Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that $ar(APB) \times ar(CPD) = ar(APD) \times ar(BPC)$.

(N/A) We have a quadrilateral $ABCD$ such that its diagonals $AC$ and $BD$ intersect at $P$. Let us draw $AM \perp BD$ and $CN \perp BD$.
$ar(\Delta APB) = \frac{1}{2} \times BP \times AM$
$ar(\Delta CPD) = \frac{1}{2} \times DP \times CN$
$ar(\Delta APB) \times ar(\Delta CPD) = (\frac{1}{2} \times BP \times AM) \times (\frac{1}{2} \times DP \times CN)$
$= \frac{1}{4} \times BP \times DP \times AM \times CN$ ... $(1)$
Similarly,
$ar(\Delta APD) = \frac{1}{2} \times DP \times AM$
$ar(\Delta BPC) = \frac{1}{2} \times BP \times CN$
$ar(\Delta APD) \times ar(\Delta BPC) = (\frac{1}{2} \times DP \times AM) \times (\frac{1}{2} \times BP \times CN)$
$= \frac{1}{4} \times BP \times DP \times AM \times CN$ ... $(2)$
From $(1)$ and $(2)$,we get
$ar(\Delta APB) \times ar(\Delta CPD) = ar(\Delta APD) \times ar(\Delta BPC)$