base BC are parallel</note>, <note>the leg AB is perpendicular to the base AE and the base BC</note>, <note>angles ∠BAE and ##∠ABC## are right angles</note>. Therefore, <note>the trapezoid ##ADCB## is a right angle trapezoid</note>.

Definition of Rectangle

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

<note>Quadrilateral ADCB is a rectangle</note>, with its interior angles <note>∠ABC, ∠AEC, ∠BCE, ∠BAE all being right angles (90 degrees)</note>, and <note>side BC is parallel and equal in length to side AE</note>, <note>side AB is parallel and equal in length to side EC</note>.

Slope

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

The slope of <note>slope CD</note> is <note>i=1:2</note>, indicating that <note>CE:DE=1:2</note>. <note>CE</note> represents the vertical height, and <note>DE</note> represents the horizontal distance. According to the definition of slope, the slope refers to the ratio of the vertical height <note>CE</note> to the horizontal distance <note>DE</note>, that is, slope = <note>CE/DE</note>.

Definition of Altitude

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

In the figure of this problem, <note>the line segment CE perpendicular to the opposite side AD from vertex C</note> is <note>the altitude from vertex C</note>. <note>The line segment CE forms a right angle (90 degrees) with side AD</note>, which indicates that <note>the line segment CE is the perpendicular distance from vertex C to the opposite side AD</note>.


Question: As shown in the figure, in rhombus ABCD, AB=15, ∠ADC=120°, find the distance between points B and D.

Answer: 15

1. Given that in rhombus ABCD, AB=15 and ∠ADC=120°.

2. Connect BD to form △ABD.

3. ##According to the definition of a rhombus##, AB=AD.

4. According to ##the property that adjacent angles of a rhombus are supplementary##, ∠A=180°-∠D=60°##.

5. ∵AB=AD=15, ∠A=60°, ##according to the theorem of equilateral triangles (the 60-degree angle criterion for isosceles triangles)##, △ABD is an equilateral triangle.

6. Since △ABD is an equilateral triangle, BD=AB=15.

7. Therefore, through the above reasoning, the distance between points B and D is 15.

Definition of Rhombus

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

<note>In quadrilateral ABCD, all sides AB, BC, CD, and DA are equal</note>, so quadrilateral ABCD is a rhombus. Additionally, <note>the diagonals AC and BD of quadrilateral ABCD are perpendicular bisectors of each other</note>, that is, <note>the diagonals AC and BD intersect at point O</note>, and <note>angle AOB is a right angle (90 degrees)</note>, and <note>AO=OC</note> and <note>BO=OD</note>.

Adjacent Angles of Rhombus are Supplementary

Any pair of adjacent angles in a rhombus are supplementary.

In <note>rhombus ABCD</note>, angle A and <note>angle D are a pair of adjacent angles</note>. According to the properties of a rhombus, any pair of adjacent angles of a rhombus are supplementary, that is, <note>angle A + angle D = 180 degrees</note>. Similarly, <note>angle B and angle C are also a pair of adjacent angles</note>, satisfying <note>angle B + angle C = 180 degrees</note>.

<note>△ABD is an isosceles triangle, with sides AB and AD equal</note>, and there is an interior angle of 60°, <note>∠A=60°</note>. According to the Equilateral Triangle Identification Theorem, if an isosceles triangle has an interior angle of 60°, then the lengths of its three sides are equal, and all three interior angles are 60°. Therefore, <note>△ABD is an equilateral triangle</note>.


Question: As shown in the figure, during an extracurricular math activity, Xiao Wen measured the angle of elevation of the top of the tree A from point C to be 37°. Given BC=20m, find the height of the tree AB. (Reference data: sin37°≈0.60, cos37°≈0.80, tan37°≈0.75)

Answer: 15m

1. It is known that Xiaowen measured the angle of elevation of the top of the tree A at point C to be 37°.

2. In △ABC, ∠ACB=37°.

3. According to the definition of the tangent function, in a right triangle, the tangent value of an acute angle is equal to the ratio of the length of the opposite side to the length of the adjacent side, so tan∠ACB=AB/BC.

