1 There is only one straight line between two points. The shortest line segment between two points is 3. The same angle or the complementary angle of the same angle is equal. 4. The same angle or the complementary angle of the same angle is equal. 5. Only one straight line is perpendicular to the known straight line. 6. Among all the line segments connected with points on a straight line, the shortest parallel axiom of a vertical line segment passes through a point outside the straight line. There is only one straight line parallel to this straight line. If both lines are parallel to the third line, the two lines are parallel to each other. The isosceles angles are equal and the two straight lines are parallel to each other. 10, the offset angles are equal, and the two straight lines are parallel to each other. 1 1 is complementary to the inner corner of the side, and the two straight lines are parallel to each other. 13, two straight lines are parallel. The internal dislocation angle is equal to 14, and the two straight lines are parallel. Theorem The sum of two sides of a triangle is greater than the third side 15. The difference between two sides of the reasoning triangle is less than the third side 17. The sum of the interior angles of a triangle is the theorem that the sum of the three interior angles of a triangle is equal to 180 18. The two acute angles of a right triangle complement each other 19. The outer corner of a triangle. It is deduced from the sum of two non-adjacent internal angles that one external angle of a triangle is larger than the corresponding side of any non-adjacent internal angle, that is, 2 1 congruent triangles, and the corresponding angle is equal to 22. The edge axiom (SAS) has two triangles with equal angles. The edge axiom (ASA) has two triangles with equal angles. The edge axiom (AAS) has two triangles with equal angles. The congruent 25 side axiom (SSS) of two triangles corresponding to the opposite sides of two angles and one angle and the congruent 26 hypotenuse and right angle axiom (HL) of two triangles corresponding to three sides. Two right-angled triangles with a hypotenuse and a right-angled side are congruent. Theorem 1 A point on the bisector of an angle is equal to the distance between two sides of the angle. Theorem 2 To a point with equal distance on both sides of an angle, on the bisector of this angle, the bisector of 29 angles is the set of all points with equal distance on both sides of this angle. The nature theorem of isosceles triangle 30 The two base angles of an isosceles triangle are equal (that is, equilateral and equilateral). 3 1 Inference 1 The bisector of the top angle of the isosceles triangle bisects the bottom and is perpendicular to the bisector of the top angle of the isosceles triangle with the bottom 32. The midline on the bottom edge coincides with the height on the bottom edge. Inference 3 All angles of an equilateral triangle are equal, and each angle is equal to 60 34 isosceles triangle. If a triangle has two equal angles, then the opposite sides of the two angles are also equal (equilateral) 35 Inference 1 A triangle with three equal angles is an equilateral triangle 36 Inference 2 An isosceles triangle with an angle equal to 60 is an equilateral triangle 37 in a right triangle. If an acute angle is equal to 30, then the right-angled side it faces is equal to half of the hypotenuse. The median line of the hypotenuse of a right triangle is equal to half of the hypotenuse. Theorem 39 A point on the vertical line of a line segment is equal to the distance between the two endpoints of this line segment. The inverse theorem and the point where the two endpoints of a line segment are equal. On the midline of this line segment, the midline of line segment 4 1 can be regarded as a set of all points with equal distance from both ends of the line segment. Theorem 42: Two graphs that are symmetrical about a straight line are congruent. Theorem 43: Two figures are symmetrical about a straight line, then the symmetry axis is the median vertical line 44 Theorem 3: Two figures are symmetrical about a straight line. If their corresponding line segments or extension lines intersect, then the intersection point is on the axis of symmetry. 45 Inverse Theorem If the straight line connecting the corresponding points of two graphs is bisected vertically by the same straight line, then the two graphs are symmetrical about this straight line. 46 Pythagorean Theorem The sum of squares of two right-angled sides A and B of a right-angled triangle is equal to the square of the hypotenuse C, that is, the inverse theorem of A 2+B 2 = C 2. 47 Pythagorean Theorem If three sides of a triangle have a relationship A 2+B 2 = C 2, then this triangle is a right-angled triangle. Theorem 48 The sum of internal angles of quadrilateral is equal to 360 49, and the sum of internal angles of polygon is equal to 360 50. Theorem The sum of internal angles of n- polygon is equal to (n-2) × 180 5 1. It is inferred that the sum of the external angles of any polygon is equal to 360 52. The property theorem of parallelogram 1 The diagonal of parallelogram is equal to 53. Theorem 2: The opposite sides of a parallelogram are equal. 54. Inference that parallel lines between two parallel lines are equal. 