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In the given figure, altitudes $AD$ and $CE$ of $∆ABC$ intersect each other at the point $P$. Show that $∆PDC \sim ∆BEC$
"
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 $∆PDC \sim ∆BEC$
Solution:
In $\Delta PDC$ and $\triangle BEC$,
$\angle PDC =\angle BEC=90^o$
$\angle PCD=\angle BCE$ (common)
Therefore, by AA criterion,
$\Delta PDC \sim \Delta BEC$
Hence proved.
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