4. It is known in the problem that BC=20 meters, sin37°≈0.60, cos37°≈0.80, tan37°≈0.75.

5. Substitute the values of BC and tan∠ACB into the definition formula of the tangent function to get AB/20=0.75.

6. Rearrange the above equation to get AB=20×0.75.

7. Calculate to get AB=15 meters.

8. Through the above reasoning, it is finally concluded that the height of the tree AB is 15 meters.

仰角; 直角三角形; 正切

Definition of Right Triangle

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

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

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.

<note>In the right triangle ABC</note>, <note>angle ∠ACB is an acute angle</note>, <note>side AB is the opposite side of angle ∠ACB</note>, <note>side BC is the adjacent side of angle ∠ACB</note>, so <note>the tangent value of angle ∠ACB is equal to the length of side AB divided by the length of side BC</note>, that is, <note>tan(∠ACB) = AB / BC</note>.


Question: As shown in the figure, a highway is being constructed in a certain area, and a tunnel needs to be built from point B to point C (B and C are on the same horizontal plane). In order to measure the distance between B and C, an engineer takes a hot air balloon from point C and ascends vertically 100m to point A. At point A, the angle of depression to point B is 30°. What is the distance between points B and C?

Answer: 100√{3}m

1. According to the problem, point A is 100 meters above point C on the ground, so the line segment AC connecting point A and point C is perpendicular to the ground BC, i.e., AC⊥BC.

2. Also, since the engineer observes the depression angle of point B from point A as 30°, ##let the line segment parallel to BC be AD, so ∠DAB=30°##.

##3. According to the definition of alternate interior angles, since AD∥BC and the transversal is AB, ∠DAB=∠ABC=30°##.

##4. According to the triangle angle sum theorem, in triangle ABC, AC⊥BC, so ∠ACB=90°, ∠CAB=180°-∠ACB-∠ABC=180°-90°-30°=60°##.

##5. According to the properties of a 30°-60°-90° triangle, side AC is opposite the 30° angle, and side BC is opposite the 60° angle, BC=√3AC=√3*100=100√3##.

####

##6. Through the above reasoning, the final answer is 100√3 m.##

直角三角形; 垂线; 正切

Angle of Depression

The angle of depression is the angle formed between the horizontal line and the line of sight when the observer looks downward from a certain position to an object below that level.

<note>In the figure of this problem, the observer stands at point A and looks down to see the object located at point B.</note> The horizontal line extends from point A to point D, forming <note>horizontal line AD</note>. The angle of depression is the angle formed when the observer looks from point A along the horizontal line towards point B, then <note>looks down towards point D</note>. This angle is represented by <note>∠DAB, which means ∠DAB is the angle of depression</note>. The problem <note>gives the angle of depression as 30°</note>.

Definition of Alternate Interior Angles

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

Triangle Angle Sum Theorem

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

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

Properties of 30°-60°-90° Triangle

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

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


Question: As shown in the figure, in △ABC, AB=AC=18, BC=12, the vertices E and F of the square DEFG are inside △ABC, the vertices D and G are on AB and AC respectively, AD=AG, DG=6, then the distance from point F to BC is ()

Answer: 6√{2}-6

1. Given AB = AC = 18, BC = 12, vertices E and F of square DEFG are inside △ABC, vertices D and G are on AB and AC respectively, AD = AG, DG = 6.

2. Through point A, draw AM perpendicular to BC at point M, intersecting DG at point N, extend GF to intersect BC at point H.

3. Since AB = AC, AD = AG, therefore AD:AB = AG:AC. Also, because ∠BAC = ∠DAG, ##according to the similarity criterion of triangles (SAS),## △ADG is similar to △ABC.

4. According to ##the definition of similar triangles##, we get ∠ADG = ∠ABC. Since DG is parallel to BC, ##quadrilateral DEFG is a square##, so FG is perpendicular to DG, and thus FH is perpendicular to BC, AN is perpendicular to DG.

5. Because AB = AC = 18, BC = 12, ##according to the theorem that the height, median, and angle bisector of an isosceles triangle coincide,## BM = 1/2 * BC = 6.

6. ##Because △ABM is a right triangle,## according to the Pythagorean theorem, AM = ##√(AB^2 - BM^2)## = 12√2.

7. ##Since AN is an altitude in △ADG, and AM is an altitude in △ABC##, and △ADG and △ABC are similar, according to the definition of similar triangles, AN = AM × (DG/BC).

