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

Given:

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

To do:

We have to show that $ar(\triangle APB) \times ar(\triangle CPD) = ar(\triangle APD) \times ar(\triangle BPC)$.

Solution:

Draw $AL$ and $CN$ perpendiculars on $BD$.

$\operatorname{ar}(\triangle \mathrm{APD}) \times \operatorname{ar}(\Delta \mathrm{BPC})=(\frac{1}{2} \times \mathrm{AL} \times \mathrm{DP}) \times(\frac{1}{2} \times \mathrm{CN} \times \mathrm{BP})$

$=(\frac{1}{2} \times \mathrm{BP} \times \mathrm{AL}) \times(\frac{1}{2} \times \mathrm{DP} \times \mathrm{CN})$

$=\operatorname{ar}(\Delta \mathrm{APB}) \times ar(\Delta \mathrm{CPD})$

Hence proved.

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Updated on: 10-Oct-2022

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