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