Question

in the given figure, AB = AC, BD = EC. Prove that ∆ABE ~ ∆ACD and AD = AE.

Answer 1

In triangles ABE and ACD :

AB=AC (given)

Therefore, angle ABE=angle ACD (isosceles property)

BD=EC(given)

BD+DE=EC+DE (when equals are added to equals the wholes are equals)

Therefore, BE=EC

ABE~ACD (SAS Congruence)

AD=AE (C.P.C.T)