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