Question

IN THE FIGURE, BD = AC, AE=EB AND ANGLE ADB = ANGLE ACB. SHOW THAT (A) TRIANGLE ABD=~ TRIANGLE BAC (B) TRIANGLE AED =~TRIANGLE BEC​

Answer 1

Answer:

BRO IT IS EASY!

Step-by-step explanation:

A)

In triangle ABD and triangle BAC

BD = AC        (Given)

If two sides of given triangles are equal then there third sides are also equal , so

AD = BC      

AB = AB          (Common)

thus, triangle ABD ≅ triangle BAC    (S.S.S.)

B)

TRIANGLE AED and TRIANGLE BEC​

Since, triangle ABD ≅ triangle BAC     (proven above)

thus, AD = BC      (By, CPCTC)

and, AE = EB          (Given)

If two sides of given triangles are equal then there third sides are also equal , so  

EC = ED  

TRIANGLE AED  ≅ TRIANGLE BEC​ (By S.S.S.)

MARK AS BRAINLIEST

Answer 2

Answer:

given,

BD=AC,

AE=EB,

/_ADB=/_ACB,

A). TO PROVE : ∆ABD~∆BAC

AB=AB. (COMMON SIDE)

BD=AC

/_ADB=/_ACB

BY S.A.S AXIOM,

∆ABD~∆BAC

B). TO PROVE : ∆AED~∆BEC

/_AED=/_BEC,

/_A=/_B,

DA=BC,

BY R.H.S AXIOM,

∆AED~∆BEC

HOPE IT'S HELPFUL TO YOU ♥♥