As far as I know, there are four ways to prove congruent triangles: SSS, SAS, ASA and AAS. The only thing that can't be used is SSA, which proved to be completely wrong in this way.

Now, I will list a topic for each proof method.

SSS means that two triangles meet with three corresponding equilateral sides.

The first question is the simplest SSS proof method.

As shown in the figure, if AB = DE, BC = EF and AF = DC are known, then ∠EFD=∠BCA, please explain the reason.

Proof: ∫AF = DC (known) E.

∴AF+FC=DC+FC

∴ AC=DF

At △ABC and △DEF A F

AC = DF (certification) c d.

AB=DE (known)

DC=EF (known)

∴△ABC≌△DEF(SSS) B

∴∠EFD =∞∠BCA (the angles corresponding to congruent triangles are equal).

This is the most basic question. .

SAS refers to the congruence of two triangles with two sides corresponding to their included angles.

The first problem is the simplest problem in SAS proof method.

As shown in the figure, AC and BD intersect at point O. It is known that OA=OC and OB=OD represent △ AOB △ COD.

It is proved that A B is at △AOB and △COD.

OA = OC (known)

∠AOB=∠COD (equal to vertex angle) o

OB=OD (known)

∴△AOB≌△COD(SAS Special Zone

This question is very simple, but it will not be so easy if the previous right-angle knowledge is not learned well. ASA refers to the congruence between two corners of two triangles and their sides.

The first problem is that ASA is relatively simple.

As shown in the figure, ∠DAB=∠CAB and ∠EBD=∠EBC are known, representing △ ABC △ Abd.

Proof: ∫∠EBD =∠EBC (known) d

∴∠ABC=∠ABD (complementary angles of equal angles are equal)

In △ABC and △ABD, A B E

∠∠DAB =∠CAB (known)

AB=AB (known)

∠ABC=∠ABD (certification) c

△ABC?△ABD(ASA)

I say this question is simple because there are many known conditions, but one condition is to remember the knowledge that the complementary angles of equal angles are equal.

The last one is to prove the problem by AAS.

As shown in the figure, it is known that ∠B=∠C and AD=AE, which means AB = AC.b

Proof: in △ABE and △ACD

∠∠B =∠C (known) d

∠A=∠A (angle) a.

AE=AD (known) e

∴△ABE≌△ACD(AAS) C

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

This is just one kind, and there is another kind of knowledge that congruent triangles is not only proved by AAS method, but also by angular bisector.

As shown in the figure, point P is a point on the bisector of pb⊥ab,pc⊥ac∠BAC, that is, PB=PC.

Prove that ∵AP is the bisector of ∠BAC (known)

∴∠ Angle =∞∠BAP (definition of angular bisector)

∵PB⊥AB,PC⊥AC (known)

∴∠ABP=∠ABP (definition of vertical line)

C in delta △APB and delta △APC

∠∠PAB =∠PAC (certification) P.

∠ABP=∠ABP (authentication)

AP=AP (male * * * side) V A B

∴△APB≌△APC(AAS)

∴PB=PC (the corresponding sides of congruent triangles are equal)

Of all these problems that prove congruent triangles, one is the most troublesome and often makes me make mistakes, like the following one:

As shown in the figure △ABC and △AB'C', AB=AB', in order to make △ ABC △ AB' c', add a condition _ _ _ _ _ _ _ _ b'

C

A

C' B

In this case, we can use SAS, ASA and AAS. The only thing that can't be proved is SSA, but I sometimes use SSA to prove that BC=B'C' is completely wrong. In this space, we can choose to fill in ∠B'=∠B or ∠ ACB = ∠ AC.

This is a problem I found in my life about proving congruent triangles.