8. Therefore, AN = 12√2 × (6/12) = 6√2.

9. Calculating, we get MN = AM - AN = 12√2 - 6√2 = 6√2.

10. Because FH = MN - GF, and the side length of square DEFG is 6, FH = 6√2 - 6.

11. Through the above reasoning, we finally conclude that the distance from point F to BC is 6√2 - 6.

等腰三角形; 正方形; 平行线; 垂线

Definition of Similar Triangles

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

<note>Triangle ADG and triangle ABC are similar triangles</note>. According to the definition of similar triangles: <note>∠DAG = ∠BAC</note>, <note>∠ADG = ∠ABC</note>, <note>∠AGD = ∠ACB</note>; <note>AD/AB = AG/AC = DG/BC</note>##=AN/AM##.

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

Definition of Perpendicular Lines

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

<note>Line AM and line BC intersect to form ∠AMB which is 90 degrees</note>, therefore according to the definition of perpendicular lines, <note>line AM and line BC are perpendicular to each other</note>; since DG is parallel to BC, the side FG of square DEFG is <note>perpendicular to DG</note>, and thus <note>FH is perpendicular to BC</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 <note>triangle ADG and triangle ABC</note>, <note>side AD corresponds to side AB</note>, <note>side AG corresponds to side AC</note>, and <note>side AD/side AB = side AG/side AC</note>, and <note>angle BAC = angle DAG</note>. Therefore, according to the <note>SAS Criterion for Similar Triangles</note>, triangle ADG is similar to triangle ABC.

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 ABM</note>, <note>∠AMB is a right angle (90 degrees)</note>, <note>sides BM and AM are the legs</note>, <note>side AB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AB^2 = AM^2 + BM^2</note>.

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

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

In the isosceles triangle <note>ABC</note>, the <note>vertex angle is angle BAC</note>, and the <note>base is side BC</note>. The angle bisector of the vertex angle <note>AM</note> not only bisects the <note>vertex angle BAC</note> but also bisects the <note>base BC</note>, making <note>BM = MC</note>, and is <note>perpendicular to the base BC</note>, forming a <note>right angle AMB (90 degrees)</note>. Therefore, the <note>line segment AM is both the angle bisector of the vertex angle, the median of the base, and the altitude</note>.

Definition of Isosceles Triangle

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

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

Definition of Right Triangle

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

Definition of Altitude

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


Question: As shown in the figure, in rhombus ABCD, the lengths of the two diagonals are AC=12 and BD=16. Find the side length of the rhombus.

Answer: 10

1. Given that quadrilateral ABCD is a rhombus, according to the property of the diagonals of a rhombus, the two diagonals AC and BD of quadrilateral ABCD are perpendicular to each other and bisect each other.

2. Let point O be the intersection of diagonals AC and BD. According to the definition of a rhombus, we have OA = 1/2AC = 1/2 × 12 = 6, OB = 1/2BD = 1/2 × 16 = 8.

3. Since AC and BD are perpendicular to each other, according to the definition of a right triangle, angle AOB is 90 degrees, and triangle △AOB is a right triangle. Based on the Pythagorean theorem, in right triangle △AOB, the side length AB = √(OA² + OB²) = √(6² + 8²) = √(36 + 64) = √100 = 10.

4. Therefore, the side length of rhombus ABCD is 10.

菱形; 直角三角形; 垂直平分线

Properties of the Diagonals of a Rhombus

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

In the rhombus ABCD, <note>the diagonals AC and BD bisect each other and are perpendicular to each other</note>. Specifically, <note>point O is the intersection point of diagonals AC and BD</note>, and <note>OA=OC and OB=OD</note>. At the same time, <note>angle AOB and angle AOD are both right angles (90 degrees)</note>, so <note>the diagonals AC and BD are perpendicular to each other</note>.

Definition of Rhombus

<note>In the diagram of this problem, all sides AB, BC, CD, and DA of quadrilateral ABCD are equal</note>, therefore <note>quadrilateral ABCD is a rhombus</note>. Additionally, <note>the diagonals AC and BD of quadrilateral ABCD are perpendicular bisectors of each other</note>, that is, <note>the diagonals AC and BD intersect at point O</note>, <note>and angle AOB is a right angle (90 degrees)</note>, and <note>AO=OC</note> and <note>BO=OD</note>.

