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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} $.]
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.
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