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In the given figure, altitudes $AD$ and $CE$ of $∆ABC$ intersect each other at the point $P$. Show that:
$∆AEP \sim ∆ADB$
"
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 $∆AEP \sim ∆ADB$.
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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