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# In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If $AD&space;\perp&space;BC$ and $EF\perp&space;AC$, prove that $\triangle&space;ABD\sim&space;\triangle&space;ECF$.

It is given that ABC is an isosceles triangle. ∴ AB = AC ⇒ ∠ABD = ∠ECF In ΔABD and ΔECF, ∠ADB = ∠EFC (Each 90°) ∠ABD = ∠ECF (Proved above) ∴ ΔABD ∼ ΔECF (By using AA similarity criterion)

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ABD = ∠ECF (Proved above)

∴ ΔABD ∼ ΔECF (By using AA similarity criterion)

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