It is given in the question that:

AC = AE

AB = AD and,

∠BAD = ∠EAC

To show: BC = DE

Proof:

∠BAD = ∠EAC (Adding ∠DAC both sides)

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠EAD

In

AC = AE (Given)

∠BAC = ∠EAD

AB = AD (Given)

Therefore,

By SAS congruence,

BC = DE (By c.p.c.t)