#### Question

In Fig. 10.22, the sides BA and CA have been produced such that: BA = AD and CA = AE.

Prove that segment DE || BC.

#### Solution

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

Is there an error in this question or solution?

Solution In Fig. 10.22, the Sides Ba and Ca Have Been Produced Such That: Ba = Ad and Ca = Ae. Prove that Segment De || Bc. Concept: Congruence of Triangles.