Pythagorean Theorem

<note>△AOB is a right triangle</note>, where <note>∠AOB is a right angle</note>, OA = 6, OB = 8. According to the Pythagorean Theorem, the side length AB = <note>√(OA² + OB²) = √(6² + 8²)</note> = √(36 + 64) = √100 = 10.

Definition of Right Triangle

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


Question: As shown in the figure, AB is the diameter of ⊙O, CD is a chord, AB ⊥ CD, the foot of the perpendicular is point E, connect OD, CB, AC, ∠DOB=60°, EB=2, then the length of CD is ()

Answer: 4√{3}

1. Given ∠DOB=60°, according to the inscribed angle theorem, we get ∠BCE=30°.

2. In the right triangle ΔBCE, given BE=2 and ∠BCE=30°.

3. According to the properties of a 30°-60°-90° triangle, BC = 2×BE=4.

4. According to the Pythagorean theorem, in the right triangle ΔBCE, CE=√(BC^2-BE^2)=√(4^2-2^2)=2√3.

5. Since AB is the diameter of ⊙O and AB ⊥ CD, according to the perpendicular diameter theorem, we get CD=2×CE=4√3.

6. Based on the above reasoning, the final answer is that the length of CD is 4√3.

圆; 垂线; 直角三角形; 弦; 圆心角

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

Definition of Right Triangle

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

Properties of 30°-60°-90° Triangle

In the <note>30°-60°-90° triangle BCE</note>, <note>angle BCE is 30 degrees</note>, <note>angle BEC is 90 degrees</note>, <note>angle CBE is 60 degrees</note>. <note>Side BC is the hypotenuse</note>, <note>side BE is opposite the 30-degree angle</note>, <note>side CE is opposite the 60-degree angle</note>. According to the properties of a 30°-60°-90° triangle, <note>side BC is twice the length of side BE</note>, <note>side CE is √3 times the length of side BE</note>. That is: <note>BC = 2 * BE</note>, <note>CE = BE * √3</note>.

Inscribed Angle Theorem

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

<note>In circle O, points B, C, D are on the circle</note>, the central angle corresponding to arc BD is <note>∠DOB</note> = 60°, and the inscribed angle is <note>∠BCD</note>. According to the Inscribed Angle Theorem, <note>∠BCD</note> is equal to half of the central angle <note>∠DOB</note> corresponding to arc BD, that is, <note>∠BCD</note> = 1/2 <note>∠DOB</note>.

Pythagorean Theorem

<note>In the right triangle △BCE</note>, <note>∠BEC is a right angle (90 degrees)</note>, <note>sides CE and BE are the legs</note>, <note>side CB is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>CE^2 = BC^2 - BE^2</note>.

Perpendicular Diameter Theorem

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

In <note>circle O</note>, the diameter <note>AB</note> is perpendicular to the chord <note>CD</note>. According to the Perpendicular Diameter Theorem, the diameter <note>AB</note> bisects the chord <note>CD</note>, that is, <note>CE = ED</note>, and the diameter <note>AB</note> also bisects the arcs subtended by the chord <note>CD</note>, that is, <note>arc BC</note> = <note>arc BD</note>.


Question: As shown in the figure, in parallelogram ABCD, if ∠B = 60°, then ∠D equals ()

Answer: 60°

1. Given that quadrilateral ABCD is a parallelogram, according to the properties of parallelograms, the opposite angles of a parallelogram are equal.

2. Therefore, ∠ABC = ∠CDA.

3. According to the given condition, ∠ABC = 60°.

4. Therefore, ∠CDA = 60°.

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

Definition of Parallelogram

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

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

Properties of Parallelogram Theorem

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

In the <note>parallelogram ABCD</note>, <note>the opposite angles ∠ABC and ∠CDA are equal</note>, <note>the opposite angles ∠BAD and ∠BCD are equal</note>.


Question: As shown in the figure, the angle of elevation from the hot air balloon at point A to the top of the building is 30°, and the angle of depression to the bottom of the building is 60°. The horizontal distance between the hot air balloon and the building is 120m. The height of the building is ()

Answer: 160√{3}m

1. Given ∠BAD=30°, ∠DAC=60°, and AD=120 meters.

2. According to the definition of the tangent function, in the right triangle ABD, tan(∠BAD)=BD/AD, therefore tan(30°)=BD/120.

