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# In the given figure, altitudes $AD$ and $CE$ of $âˆ†ABC$ intersect each other at the point $P$. Show that:

$âˆ†AEP \sim âˆ†ADB$

"

Given:

In the given figure, altitudes $AD$ and $CE$ of $âˆ†ABC$ intersect each other at the point $P$.

To do:

We have to show that $âˆ†AEP \sim âˆ†ADB$.

Solution:

In $\Delta AEP$ and $\triangle ADB$,

$\angle AEP =\angle ADB=90^o$

$\angle A=\angle A$ (common)

Therefore, by AA criterion,

$\Delta AEP \sim \Delta ADB$

Hence proved.

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