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Question

Diagonals $$AC$$ and $$BD$$ of a quadrilateral $$ABCD$$ intersect each other at $$P$$ show that 
$$ar. (\triangle APB)\times ar. (\triangle CPD)=ar. (\triangle APD)\times ar. (\triangle BPC)$$
1869552_7ad2bb3e8fac4a0cb65c25f72efff9c6.png


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

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