Question

in the figure, AC = AE, AB = AD and <BAD = <EAC. show that BC = DE

Answer 1

It is given that ∠BAD = ∠EAC

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

In ΔBAC and ΔDAE,

AB = AD (Given)

∠BAC = ∠DAE (Proved above)

AC = AE (Given)

∴ ΔBAC ≅ ΔDAE (By SAS congruence rule)

∴ BC = DE (By CPCT)

Answer 2

in ABC and DEF
Ac=Ae (given)
Ab=Ad (given)
angle<BAD=<EAC (angles opposite to equal Side)
therefore ABC is congurance to DEF (by side angle side)
so, BC=DE(by CPCT)
proved..
hope it will help u..