Math Genius Advance! Elementary 2 Geometry Congruent Triangle Proof Questions! To process!! Very urg

The triangle ACB and the triangle ADB can be found congruence, so the angle cab=angle bad AC=AD, so the triangle ace is all equal to the triangle ADE, so CEA= DEA

Proof: in ABE and DBA.

bae= d, b is the common angle of these two triangles.

ABA (two triangles are similar if two corners correspond to each other).

According to the nature of similar triangles: if two triangles are similar, the corresponding sides are proportional.

ad/ae=ab/be...

and bac acb, abc is an isosceles triangle, ab=bc and e is the midpoint of bc be=ec=bc 2, i.e. be=ab 2....

Substituting the formula into the formula yields: ad=2ae. And thus it was proven.

Because bae d, bac acb

So, ABE is similar to DBA

Because of BAC ACB

ABC is an isosceles triangle.

ab=BCE is the midpoint of BC.

2be=ab

So ad=2ea

Angle bae d, angle bac = acb

Horn bae cae bac

d＋∠cad＝∠acb

cae=∠cad

Prove cad= d again (I won't, you think for yourself).

dae=2∠d

ad=2ea

Upstairs is correct, but the process is not detailed enough to almost faint.

The one upstairs admired it very much.

Proof:

PBC and QCD are both equilateral triangles, and quadrilateral ABCD is rectangular, PB=PC, CQ=CD=AB

PBC=PCB=60°, the quadrilateral ABCD is rectangular, and PB=PC, CQ=AB

pab≌△pqc．

ap=pq．

The quadrilateral ABCD is a rectangle, PBC, QCD are equilateral triangles (known) AB=OC=QC,BPP=PC, QC= CBP= BCP=60°, ABC= BCO=90°

abc- cbp= bco- qco=30°, i.e. abp= bcq=30°

BCP=60°

qcp = bcp - bcq = abp = 30° and ab = cq, bp = pc (verified).

abp≌△qcp （sas)

Pa=pq (congruent triangles correspond to equal sides).

The first question has no pictures, so you can only do the second Qingyin question!

Two (1) abc is equal to def => scrambling a= d and a+ b=90° d+ b=90°

bpd=90° ∴ab⊥ed

2) PB=BC BPC= BCP with positive CPD=90- BPC=

pca=90-∠bcp∴∠cpd=∠pca∠cpd+90=∠pca+90=>∠cpa=∠pcd ec=ce△cpa≡△pcd

ae=cf bf=de quadrilateral, abcd is a parallelogram.

ed bf parallel angle aed, angle bfc equal angle ade cbf equal.

Angles abf edc equal...

The right-angled edge of a knowledge point 30 degrees is half of the hypotenuse, and you understand this and make it.

This is a hassle to say. You need to verify the angles of the 123 corners. And what kind of shape is the triangle where these corners are located. Whether it's the same or not is the answer to the question.