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(Please refer to the pics which i have attached with the answer)

Given ; Diagonal AC and BD of a qua intersect each other at O
We need to PT;  ar (AOB) x ar(COD) = ar(AOD)x ar(BOC)
Construction: From A and C draw a perpendiculars AE and CF respectively to BD (fig 2 is attached to the answer)
Area of a triangle = base x altitude / 2
Area of AOB and COD = (OB) x \frac{OP}{AE} / 2 x  \frac{DO}{CF} / 2

= 1/4 (OB)(AE)(DO)(CF) --------------------------- (1)

ar((AOD) x ar(BOC) = 1/4 (OB)(AE)(DO)(CF) --------------------- (2)

From (1) and (2)

ar(AOB) x ar(COD) = ar(AOD) x ar(BOC)

1 5 1
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