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

Answer :

Given, ABC is an isosceles triangle.

AB = AC
$$\angle$$ ABD = $$\angle$$ ECF

In $$\triangle$$ ABD and $$\triangle$$ ECF,
$$\angle$$ ADB = $$\angle$$ EFC (Each 90°)
$$\angle$$ BAD = $$\angle$$ CEF (Already proved)
$$\triangle$$ ABD ~ $$\triangle$$ ECF (using AA similarity criterion)