In the second year of junior high school, find the triangle congruence problem and solve it with a h

Hope. Proof of the following: Extend AE, cross the BC extension line to F, AD BC, 1= 2, 3= 4 AEB= 2 3=90 , be af ABF is an isosceles triangle, AE=EF isosceles triangle is a three-in-one, AB=BF is in ADE and FCE.

1=∠f，aed=∠cef

ae=ef⊿ade≌⊿fce

ad＝cfad+bc=cf+bc=bf=abad+bc=ab.

Pass e as an ad parallel line, cross ab to f, you can get: ad ef bcad ef

1= aef is also known as 1= 2

AEF= 2 (triangle AEF is an isosceles triangle) AF=EF The same can be done for BF=EF

af=bf (f is the midpoint of ab).

EF is the midline of trapezoidal ABCD.

ad+bc=2ef

and af=ef bf=ef

ad+bc=af+bf=ab

You are the first year of junior high school to solve the problem easy not the two upstairs: Person: Know 905 Intercept FB=CB to connect EF Person: Know 905

The first person who can prove the CBE congruent FBE by SAS: knows the 905AD EF: knows the 905

ADC+ C=180°: Know 905

C= EFB , EFB+ AFE=180° Person: Someone who knows 905 ADC= AFE (see for yourself) Person: Someone who knows that 905 DAE is a DAE congruent FAE that can be obtained from AAS:

Those who know 905ad=af, bc=bf: know 905

ad+bc=ab: know 905

Don't plagiarize the people in the back Look at the world.

Diagrams are very important when doing math problems, and you try to draw standard points as much as possible. Such as AEB is a right angle.

Do EF AB over point E to F

then afe= bfe=90°

In ADE and AFE.

d=∠afe=90°

ae=ae△ade≌△afe

AD=AF CEB FEB

cb=fbad+bc=af+bf

i.e. AD+BC=AB.

In the triangle ABC, DF is perpendicular EF, F is the midpoint of BC, verify the relationship between BE+CD and ED (don't guess, verify it, don't say that the title is wrong, I didn't think the title is wrong, but it's really right) tip: 3 auxiliary lines.

Ask your teacher to go, he will be very happy...

1.A congruent triangle is two congruent triangles in which the three sides and three corners are correspondingly equal. Congruent triangles are a type of congruence in geometry.

Depending on the congruence transformation, two congruent triangles can be translational, rotated, axisymmetric, or overlapping, and so on. When the corresponding sides and corners of two triangles are completely opposite, the two triangles are congruent triangles. Normally, two congruent triangles are verified with three equal parts, and the result is finally obtained.

2.(1) The side of the corresponding angle of the congruent triangle is the corresponding edge, and the edge sandwiched by the two corresponding angles is the corresponding edge; (2) The angles of the corresponding sides of the congruent triangle are the corresponding angles, and the angles between the two corresponding sides are the corresponding angles; (3) If there is a common edge, the common edge must be the corresponding edge; (4) If there is a common angle, the angle must be the corresponding angle; (5) If there is a pair of apex angles, the pair of apex angles must be corresponding angles;

3.1. Three groups of two triangles with equal sides (SSS or "edge-edge-edge") also explains the reason for the stability of triangles.

2. There are two triangles with equal sides and their angles corresponding to the congruence (SAS or "corner edges").

3. There are two corners and their sandwich edges corresponding to two equal triangles congruence (ASA or "corner corners"). Pushable by 3.

4. There are two corners and the opposite side of one of the corners corresponds to two equal triangles congruence (AAS or "corner edges").

5. The congruence conditions of a right triangle are: the hypotenuse and the straight angle side correspond to the equal congruence of two right triangles (hl or "hypotenuse, right angled edge") Therefore, sss, sas, asa, aas, hl are all theorems to determine the congruence of triangles.

This is the simplest geometry, and the geometry learned in the first year of junior high school is fundamentally pediatric in the second year of junior high school.

In fact, some congruent triangles can be seen at a glance, and then find conditional proofs, but not every question is like this, maybe the drawing is not standard.

Congruent triangles, the key is to use them flexibly, so it's OK.

It should be AD BC, right? Words like this.

ad//bc

1= f, a= 3 (two straight lines are parallel, the inner misalignment angles are equal) e is the midpoint of ab ae=be ade bfe (aas).

ade≌△bfe

de=fege is the midline of GDF.

1 = 2 (known).

Again 1= f (proven).

2= fgd=gf, i.e. gdf is an isosceles triangle.

e.g. df (the high and mid lines on the base edge of the isosceles triangle coincide).

Solution: 1).

ad//bc

and e is the midpoint of ab.

AE=BE in ADE and BFE.

aed=∠bef

ae=be△ade≌△bfe

2).∵gdf=∠adf

2=∠gdf ∴dg=fg

DGF is an isosceles triangle.

and fe=de

EG is the midline of DGF.

and the triangle is three lines in one.

EG is also high for DGF.

EG is perpendicular to DF

If you change AB BC to AD BC

Solution: 1).Because of AD BC

So 1= 2

Because point e divides ab

So ab=be

In the triangular ADE and triangular BFE.

aed=∠bef

AE=BE, so the triangle ADE is all equal to BFE

2).Because gdf= adf

So 2= GDF

Because the triangle ADE is all equal to BFE

So ef=de

So g is perpendicular to df

There is a problem with your inscription: AB cannot be parallel to BC

If you don't say where m and n are next to ab and ac, then you can do this:

Suppose you make an angle Mdn 60 degrees at point D, and the triangle Mdn is an isosceles triangle, then the problem is not difficult.

Obviously, the triangle MDN is an equilateral triangle, first find DC, then you can know that the angle ACD is a right angle, and then find the length of DN, that is, DN mn 2 3, because the triangle AMN is also an equilateral triangle, so you can find the circumference of the triangle AMN is 2 (if you can't type the root number, you can calculate it).

This question lacks conditions and cannot find the specific value.

Because the perimeter of the AMN ranges from: 3 2 to (3 + root number 3) 2

Proof: If the bisector of the two corners of a triangle is equal, try to prove that the triangle is an isosceles triangle.

Proof: Angle ACD = Angle ECB = 60 degrees, then: Angle ACE = Angle Book Eggplant DCB;

and ac=dc, ec=bcHence ace δdcb(sas)

Proof: ace then picoδdcb, then: angle cam = angle cdn;

and angle ACM = angle NCB = 60 degrees, then angle ACM = angle DCN = 60 degrees state inspection;

and ac=dcTherefore ACM δdcn(asa), yield: cm=cn

ac=dn is not necessarily true, so it cannot be proven!

In fact, it should be: AM=DN, which can be proved by ACM δdcn.