3. Thus, we get BD=120 * tan(30°)=120 * (√3/3)=40√3.

4. Similarly, in the right triangle ADC, tan(∠DAC)=CD/AD, therefore tan(60°)=CD/120.

5. Thus, we get CD=120 * tan(60°)=120 * √3=120√3.

6. The height of the building BC equals BD plus CD, i.e., BC=BD+CD=40√3+120√3=160√3.

7. Through the above reasoning, the final answer is 160√3 meters.

仰角; 直角三角形; 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.

<note>In the figure of this problem, in the right triangle ABD, angle ∠BAD is an acute angle</note>, <note>side BD is the opposite leg of angle ∠BAD</note>, <note>side AD is the adjacent leg of angle ∠BAD</note>, so the tangent value of angle ∠BAD is equal to the length of side BD divided by the length of side AD, that is <note>tan(∠BAD) = BD / AD</note>. Similarly, <note>in the right triangle ADC, angle ∠DAC is an acute angle</note>, <note>side CD is the opposite leg of angle ∠DAC</note>, <note>side AD is the adjacent leg of angle ∠DAC</note>, so the tangent value of angle ∠DAC is equal to the length of side CD divided by the length of side AD, that is <note>tan(∠DAC) = CD / AD</note>.

Angle of Depression

The angle of depression is the angle formed between the horizontal line and the line of sight when the observer looks downward from a certain position to an object below that level.

The observer is at <note>point A</note>, looking down to see the bottom of the building at <note>point C</note>. <note>A horizontal line extends from point A to point D, forming the horizontal line AD</note>. The angle of depression is the angle formed when the observer looks from <note>point A along the horizontal line to point C</note>, and then looks up to <note>point D</note>. This angle is represented by <note>∠DAC</note>, i.e., <note>∠DAC is the angle of depression = 60°</note>.

Angle of Elevation

The angle formed between the horizontal line and the observer's line of sight when the observer looks upward towards an object is referred to as the angle of elevation.

<note>The observer is at point A</note>, looking forward from point A along the horizontal line forms <note>line segment AD</note>. When the observer looks up from point A at an object located at <note>point B</note>, the line of sight forms <note>line segment AB</note>. At this time, the angle formed between the line of sight <note>AB</note> and the horizontal line <note>AD</note>, <note>angle ∠BAD</note>, is the angle of elevation.

Definition of Right Triangle

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

<note>In the figure of this problem, in triangle ABD, angle ADB is a right angle (90 degrees)</note>, therefore triangle ABD is a right triangle. <note>Sides AD and BD are the legs, and side AB is the hypotenuse</note>. Similarly, <note>in triangle ADC, angle ADC is a right angle (90 degrees)</note>, therefore triangle ADC is a right triangle. <note>Sides AD and CD are the legs, and side AC is the hypotenuse</note>.


Question: As shown in the figure, in trapezoid ABCD, AD∥BC, diagonals AC and BD intersect at point O. If S~△AOD~:S~△OCD~=1:2, then S~△AOD~:S~△BOC~=()

Answer: \frac{1}{4}

1. Let the distance from point D to side AC be h. According to the triangle area formula, the area ratio of triangle AOD to triangle OCD is (1/2 × OA × h) : (1/2 × OC × h) = OA : OC.

2. Given S△AOD : S△OCD = 1:2, it can be concluded that OA : OC = 1:2.

3. Since AD ∥ BC, ##by the parallel axiom 2 of parallel lines##, ∠DAO = ∠BCO, ∠ADO = ∠CBO.

4. ##According to the similarity triangle determination theorem (AA)##, △AOD ∽ △COB.

5. Since the area ratio of similar triangles is equal to the square of the similarity ratio, S△AOD : S△BOC = ##(OA : OC)^2 = (1:2)^2## = 1:4.

6. Through the above reasoning, the final answer is 1:4.

梯形; 平行线; 普通三角形

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

In this problem diagram, <note>triangle AOD</note> and <note>triangle COB</note> are similar because <note>∠DAO = ∠BCO</note> and <note>∠ADO = ∠CBO</note>, therefore <note>△AOD ∽ △COB</note>. According to the theorem that the area ratio of similar triangles equals the square of the similarity ratio, if the side length ratio of two similar triangles is k, then their area ratio equals k squared. Thus, <note>S△AOD : S△BOC = (OA : OC)^2</note>.

