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

Solution

## Data: Diagonals $$AC$$ and $$BD$$ of a quadrilateral $$ABCD$$ intersect each other at $$P$$.To Prove: $$ar.(\triangle APB)\times ar.(\triangle CPD)=ar.(\triangle APD)\times ar.(\triangle BPC)$$Construction: Draw $$AM\bot DB, CN\bot DB$$Proof: $$ar.(\triangle APB)\times ar.(\triangle CPD)=$$$$=\left(\dfrac 12\times PB\times AM\right) \times \left(\dfrac 12\times PD\times CN\right)$$$$=\dfrac 14 \times PB \times AM\times PD\times CN....(i)$$$$ar.(\triangle APD)\times ar.(\triangle BPC)=$$$$=\left(\dfrac 12\times PD\times AM\right)\times \left(\dfrac 12\times PB\times CN \right)$$$$=\dfrac 14 \times PD\times AM\times PB\times CN....(ii)$$From (i) and (ii)$$ar.(\triangle APB)\times ar.(\triangle CPD)=ar.(\triangle APD)\times ar.(\triangle BPC)$$Maths

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