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

"

Given:

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

To do:

We have to show that $∆ABD \sim ∆CBE$

Solution:

In $\Delta ABD$ and $\triangle CBE$,

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

$\angle ABD=\angle CBE$ (common)

Therefore, by AA criterion,

$\Delta ABD \sim \Delta CBE$

Hence proved.

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