Question

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

Answer 1

Given,

AC = AE, AB = AD and ∠BAD =∠EAC

To prove:

BC = DE

Proof: We have

∠BAD =∠EAC

(Adding ∠DAC to both sides) ∠BAD +∠DAC =∠EAC +∠DAC

⇒ ∠BAC = ∠EAD

In ΔABC and ΔADE,

AC = AE (Given)

∠BAC =∠EAD (proved above)

AB = AD (Given)

Hence, ΔABC ≅ ΔADE (by SAS congruence rule)

Then,

BC = DE ( by CPCT.)

Answer 2

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.