Question

In the given figure, E is the point on side CB produced of an isosceles ΔABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF. Hence, prove that `(AC)/(EC) = (BD)/(CF)`.

Answer 1

Given:

ΔABC is an isosceles triangle where AB = AC.

AD ⊥ BC (∠ADB = 90°)

EF ⊥ AC (∠EFC = 90°)

To Prove:

ΔABD ∼ ΔECF

`(AC)/(EC) = (BD)/(CF)`

Proof: 

Part 1: Proving similarity (ΔABD ∼ ΔECF)

In ΔABC, it is given that AB = AC.

Therefore, ∠ABD = ∠ACD.   ...(Angles opposite to equal sides of a triangle are equal)

We can also write this as ∠ABD = ∠ECF since D, C and B lie on the same line. – (Equation 1)

Now, let’s compare ΔABD and ΔECF:

∠ADB = ∠EFC = 90°   ...(Given that AD ⊥ BC and EF ⊥ AC)

∠ABD = ∠ECF   ...(From equation 1)

By the AA (Angle-Angle) Similarity Criterion, we have:

ΔABD ∼ ΔECF

Part 2: Proving the ratio `((AC)/(EC) = (BD)/(CF))`

Since the triangles are similar (ΔABD ∼ ΔECF), their corresponding sides must be proportional:

`(AB)/(EC) = (BD)/(CF) = (AD)/(EF)`

From the first two parts of the ratio, we get:

`(AB)/(EC) = (BD)/(CF)`

Since it is given that AB = AC, we can substitute AC for AB in the equation:

`(AC)/(EC) = (BD)/(CF)`

Hence Proved.