Question

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

Answer 1

Here, $ABCD$ is a quadrilateral where diagonals $AC$ and $BD$ intersect at $O$.

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

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

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

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

Now, $△ADC$ and $△BDC$ lie on the same base $DC$ and are equal in area and they lie between the lines $AB$ and $DC$

$\Rightarrow$ $AB\parallel DC$ ( Two triangles having the same base and equal areas lie between the same parallels )

In $ABCD,$

$\Rightarrow$ $AB\parallel DC$

So, one pair of opposite sides is parallel,

$\therefore$ $ABCD$ is trapezium.