Question

diagonals AC & BD of a quad.ABCD intersect each other at o. then show that,Ar.aob*Ar.cod=Ar.boc*Ar.doa

Answer 1

(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)
.
Proof:
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)

Similarly 
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)