55. parallelogram property theorem 3. The parallelogram bisects 56 diagonally. Parallelogram judgment theorem 1. Two groups of parallelograms with equal diagonals are parallelograms. 57. parallelogram judgment theorem 2. Two groups of parallelograms with equal opposite sides are parallelograms. 58. parallelogram judgment theorem 3. Diagonal lines are mutual. The bisected quadrilateral is a parallelogram 59. Parallelogram Decision Theorem 4. A set of parallelograms with equal opposite sides is parallelogram 60. Rectangular property theorem 1. All four corners of a rectangle are right angles 6 1. Theorem of rectangle properties II. The diagonal lines of the rectangle are equal to 62. Rectangular decision theorem 1. A quadrilateral with three right angles is a rectangle 63. Rectangular decision theorem ii. A parallelogram with equal diagonals is a moment. Form 64 diamond property theorem 1 all four sides of the diamond are equal. 65 Diamond Property Theorem 2 Diagonal lines of diamonds are perpendicular to each other, and each diagonal line bisects a set of diagonal lines. 66 diamond area = half of diagonal product. That is, S=(a×b)÷2 67 rhombus decision theorem 1 A quadrilateral with four equal sides is a rhombus 68 rhombus decision theorem 2 A parallelogram with diagonal lines perpendicular to each other is a rhombus 69 square property theorem 1 All four corners of a square are right angles, and all four sides are equal to 70 square property theorem 2 Two diagonal lines of a square are equal and divided vertically. Each diagonal bisects a set of diagonals 7 1 theorem 1 congruence of two figures symmetrical about the center 72 Theorem 2 For two figures symmetrical about the center, the straight line of the symmetrical point passes through the symmetrical center and is bisected by the symmetrical center 73 Inverse Theorem If the straight line of the corresponding point of two figures passes through a point and is bisected by the point, the two figures are symmetrical about the point. Property theorem of isosceles trapezoid. The two angles of an isosceles trapezoid on the same base are equal. The two diagonals of an isosceles trapezoid are equal. 76 isosceles trapeziums have equal angles on the same base, which is an isosceles trapezoid. The diagonal trapezoid is an isosceles trapezoid. Theorem of bisecting line segments by 78 parallel lines. If a set of parallel lines cut on a straight line are equal, then the line segments cut on other straight lines are also equal. 79 Inference 1 Through a straight line parallel to the bottom of the trapezoid, the other waist 80 must be equally divided. Inference 2 Inference 2 Through a straight line parallel to the other side of the triangle, the third side must be bisected. 8 1 The midline theorem of the triangle is parallel to the third side. And equal to half of it. The trapezium midline theorem is parallel to the two bottoms and is equal to half the sum of the two bottoms. Basic properties of L=(a+b)÷2 S=L×h 83 (1) If a:b=c:d, then ad=bc If ad=bc, then A: B = C: D. Then (A B)/B = (C D)/D. Then the corresponding line segment obtained by (A+C+…+M)/(B+D) infers that the straight line parallel to one side of the triangle cuts the other two sides (or the extension lines on both sides), and the corresponding line segment obtained is proportional to Theorem 88. If the corresponding line segments obtained by cutting two sides (or extension lines of two sides) of a triangle with a straight line are proportional, then this straight line is parallel to the third side 89 of the triangle, parallel to one side of the triangle and intersects with the other two sides. The three sides of the cutting triangle correspond to the three sides of the original triangle in proportion. Theorem 90 A straight line parallel to one side of a triangle intersects the other two sides (or extension lines of both sides), and the triangle formed is similar to the original triangle. Theorem 1 similar triangle judgment theorem 1 two angles are equal. Similarity between two triangles (ASA) 92 A right triangle divided by the height on the hypotenuse is divided into two right triangles. Similarity with the original triangle 93 Judgment Theorem 2. Two sides are proportional and the included angles are equal. Similarity between two triangles (SAS) 94 Judgment Theorem 3. Three sides are proportional. Two triangles are similar (SSS) Theorem 95 If the hypotenuse and a right-angled side of a right-angled triangle are proportional to the hypotenuse and a right-angled side of another right-angled triangle, then the two right-angled triangles are similar. Theorem 1 similar triangles corresponding height ratio. The ratio of the corresponding median line to the bisector of the corresponding angle is equal to the similarity ratio 97 Property Theorem 2 The ratio of similar triangles perimeter is equal to the similarity ratio 98 Property Theorem 3 The ratio of similar triangles area is equal to the square of the similarity ratio 99. The sine value of any acute angle is equal to the cosine value of the remaining angles, and the cosine value of any acute angle is equal to the sine value of the remaining angles 100. The tangent of any acute angle is equal to the cotangent of the other angles. The cotangent value of any acute angle is equal to the tangent