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In Figure 1, $E$ is a point on $CB$ produced of an isosceles $\Delta ABC$ , with side $AB=AC$ . If $AD\perp BC$ and $EF\perp AC,$ prove that $\Delta ABD\sim \Delta ECF$ .

Answers (1)

Given - $ABC$ is isosceles $\Delta$ where $AB=AC$ and $AD\perp BC\; \; ;\; EF\perp AC$

To prove - $\Delta ABD\sim \Delta ECF$

Proof - As $AC=AB$ hence $\angle C=\angle B$ (angles opp to equal sides of an isocelus triangle are equal)

In $\Delta ABD$ and $\Delta ECF$

$\angle ABD=\angle ECF$ ____(i)

$\Rightarrow \angle ADB=\angle EFC\; \; \; \; (\therefore both\; 90^{\circ})$ _____(ii)

From (i) and (ii) we can say by AA similarity,

$\Rightarrow \Delta ABD\sim \Delta ECF$

Hence proved

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Safeer PP

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