# Q: 6    Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show   that  $ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).$            [Hint: From A and C, draw perpendiculars to BD.]

M manish painkra

Given that,
A quadrilateral ABCD such that it's diagonal AC and BD intersect at P. Draw $AM \perp BD$ and $CN \perp BD$

Now, ar($\Delta$APB) = $\frac{1}{2}\times BP\times AM$ and,
$ar(\Delta CDP)=\frac{1}{2}\times DP\times CN$
Therefore, ar($\Delta$APB) $\times$ ar($\Delta$CDP)=
$\\=(\frac{1}{2}\times BP\times AM).(\frac{1}{2}\times DP \times CN)\\ =\frac{1}{4}\times BP \times DP\times AM\times CN$....................(i)

Similarly, ar($\Delta$ APD) $\times$ ar ($\Delta$ BPC) =
$=\frac{1}{4}\times BP \times DP\times AM\times CN$................(ii)

From eq (i) and eq (ii), we get

$ar (APB) \times ar (CPD) = ar (APD) \times ar (BPC).$
Hence proved

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