# Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a quadrilateral $\mathrm{ABCD}$ intersect each other at $\mathrm{P}$. Show that ar $(\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\operatorname{ar}(\mathrm{APD}) \times \operatorname{ar}(\mathrm{BPC})$.[Hint: From $\mathrm{A}$ and $\mathrm{C}$, draw perpendiculars to $\mathrm{BD}$.]

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Given:

Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a quadrilateral $\mathrm{ABCD}$ intersect each other at $\mathrm{P}$.

To do:

We have to show that ar $(\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\operatorname{ar}(\mathrm{APD}) \times \operatorname{ar}(\mathrm{BPC})$.

Solution:

Draw $AM$ perpendicular to $BD$ and $CN$ perpendicular to $BD$

$ar(\triangle ABP) = \frac{1}{2}\times BP \times AM$…………..(i)

$ar(\triangle APD) = \frac{1}{2}\times DP \times AM$…………..(ii)

Dividing (ii) by (i), we get,

$\frac{\operatorname{ar}(\triangle \mathrm{APD})}{\operatorname{ar}(\triangle \mathrm{ABP})}=\frac{\frac{1}{2} \times \mathrm{DP} \times \mathrm{AM}}{\frac{1}{2} \times \mathrm{BP} \times \mathrm{AM}}$

$\frac{ar(APD)}{ar(ABP)}= \frac{DP}{BP}$…….....(iii)

Similarly,

$\frac{ar(CDP)}{ar(BPC)} = \frac{DP}{BP}$……. (iv)

From (iii) and (iv), we get,

$\frac{ar(APD)}{ar(ABP)} = \frac{ar(CDP)}{ar(BPC)}$

$ar(APD) \times ar(BPC) = ar(ABP) \times ar (CDP)$

Hence proved.

Updated on 10-Oct-2022 13:46:30