It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

To prove:

The line segment BC and DE are similar i.e. BC = DE

Proof:

We know that BAD = EAC

Now, by adding DAC on both sides we get,

BAD + DAC = EAC +DAC

This implies, BAC = EAD

Now, ΔABC and ΔADE are similar by SAS congruency

(i) AC = AE (As given in the question)

(ii) BAC = EAD

(iii) AB = AD (It is also given in the question)

∴ Triangles ABC and ADE are similar i.e. ΔABC ΔADE.

So, by the rule of CPCT, it can be said that BC = DE.