In Fig. 10.22, the sides BA and CA have been produced such that: BA = AD and CA = AE.
Prove that segment DE || BC.
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
BA = ADand CA = AE [∵ given in the data]
And also ∠BAC=∠DAE [ ∵vertically opposite
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