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,
The sides BA and CA have been produced, such that:
BA = AD
And, CA = AE
We have to prove that,
DE ‖ BC
Consider and
, we have
BA = AD and CA = AE (Given)
∠BAC = ∠DAE (Vertically opposite angle)
So, by SAS congruence rule we have:
Therefore, BC = DE and
∠DEA = ∠BCA,
∠EDA = ∠CBA (By c.p.c.t)
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