To Prove: ∆ACB ∼ ∆DCE

Proof:

Given that, ∠1 = ∠2

⇒ ∠DBC = ∠DCE

Also in ∆ABC & ∆DCE, we get

∠DCE = ∠ACB [they are common angles to both triangles]

And

Or

Or [∵ BD = DC as ∠1 = ∠2]

Thus by SAS-similarity criteria, we get

∆ACB ∼ ∆DCE

Hence, proved.