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AAA Similarity

In fig 1.41, ...

Question

In fig 1.41, E is a point on side CB (that is produced) of isosceles $△ ABC$ . AB=AC.
If AD $⊥$ BC and EF $⊥$ AC, prove that $△ ABD ~ △ ECF$ .

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Solution

Given : AB = AC
AD $⊥$ BC and EF $⊥$ AC.
To prove : $△ ABD ~ △ ECF$

Proof :
$In △ ABD & △ ECF, ∠ ABD ≅ ∠ ECF (Angles opposite equal sides are equal) ∠ ADB ≅ ∠ EFC (90 ° each) Therefore, by the AA similarity test, △ ABD ~ △ ECF Hence proved .$

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