In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta AEP \sim \Delta ADB$.
(N/A) In $\triangle AEP$ and $\triangle ADB$:
$1. \angle AEP = \angle ADB = 90^{\circ}$ (Given that $AD$ and $CE$ are altitudes).
$2. \angle PAE = \angle DAB$ (Common angle).
Therefore,by the $AA$ (Angle-Angle) similarity criterion,we have:
$\triangle AEP \sim \triangle ADB$.
$1. \angle AEP = \angle ADB = 90^{\circ}$ (Given that $AD$ and $CE$ are altitudes).
$2. \angle PAE = \angle DAB$ (Common angle).
Therefore,by the $AA$ (Angle-Angle) similarity criterion,we have:
$\triangle AEP \sim \triangle ADB$.
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