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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