value of other angles 10 1. A circle is a set of points whose distance from a fixed point is equal to the fixed length 102. The interior of a circle can be regarded as a set of points whose distance from the center of the circle is less than the radius 103. The outer circle of a circle can be regarded as a group of points whose distance from the center of the circle is greater than the radius 104. The radius of the same circle or the same circle is equal to 13. The locus of a point with a distance equal to a fixed length is the locus of a fixed length circle with a radius of 106 and two endpoints of a known line segment with the same distance, the locus of a point with the same distance from the middle vertical line of a line segment 107 to both sides of a known angle, and the locus of a bisector of this angle 108 to a point with the same distance between two parallel lines. 1 10 Vertical Diameter Theorem bisects the chord perpendicular to the chord diameter and bisects the two arcs opposite to the chord11inference 1 ① bisects the diameter (not the diameter) of the chord perpendicular to the chord, and the midpoints of the two arcs opposite to the chord pass through the center of the circle, and the two arcs opposite to the chord. And the other arc bisecting the chord is 1 12. It is inferred that the arcs sandwiched by two parallel chords of circle 2 are equal. The circle 1 13 is a centrosymmetric figure with the center of the circle as the symmetry center. Theorem 1 14 In the same circle or in the same circle, equal central angles subtend equal arcs and equal chords. The distance between chords of a pair of chords is equal. 1 15 It is inferred that in the same circle or the same circle, if the distances between two central angles, two arcs, two chords or two chords are equal, the corresponding other components are equal. 1 16 Theorem: The circumferential angle of an arc is equal to half of its central angle. In the same circle or equal circle, the arc opposite to the equal circle angle is also equal. 1 18 infers that 2 semicircles (or diameters) are right angles; The chord subtended by the circumferential angle of 90 is 1 19 Inference 3 If the median line of one side of a triangle is equal to half of this side, then this triangle is the diagonal complement of the inscribed quadrilateral of the right triangle 120 theorem circle. And any outer angle is equal to the intersection point of the inner diagonal line 12 1① and ⊙O D < R2, and the tangent judgment theorem of ⊙O D = R3 and ⊙O D > R 122 passes through the outer end of the radius, and the straight line perpendicular to this radius is the tangent of the circle. 438+024 Inference 1 A straight line passing through the center and perpendicular to the tangent must pass through the tangent point 125 Inference 2 A straight line passing through the tangent point and perpendicular to the tangent must pass through the center 126 The tangent length theorem leads to two tangents from a point outside the circle, and their tangent lengths are equal. The line between the center of the circle and this point bisects the included angle of the two tangents. The sum of two opposite sides of the circumscribed quadrangle of a circle is equal. The tangent angle theorem is equal to the circumferential angle of the arc pair it clamps. It is deduced that if the arcs sandwiched by two chord tangent angles are equal, then the two chord tangent angles are equal to the two intersecting chords in the chord theorem circle. The product of the length of two lines divided by the intersection is equal to 13 1. It is deduced that if the chord intersects the diameter vertically, then half of the chord is the tangent and secant of the circle, which is drawn by the middle term 132 according to the ratio of two line segments formed by a point outside the circle. The tangent length is the ratio of the lengths of two lines from this point to the intersection of the secant and the circle. 133 This item infers that two secant lines are drawn from a point outside the circle, and the product of the lengths of the two lines from this point to the intersection of each secant line and the circle is equal to 134. If two circles are tangent, then the tangent point must be on the line 135① two circles are tangent to D > R+R ② two circles are tangent to d=R+r ③ two circles intersect R-R < D+R (R > R) ④ two circles are inscribed with D = R-R (R > R) ⑤ two circles contain D < R. The chord 137 theorem divides a circle into n (n ≥ 3): (1) The polygon obtained by connecting points in turn is an inscribed regular N polygon of the circle; (1) The circle passes through the tangents of each point, and the polygon whose vertices are the intersections of adjacent tangents is an circumscribed regular N polygon of the circle. These two circles are concentric circles 139. Every inner angle of a regular N-polygon is equal to the radius and area of the regular N-polygon in theorem (n-2) × 180/N 140, where apome divides the regular N-polygon into 2n congruent right-angled triangles 14 1. A/4A refers to the side length 143. If there are K positive N corners around a vertex, since the sum of these angles should be 360, k × (n-2) 180/n = 360 is converted into (n-2)(k-2)=4 144. Arc length calculation formula: L=n R/ 180 145. Sector area formula: S.

