In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta PDC \sim \Delta BEC$.
(N/A) In $\Delta PDC$ and $\Delta BEC$:
$\angle PDC = \angle BEC = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle PCD = \angle BCE$ (Common angle)
Therefore,by using the $AA$ similarity criterion,
$\Delta PDC \sim \Delta BEC$.
$\angle PDC = \angle BEC = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle PCD = \angle BCE$ (Common angle)
Therefore,by using the $AA$ similarity criterion,
$\Delta PDC \sim \Delta BEC$.
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