Given that, the sides BA and CA have been produced such that BA= AD and CA=AE and given to prove DE|| BC

Consider triangle BAC and DAE,

We have

BA = ADand CA = AE [∴ given in the data]

And also ∠BAC =∠DAE [∴ vertically opposite angles]

So, by SAS congruence criterion, we have ΔBAC ≅ ΔDAE

⇒ BC = DE and ∠DEA=∠BCA,∠EDA=∠CBA

[Corresponding parts of congruent triangles are equal]

Now, DE and BC are two lines intersected by a transversal DB such that ∠DEA=∠BCA,

i.e., alternate angles are equal

Therefore, DE || BC