# Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

Question. Diagonals $A C$ and $B D$ of a quadrilateral $A B C D$ intersect each other at $P .$ Show that $\operatorname{ar}(A P B) \times \operatorname{ar}(C P D)=\operatorname{ar}(A P D) \times \operatorname{ar}(B P C)$

Solution:

Let us draw $A M \perp B D$ and $C N \perp B D$

Area of a triangle $=\frac{1}{2} \times$ Base $\times$ Altitude

$\operatorname{ar}(\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\left[\frac{1}{2} \times \mathrm{BP} \times \mathrm{AM}\right] \times\left[\frac{1}{2} \times \mathrm{PD} \times \mathrm{CN}\right]$

$=\frac{1}{4} \times \mathrm{BP} \times \mathrm{AM} \times \mathrm{PD} \times \mathrm{CN}$

$\operatorname{ar}(\mathrm{APD}) \times \operatorname{ar}(\mathrm{BPC})=\left[\frac{1}{2} \times \mathrm{PD} \times \mathrm{AM}\right] \times\left[\frac{1}{2} \times \mathrm{CN} \times \mathrm{BP}\right]$

$=\frac{1}{4} \times \mathrm{PD} \times \mathrm{AM} \times \mathrm{CN} \times \mathrm{BP}$

$=\frac{1}{4} \times \mathrm{BP} \times \mathrm{AM} \times \mathrm{PD} \times \mathrm{CN}$

$\therefore \operatorname{ar}(\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\operatorname{ar}(\mathrm{APD}) \times \operatorname{ar}(\mathrm{BPC})$