Sides BA and CA have been produced such that BA = AD and CA = AE.
To prove: DE ∥ BC
Consider △BAC and △DAE,
BA = AD and CA= AE (Given)
∠BAC = ∠DAE (vertically opposite angles)
By SAS congruence criterion, we have
△BAC ≃ △DAE
We know, corresponding parts of congruent triangles are equal
So, BC = DE and ∠DEA = ∠BCA, ∠EDA = ∠CBA
Now, DE and BC are two lines intersected by a transversal DB s.t.
∠DEA=∠BCA (alternate angles are equal)
Therefore, DE ∥ BC. Proved.