Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that$ar\left(AOD\right)=ar\left(BOC\right)\dot{}$Prove that ABCD is a trapezium.

Solution: $area△ABD=area△AOD+area△AOB$ -(1)
$area△ABC=area△AOB+area△COB$ -(2)
subtracting equation 2 from equation 1
$area△ABD-area△ABC=0$
$area△ABD=area△ABC$
base is common of both trianglei.e.AB
so,AB||CD
This means ABCD is a trapezium