1. As shown in the figure, it is known that EB⊥AD is in B, FC⊥AD is in C, EB=FC, AB=CD. Proof: AF=DE.

Analysis: Find the triangle where AF and DE are located, and first prove δ AFC δ Deb. Then prove that AF=DE.

Proof: ∵EB⊥AD (known)

∴∠ EBD = 90 (vertical definition) fca = 90,

∴∠EBD=∠FCA，

AB = CD，BC=BC，

∴ AC=AB+BC=BC+CD=BD

In δδACF and δδDBE,

∴δacf≌δdbe(sas)，

∴AF=DE (the corresponding sides of congruent triangles are equal).

Example 2, as shown in the figure, it is known that AB and CD are equally divided into O, and the straight line passing through O intersects with AD and BC at E and F respectively, which proves AE=BF.

Analysis: We can analyze the concept of proof from two directions:

(1) "Causality leads to effect": From the known conditions, which pairs of triangles can be proved to be congruent? Which line segments or angles are equal? Can you draw a verified conclusion from this?

In this example, from the known conditions of AO=BO, ∠AOD=∠BOC and DO=CO, we can quickly prove △ AOD △ BOC by SAS, and then get AD=BC, ∠A=∠B, ∠ according to the properties of congruent triangles. From these three intermediate results, the most effective condition for transforming into a new triangular congruence is selected: AE and BF are in △AOE and △BOF respectively, so ∠A=∠B is selected as the new condition, △ AOE △ BOF is proved by (ASA), and AE=BF is obtained from the properties of congruence triangles.

(1 1) "Causality": according to the verification target, which pair of triangles need to be proved to be congruent; If the condition is not enough, can you provide the condition by proving the congruence of another pair of triangles?

In this example, in order to prove AE=BF, since AE and BF are in △AOE and △BOF respectively, we can first consider proving that △ AOE △ BOF has OA=OB, ∠AOE=∠BOF, and the missing condition is ∠A=∠B or OE.

Of these two methods, the first represents the idea of "pushing forward" and the second represents the idea of "pushing back", but in either case, the proof must be written in the way of pushing forward.

Prove that ∵AB and CD are equally divided into O (known)

∴AO=BO, OC=OD (definition of midpoint of line segment)

In lines △AOD and △,

∵

∴△AOD≌△BOC(SAS)

∴∠A=∠B (the corresponding angles of congruent triangles are equal).

In delta △AOE and delta △BOF

∵

∴△AOE≌△BOF (American Soybean Association)

∴AE=BF (the corresponding sides of congruent triangles are equal)

Example 3, as shown in the figure, AC and BD intersect at E, AC=BD, AB=DC, and verification: BE=CE.

Analysis: in order to prove BE=CE, we only need to prove △ Abe △ DCE. In these two triangles, AB=DC and ∠AEB=∠DEC, one angle and the opposite side are equal respectively. There is still a condition missing, so we can only find another diagonal equality condition, which naturally makes us turn our attention to proving ∠ A. △ ACD △ DBA, and find a complete proof idea from it.

The verified route is as follows:

Proof: Linked advertising.

In delta delta △△ACD and delta delta delta △△DBA.

∵

∴△ACD≌△DBA(SSS)

∴∠B=∠C (the corresponding angles of congruent triangles are equal).

In △ Abe and △DCE.

∵

∴△ABE≌△DCE(AAS)

∴BE=CE (the corresponding sides of congruent triangles are equal)

Example 4 proves that the bisectors of the corresponding angles of congruent triangles are equal.

