Question

In given figure $ABCD$ is a parallelogram. The diagonals $AC$ and $BD$ intersect each other at ${}^{\prime }{O}^{\prime }$. Prove that $area\left(△AOD\right)=area\left(△BOC\right).$

Answer 1

Consider $△AOD$ and $△BOC$,

$\angle AOD=\angle BOC$ [Vertically opposite angles]

$DO=OB$ and $AO=OC$ [diagonals in parallelogram $AC$ and $BD$ intersect and bisect at $O$]

$\therefore △AOD\cong △BOC$ [By $SAS$ congruence]

$\Rightarrow area\left(△AOD\right)=area\left(△BOC\right)$ $\left[henceproved\right]$