Question

In the figure, altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta PDC~\Delta BEC$

Answer 1

To prove that $△PDC~△BEC$

Given that altitudes $AD$ and $CE$ of$△ABC$ intersect each other at the point $P$

Consider $△PDC$ and $△BEC$

$\angle CEB=\angle CDP$ (Each angle equal to $90°$)

$\angle BCE=\angle PCD$ (Common angle)

$\therefore$By AA similarity criteria $△PDC~△BEC$

Hence, it is proved that $△PDC~△BEC$.