Question

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF.

Answer 1

Given: ∆ABC in which AB = AC and AD ⊥ BC. Side CB is produced to E and EF ⊥ AC.

To prove ∆ABD ~ ∆ECF,

Proof: we know that the angles opposite to equal sides of a triangle are equal.

∠B = ∠C              ...[∵ AB = AC]

Now, in ∆ABD and ∆ECF, we have

∴ ∠B = ∠C            ...[proved above]

∠ADB = ∠EFC = 90°

∴ ∆ABD ~ ∆ECF      ...[By AA-similarity]