Two congruent triangle questions will add points

1)ac=a'c',cd=c'd'Therefore, HL proves that the right triangle ACD is equal to the right triangle A'c'd', so the angle a=a', so angle b = angle b'，ac=a'c', angle a=a',b=b', AAS is proof.

Take the midpoint p of the ad and connect it to the MP

then MA AP

So the angle apm=45 degrees.

So the angle MPD = 135 degrees = the angle MBN

Because the angle NME + angle DMA = 90 degrees.

Angular DMA + angular MDP = 90 degrees.

So angular mdp = angular nmb

And because dp=bm

So mdp is all equal to mbn

So md=mn

Happy studying.

The intersection of MN and BC is O, and the perpendicular line of ME is made over N.

MN MD proves that the triangle DAM is similar to the triangle MBO.

It is concluded that o is 1 4 points.

The triangle DAM and the triangle MNE can then be proved to be congruent.

md=mn

1 Easy Certificate; CDF and BEF, ACE and ADB2 because; cdf and bef, ace and adb are congruent, so be=cd

Because abe= acd, ae=ad, a= a= a, so acd= abe ac=ab

ae=ad

So db=ec

efc=∠dfb，∠abe=∠acd

So fce and fdb are congruent.

So df=ef

As shown in the figure, in isosceles RT ABC, P is the midpoint of the hypotenuse BC, and the two sides of the right angle with P as the vertex intersect with the edge AB and AC respectively at the points E, F, and connect EF When the EPF rotates around the vertex (the point E does not coincide with A and B), PEF is always an isosceles right triangle, please explain the reason

1. There are 4 congruent right-angled cardboards, can you use them to spell the Pythagorean theorem? If you can, explain your ideas and methods, the more ways the better (write at least four methods).

2. Divide the equilateral triangle into three congruent figures, please draw three different division methods

3. Read the following questions and analysis process, and prove them as required

Verification: ab=cd

Analysis: To prove that two line segments are equal, the common general method is to apply the judgment and properties of congruent triangles or isosceles triangles, observe the two line segments to be proved in this problem, they are not in the same triangle, and the two triangles in which they are located are not congruent Therefore, to prove ab=cd, it is necessary to add appropriate auxiliary lines to construct a congruent triangle or isosceles triangle

Here are three ways to add auxiliary lines, please choose any one of them to prove the original question

Proof: Method 1: Make bf de at point f, cg de at point g

f=∠cge=90°．

bef= ceg, be=ce, bfe cge

bf=cg．

In ABF and DCG, F= DGC=90°, BAE= CDE, BF=CG, ABF DCG

ab=cd．

Method 2: Make CF AB and cross the extension line of DE at the point F

f=∠bae．

and abe= d, f= d

cf=cd．

f=∠bae，∠aeb=∠fec，be=ce，△abe≌△fce．

ab=cf．

ab=cd．

Method 3: Extend de to point F so that ef=de

be=ce, bef= ced, bef ced

bf=cd，∠d=∠f．

and bae= d, bae= f

ab=bf．

ab=cd．

Is AB ABE congruent with ADF?

Solve it yourself.

ab=cd,bc=da

ABCD is a parallelogram.

dac=∠acb

AD=BC, AE=CF

ADE congruent triangle CBF [corner edge].

bf=de, I hope you have some help to envy the friends hall.

ab=cd bc=da ac=ac

ADC is fully equal to CBA

dca=∠cab

ae=cfae+ef=ce+ef

af=ceab=cd

The difference in the hall abf is all equal to the virtual and hidden cde(sas)bf=de