Question

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

Answer 1

Given :

For a quadrilateral $ABCD,$ diagonals $AC$ and $BD$ intersect at $O$

And $ar\left(△AOD\right)=ar\left(△BOC\right)$

To prove:

$ABCD$ is a trapezium.

Adding $ar\left(ODC\right)$ on both sides ,

$ar\left(△AOD\right)+ar\left(△ODC\right)=ar\left(△BOC\right)+ar\left(△ODC\right)$

$\Rightarrow ar\left(△ADC\right)=ar\left(△BDC\right)$

$\Rightarrow \frac{1}{2}\times DC\times AL=\frac{1}{2}\times DC\times BH$

$\Rightarrow AL=BH$

$\Rightarrow AB\parallel DC$

So, $ABCD$ is a trapezium.