Area Formula of a Triangle

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

In this problem, <note>in triangle AOD, side OA is the base, and the perpendicular distance of line segment OD, h, is the height</note>. According to the area formula of a triangle, <note>the area of triangle AOD is equal to the base OA multiplied by the height h and then divided by 2, i.e., area = (OA * h) / 2</note>. Similarly, <note>in triangle OCD, side OC is the base, and the perpendicular distance of line segment OD, h, is the height</note>. According to the area formula of a triangle, <note>the area of triangle OCD is equal to the base OC multiplied by the height h and then divided by 2, i.e., area = (OC * h) / 2</note>. Therefore, <note>the area ratio of triangles AOD and OCD is OA : OC</note>.

Parallel Postulate 2 of Parallel Lines

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

<note>In the diagram of this problem, two parallel lines AD and BC are intersected by a third line AC</note>, forming the following geometric relationships:
1. <note>Alternate interior angles</note>: <note>Angle DAO and angle BCO are equal</note>.
2. Alternate interior angles: <note>Angle ADO and angle CBO are equal</note>.

Definition of Similar Triangles

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

<note>In the figure of this problem, triangle AOD and triangle COB are similar triangles</note>. According to the definition of similar triangles: <note>angle AOD = angle BOC</note>, <note>angle DAO = angle BCO</note>, <note>angle ADO = angle CBO</note>; <note>AO/CO = DO/BO = AD/BC</note>.

Similarity Theorem for Triangles (AA)

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

In the original text: <note>△AOD and △COB</note>, <note>∠DAO = ∠BCO</note>, <note>∠ADO = ∠CBO</note>, so <note>△AOD ∽ △COB</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.

Original: <note>在本题图中，两条平行直线AD和BC被直线AC和BD截交</note>，其中<note>∠DAO 和 ∠BCO位于两平行线之间，∠ADO 和 ∠CBO位于两平行线之间</note>，且分别<note>在AC和BD的对侧</note>，因此<note>∠DAO 和 ∠BCO是内错角，∠ADO 和 ∠CBO是内错角</note>。内错角相等，即<note>∠DAO = ∠BCO，∠ADO = ∠CBO</note>。

Translation: <note>In the diagram of this problem, two parallel lines AD and BC are intersected by lines AC and BD</note>, where <note>∠DAO and ∠BCO are between the two parallel lines, and ∠ADO and ∠CBO are between the two parallel lines</note>, and respectively <note>on opposite sides of AC and BD</note>, therefore <note>∠DAO and ∠BCO are alternate interior angles, and ∠ADO and ∠CBO are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>∠DAO = ∠BCO, ∠ADO = ∠CBO</note>.

Definition of Altitude

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

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


Question: As shown in the figure, in △ABC, BF bisects ∠ABC, AF is perpendicular to BF at point A with F as the foot, and extended to intersect BC at point G. D is the midpoint of AB, and DF is connected and extended to intersect AC at point E. If AB=12 and BC=20, then the length of segment EF is ()

Answer: 4

1. In △BFA and △BFG, according to the definition of angle bisector, we have: ∠ABF = ∠GBF (because BF bisects ∠ABC), BF = BF (common side), ∠BFA = ∠BFG (both are 90 degrees), therefore ΔBFA ≌ ΔBFG (angle-side-angle criterion).

2. Since △BFA and △BFG are congruent, we get BG = AB = 12, AF = FG.

3. Because BC = 20, it follows that GC = BC - BG = 20 - 12 = 8.

4. According to the problem statement, D is the midpoint of AB, F is the midpoint of AG, according to the midline theorem of triangles, DF is parallel to BG, since E is on the extension line of DF, G is a point on BC, therefore EF is parallel to CG.

5. According to the theorem of parallel lines dividing segments proportionally, AE/CE = AF/FG, substituting the values we get AE = CE, that is, E is the midpoint of AC, so EF is the midline of triangle ACG.

6. Therefore, according to the midline theorem of triangles, EF is equal to half of GC, that is, EF = 1/2 * GC = 4.

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

垂线; 中点; 普通三角形; 直角三角形; 线段

Definition of Angle Bisector

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

The vertex of <note>angle ABG</note> is point B. A line BF is drawn from point B, which divides <note>angle ABG</note> into two equal angles, namely angle ABF and <note>angle GBF</note>. Therefore, line BF is the angle bisector of <note>angle ABG</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 lines <note>AF and BF</note> intersect to form an angle <note>∠AFB of 90 degrees</note>, so according to the definition of perpendicular lines, <note>lines AF and BF are perpendicular to each other</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 AB is point D</note>. According to the definition of the midpoint of a line segment, <note>point D divides line segment AB into two equal parts</note>, that is, <note>the lengths of line segment AD and line segment DB are equal</note>. That is, <note>AD = DB</note>.

