What are the problems of congruent triangles in the first volume of eighth grade mathematics published by People's Education Press? The more the better! ! !

In triangle ABC, angle CAE= angle B, e is the midpoint of CD, and the bisecting angle BAE of AD proves that BD=AC.

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If we make a point F on AB, so that AF=AE, we can know from the bisector of angle BAE that triangle ADE is all equal to triangle ADF, from the midpoint of CD that DF=DE, angle DEA= angle DFA, CE=DE=DF, angle DFB= angle AEC (the complementary angles of equal angles are also equal), and then from angle CAE= angle B that triangle CAE is all equal to triangle DBF.

2.？

Q: As shown in the figure, in the triangle ABC, the bisectors AD, BE and CF intersect at point H, the intersection point H is HG perpendicular to AC, and the vertical foot is G, so is the triangle AHE equal to the triangle CHG? Why? ?

∵∠AHE=∠ABH+∠BAH, (one outer angle of a triangle is equal to the sum of two non-adjacent inner angles);

∠ABH= 1/2∠ABC，

Bah = 1/2 ∠ BAC (meaning angular bisector),

∴∠AHE= 1/2(∠BAC+∠ABC) (equivalent substitution).

∠∠BAC+∠ABC = 180-∠C (the sum of the interior angles of the triangle is equal to 180).

∴∠ahe=( 180-∠ACB)/2 = 90- 1/2∠ACB，

At Rt△HGC,

∠GHC=90 -∠HCG=90 - 1/2∠ACB，

∴? ∠AHE=∠CHG。

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3.？ It is known that when △AB=AC, ∠ A = 90, AB=AC, AE⊥BD is in E, ∠ ADB = ∠ CDF, AE is extended from BC to F, which proves that D is the midpoint of AC.

Let d be the symmetry point g of BC connecting FG and CG.

Because angle ADB= angle BAF and angle FDC= angle BAF.

Angle B= angle c = 45.

So angle AFB = 180- angle B- angle BAF = 180- angle C- angle CDF= angle DFG.

So angle AFD+ angle DFG= angle AFD+ angle DFC+ angle AFB = 180.

So lines a, f and G***

And because angle CAG= angle ABD.

Angle ACG = 2 * 45 = 90 = bad angle.

So all the bad triangles are equal to triangle ACG.

So CG=AD

CG=DC

So AD=DC

4.？ In the known triangle ABC, AD is the center line of BC side, E is the upper point of AC, BE and AD intersect at F, and if AE=EF, it is verified that AC=BF.

Expand AD to m so that DM = ad, even BM, CM.

AD=DM，BD=CD

∴ABMC is a parallelogram (diagonal bisection)

∴AC‖BM,AC=BM (equal to the last use)

∴∠DAC=∠DMB(∠DAC is ∠EAF, ∠DMB is ∝∞∠bmf used below) (the internal angles are equal) ...

In the triangle AEF,

AE = EF

∴∠EAF=∠EFA (isosceles triangle) ... ②

∫∠EFA =∠BFM (relative vertex angle is equal) ... ③

From ① ② ③, we get ∠EAF=∠EFA=∠BFM=∠BMF.

In a triangular BFM,

∠∠BFM =∠BMF