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 ∆ABD ∼ ∆ECF

Answer 1

It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ $\angle$ABD = $\angle$ECF

In ∆ABD and ∆ECF,
$\angle$ADB = $\angle$EFC (Each 90°)
$\angle$BAD = $\angle$CEF (Proved above)
∴ ∆ABD ∼ ∆ECF (By using AA similarity criterion)

Hence Proved