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