Definition of Congruent Triangles

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

<note>Triangles BFA and BFG are congruent triangles</note>, the corresponding sides and angles of triangle BFA are equal to those of triangle BFG, namely: <note>side BF = side BF</note>, <note>side AB = side BG</note>, <note>side AF = side GF</note>, and the corresponding angles are also equal: <note>angle ABF = angle GBF</note>, <note>angle AFB = angle GFB</note>, <note>angle BAF = angle BGF</note>.

Angle-Side-Angle (ASA) Criterion for Congruence of Triangles

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

In the figure of this problem, in triangles BFA and BFG, <note>angle ABF is equal to angle GBF (property of angle bisectors)</note>, <note>angle AFB is equal to angle GFB (both are 90°)</note>, and <note>side BF is equal to side BF (common side)</note>. Since the two angles and the included side of these two triangles are respectively equal, according to the Angle-Side-Angle (ASA) criterion for congruence of triangles, it can be concluded that <note>triangle BFA is congruent to triangle BFG</note>.

Triangle Midline Theorem

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

<note>Point F</note> is the midpoint of <note>side AC</note>, and point E is the midpoint of side AC. <note>Line segment EF</note> connects these two midpoints. According to the Triangle Midline Theorem, <note>line segment EF</note> is parallel to <note>the third side GC</note> and is equal to half of <note>the third side GC</note>, i.e., <note>EF || GC</note>, and <note>EF = 1/2 * GC</note>. Similarly, in triangle ABG, point F is the midpoint of side AG, and point D is the midpoint of side AB. Line segment DF connects these two midpoints. According to the Triangle Midline Theorem, line segment DF is parallel to the third side BG and is equal to half of the third side BG, i.e., <note>DF || BG</note>, and <note>DF = 1/2 * BG</note>.

Proportional Segments Theorem

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

In <note>triangle ACG</note>, the line <note>EF is parallel to side CG</note>, and it intersects the other two sides <note>AG and AC at points E and F</note>. According to the Proportional Segments Theorem, we have: <note>AF/AG = AE/AC</note>, which means the intercepted segments are proportional to the corresponding segments of the original triangle.


Question: Point B is on ⊙O, point C is a point on ⊙O different from A and B. If ∠AOB=50°, then the degree of ∠ACB is ()

Answer: 25°

####

##1.## Given ∠AOB = 50°.

##2.## According to the inscribed angle theorem, the inscribed angle subtended by the same arc is equal to half of the central angle subtended by that arc. The inscribed angle theorem states that if the vertex of an angle is on the circle and both sides intersect the circle, then the angle is an inscribed angle.

##3.## In this problem, ∠ACB is the inscribed angle subtended by arc AB, corresponding to the central angle ∠AOB on arc AB.

##4.## Therefore, according to the inscribed angle theorem, ∠ACB = 1/2 * ∠AOB.

##5.## Substituting the given condition, ∠ACB = 1/2 * 50° = 25°.

##6.## Thus, through the above reasoning, the final answer is 25°.

圆; 圆周角; 圆心角; 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.

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

Definition of Central Angle

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

In the figure of this problem, <note>in circle O</note>, point A and point B are two points on the circle, <note>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 a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

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


Question: As shown in the figure, in parallelogram ABCD, AB=4, BC=6, the perpendicular bisector of AC intersects AD at point E. Then the perimeter of △CDE is ()

Answer: 10

1. Given that the perpendicular bisector of AC intersects AD at point E, according to the properties of the perpendicular bisector, we get AE = EC.

2. Since quadrilateral ABCD is a parallelogram, according to the properties of a parallelogram, side AB is parallel and equal to side DC, and side BC is parallel and equal to side AD, therefore AB = DC = 4, BC = AD = 6.

