In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

 

Solution:

It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

To prove:

The line segment BC and DE are similar i.e. BC = DE

Proof:

We know that ∠BAD = ∠EAC

Now, by adding ∠DAC on both sides we get,

∠BAD + ∠DAC = ∠EAC +∠DAC

This implies, ∠BAC = ∠EAD

Now, ΔABC and ΔADE are similar by SAS congruency since:

(i) AC = AE (As given in the question)

(ii) ∠BAC = ∠EAD

(iii) AB = AD (It is also given in the question)

∴ Triangles ABC and ADE are similar i.e. ΔABC ≅ ΔADE.

So, by the rule of CPCT, it can be said that BC = DE.