Analysis: First of all, we should distinguish the topic and conclusion in the proposition. Formally, it seems that there is only a conclusion in the topic. I don't know what the topic should be written. In fact, any mathematical proposition is a complete narrative and a sentence to judge something.

Then there must be an object of judgment and the result obtained after judgment in this sentence, then the object of judgment is the condition (topic setting) of the proposition, and the result is the conclusion of the proposition. According to this standard, the problem in the example should be: bisectors of two congruent triangles and their corresponding angles. The conclusion is that the bisectors of the corresponding angles are equal.

By distinguishing the topic setting and conclusion of the proposition, the content of the proposition can be drawn into corresponding geometric figures, so that simple symbols can be used instead of text narration. This example can be drawn like this. Draw two congruent triangles △ABC and △A'B'C, and then make a pair of bisectors AD and A'D' corresponding to the angle ∠A∠A'

When drawing, you must pay attention to two points: (1) Don't draw unnecessary conditions that are not in the topic. According to the requirements of this problem, triangles can only be drawn as arbitrary triangles, and isosceles triangles or equilateral triangles cannot be drawn, so as not to interfere with thinking. (2) Don't ignore the inherent nature of the figure referred to in the question. If two triangles are the same, don't draw one big and one small, or two triangles with different shapes.

Then combined with graphics, write the known content and verify it according to the exact description of each concept.

It is known that △ ABC △ a 'b 'c', AD and A'D' are bisectors of ∠BAC and ∠B'A'C, respectively. Proof: AD=A'D'.

The verified route is as follows:

Proof: ∫△ ABC△ a 'b 'c' (known)

∴∠B=∠B', ∠BAC=∠B'A'C' (the corresponding angles in congruent triangles are equal).

AB=A'B' (the corresponding sides of congruent triangles are equal)

And ∵AD and A'D' are the bisectors of ∠BAC and ∠B'A'C' respectively (known).

∴∠ 1=∠BAC, ∠2=∠B'A'C' (definition of angular bisector)

∴∠∠∠∠∠∠∠∠∠∠∠∠∠ 1 =∠∠∠∠∠∠∠∠∠∠∠∠∠8736

In △ABD and △A'B'D'

∵

∴△ABD≌△A'B'D'(ASA)

∴AD=A'D' (the corresponding sides of congruent triangles are equal)

Example 5, verification: the median lines of two sides and the third side of two triangles are equal, and the third side of two triangles is equal.

Analysis: The topic is that the median lines of two sides and the third side of two triangles are equal, and the conclusion is that the third side of these two triangles is equal.

δ△ABC and δA' B'C, AB=A'B', AC=A'C', D is the midpoint of BC, D' is the midpoint of B' c', and AD=A'D'. Verification: BC=B'C'

Analysis: According to the topic, the given known conditions are not in the same triangle. If we want to make full use of them.

In view of these conditions, we must find a way to concentrate these scattered conditions in a triangle. Because there is a midline in the topic, we often use an auxiliary line with twice the length of the midline to create a congruent triangles, and then use the properties of congruent triangles. In this way, the scattered conditions are concentrated in a triangle and the problem is transformed into a solvable direction.

Prove: extend AD to e, make DE=AD, connect BE,

Expand A'D' to e' so that D'E'=A'D' and connect B'E'

Ad = a'd' (known) ∴DE=D'E' (equivalent substitution)

∫D is the midpoint of BC, and d' is the midpoint of B'C' (known).

∴BD=DC, B'D'=D'C' (defined as the midpoint of a line segment)

In △ACD and △EBD, in △A'C'D' and △E'B'D'

∵ ∵

∴△acd≌△ebd(sas)∴△a'c'd'≌△e'b'd'(sas)

∴AC=BE (the corresponding sides of congruent triangles are equal)

∴A'C'=B'E' (the corresponding sides of congruent triangles are equal).

∴∠E=∠5 (the corresponding angle of congruent triangles, etc. )

∴∠E'=∠6 (the corresponding angle of congruent triangles, etc. )

AC = A 'C' (known)

∴BE=B'E' (equivalent substitution)

∴2AD=AE, 2A'D'=A'E' (equal property)

∴AE=A'E' (equivalent substitution)

In △ABE and △A'B'E

∵

∴△ABE≌△A'B'E'(SSS)

∴∠7=∠8

∴∠ e =∠∠ e' (the corresponding angles of congruent triangles are equal).