3. Because AE = EC, and E is on AD, we have DE = DA - AE = 6 - AE = 6 - EC.

4. Therefore, ##according to the perimeter theorem of a triangle,## the perimeter of triangle CDE is EC + CD + DE.

5. Based on the previous conclusions, EC + CD + DE = EC + 4 + (6 - EC) = 4 + 6 = 10.

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

平行四边形; 垂直平分线; 等腰三角形; 普通三角形

Properties of the Perpendicular Bisector

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

<note>In the figure of this problem, the perpendicular bisector of segment AC intersects AD at point E</note>. According to the properties of the perpendicular bisector, <note>the distance from point E to the endpoints A and C of segment AC is equal, i.e., EA = EC</note>.

Properties of Parallelogram Theorem

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

In the diagram of this problem, in parallelogram ABCD, <note>∠BAD</note> and <note>∠BCD</note> are equal, opposite angles <note>∠B</note> and <note>∠D</note> are equal; sides <note>AB</note> and <note>DC</note> are equal, sides <note>BC</note> and <note>AD</note> are equal. Therefore, AB = DC = 4 and BC = AD = 6.

Perimeter Theorem of Triangle

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

The three sides of <note>triangle △CDE</note> are <note>EC, CD, DE</note> respectively. According to the Perimeter Theorem of Triangle, the perimeter of the triangle is equal to the sum of the lengths of its three sides, i.e., <note>perimeter P = EC + CD + DE</note>.


Question: Definition: The minimum distance between a fixed point A and any point on ⊙O is called the distance between point A and ⊙O. There is a rectangle ABCD (as shown in the figure), AB = 14 cm, BC = 12 cm, ⊙K is tangent to the sides AB, BC, CD of the rectangle at points E, F, G respectively. Then the distance between point A and ⊙K is ()

Answer: 4cm

1. Given the side lengths of rectangle ABCD are AB=14cm, BC=12cm, ⊙K is tangent to the sides AB, BC, CD at points E, F, G respectively.

2. Connect the segments KE, KF, KG, AK, and intersect ⊙K at point H.

3. Since ⊙K is tangent to the sides AB, BC, CD of the rectangle, points E, F, G are the tangent points on these sides, and we have EK=FK=KG.

4. ##Since points E and F are tangent points, ∠BFK=∠BEK=90°, and since ∠B=90°, according to the rectangle determination theorem, quadrilateral BEKF is a rectangle, so the side lengths are BE=FK=radius=6cm##.

5. Since the side length of rectangle ABCD is AB=14cm, then AE=AB-BE=14cm-6cm=8cm. Therefore, AK=√(AE^2+EK^2)=√(8cm^2+6cm^2)=√(64cm##^2##+36cm##^2##)=√100 cm##^2##=10cm.

6. Point H is one of the intersection points of segment AK and circle ⊙K, point ##H ##is the ##minimum distance## from point A to circle ⊙K, ##so the distance between point A and circle K## is the length of AK ##minus the radius part of EK##, therefore, AH = AK - KH = 10cm - 6cm = 4cm.

7. The distance between point A and ⊙K is 4cm.

矩形; 圆; 切线; 垂线; 对称

Definition of Tangent to a Circle

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

<note>Circle ⊙K and line AB have only one common point E</note>, this common point is called the point of tangency; <note>Circle ⊙K and line BC have only one common point F</note>, this common point is called the point of tangency; <note>Circle ⊙K and line CD have only one common point G</note>, this common point is called the point of tangency. Therefore, <note>lines AB, BC, CD are tangents to circle ⊙K</note>.

Definition of Rectangle

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

<note>Angle B</note>, <note>Angle BEK</note>, and <note>Angle BFK are all right angles (90 degrees)</note>. According to the definition of a rectangle, if a quadrilateral has <note>three right angles</note>, then the quadrilateral is a <note>rectangle</note>. Therefore, <note>quadrilateral BEFK is a rectangle</note>.

Tangent-Segment Theorem

If two tangent segments are drawn to a circle from an external point, then the lengths of these tangent segments are equal. Moreover, the line segment connecting the center of the circle to the external point bisects the angle formed by the two tangent segments.

Original text: <note>Points E, F, and G are the points of tangency of circle ⊙K</note>, therefore <note>KE=KF=KG=r (the radius of the circle)</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 AEK, ∠AEK is a right angle (90 degrees)</note>, <note>sides AE and EK are the legs</note>, <note>side AK is the hypotenuse</note>, so according to the Pythagorean Theorem, <note>AK##^2## = AE##^2## + EK##^2##</note>.