In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta AEP \sim \Delta CDP$.
(N/A) In $\Delta AEP$ and $\Delta CDP$:
$\angle AEP = \angle CDP = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle APE = \angle CPD$ (Vertically opposite angles)
Therefore,by the $AA$ similarity criterion,
$\Delta AEP \sim \Delta CDP$.
$\angle AEP = \angle CDP = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle APE = \angle CPD$ (Vertically opposite angles)
Therefore,by the $AA$ similarity criterion,
$\Delta AEP \sim \Delta CDP$.
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