∠∠E =∠E' (authentication) ∠E=∠5, ∠E'=∠6 (authentication)

∴∠5=∠6 (equivalent substitution)

∠∠7 =∠8 (certification)

∴∠ 7 +∠ 5 =∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠8

That is, ∠BAC=∠B'A'C'

In △BAC and △B'A'C,

∵

∴△BAC≌△B'A'C'

∴BC=B'C'

Third, the practice of auxiliary lines: congruent triangles's proof needs to add auxiliary lines. The basic idea of adding auxiliary lines is to add auxiliary lines and construct congruent triangles. Now let's introduce some methods to add auxiliary lines for everyone to learn.

1. According to the principle of "central symmetry", the congruent triangles is constructed and auxiliary lines are added.

Rotate a triangle around one of its vertices 180 to get another triangle. Such a pair of triangles is called centrosymmetric congruent triangles (or, rotate a triangle around a certain point 180 to get another triangle, and such a pair of triangles is called centrosymmetric congruent triangles), as shown in the following basic figure.

Note: When there are two equal line segments on both sides of a pair of vertex angles in a geometric problem and they are on a straight line, you can add a centrosymmetric congruent triangles to prove it. Addition is to make parallel lines through endpoints, or to cut equal line segments according to the method in Example 5 above.

Example analysis: As shown in the figure, it is known that Δ Δ AB = AB in AC = AC, BD = CF Verification: DE = EF.

The conclusion to be proved by analyzing this topic is de = ef. As shown in the figure, there are two equal line segments on both sides of a set of vertex angles, and they are on a straight line. In this case, we can add a pair of centrosymmetric congruent triangles to prove it. The method of addition is to cross D as DG//AC and BC as G, as shown in the figure, then Δ δDGE and Δ δFCE must be a pair of central symmetric congruent triangles. To prove that these two triangles are congruent, we need to grasp the condition that one set of edges is equal, and the conclusion that DE=EF is unusable needs to prove the other set of edges.

The known conditions tell us that CF=BD, so it is necessary to prove that CF is equal to its corresponding side DG, as shown in the figure, that is, it is proved that DB=DG and DG//AC, so ∠ 1=∠2, AB=AC is known, so ∠2=∠B, so ∠.

Prove that 1: D is DG//AC and BC is G.

∫DG//AC

∴∠ 1=∠2; ∠3=∠4

AB = AC (triangle ABC is an isosceles triangle)

∴∠B=∠2

∴∠ 1=∠B,∴DG=DB=CF

At △DGE and △FCE.

∴△DGE≌△FCE

∴DE=EF

Analysis 2: As shown in the following figure, this problem can also be extended from the end point F to FH//AB to BC at H, supplemented by a pair of central symmetric congruent δδBDE and δδHFE.

Proof 2: It is basically the same as the previous proof. Prove DE=EF by proving △ efh △ EDB. Pay attention to the relationship between BD//FH and angle. Prove it briefly.

Note: The two angles of an isosceles triangle are equal. I learned it in primary school and I will learn it in the future.

2. According to the principle of "axis symmetry", the congruent triangles is constructed and auxiliary lines are added.

Rotate a triangle along a straight line and then overlap another triangle, then this pair of triangles is called axisymmetric congruent triangles.

Basic graphics:

When two equal line segments or two equal angles are symmetrical about a line segment or a straight line in geometric problems, an axisymmetric congruent triangles can be constructed to prove it.

Example analysis: As shown in the figure, AE=AB is intercepted on the diagonal AC of the square ABCD, and EF⊥AC passes through BC to F for verification: EF=FB.

Analysis: the conclusion to be proved in this question is EF=FB. It is known that there are AE=AB and ∠ AEF = ∠ B = 90 in this question, so the connecting AF structures △AEF and △ABF are congruent and easy to prove.

It is proved that the connection AF is ∠ b = 90 in the square ABCD,

* ef⊥ac

∴∠AEF = 90°

At RT△AEF and RT△ABF,

AE=AB

AF=AF

∴ RT△AEF≌RT△ABF (HL)

∴EF=BF.

I wish you success!