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Non-mathematical form
1. Let's prove the congruence of △ABF and the hypotenuse and right angle of △CED right triangle.

Get BF=DE

AAS should only prove that EGD FGB is congruent.

2. Prove that △ACD is congruent △AED [it is not necessary to prove congruence ... the distance from the point on the bisector of the angle to both sides of the angle is equal ~]

Get CD=DE AC= AE.

AB=AE+EB=8

So AB=CB+EB=8

CB=CD+DB

So AB=ED+BD+BE

Perimeter ~ See you later ~ ~

3.∠ 1+∠AEH+∠AHE= 180

∠2+∠BHD+∠ ADB = 180

There are two right angles ... a set of diagonal angles.

De ∠ 1=∠2 Open

Prove AEH congruence BEC ..

Get AH=BC

An isosceles triangle with three lines in one .. d is the midpoint of BC. ...

2BD=BC

∴2BD=AH

Ho ~ I'm finished ... I'm so tired ... I'm just making a rough call ... I'm sorting out the process myself ~ Hehe ~ Come on!