Question

In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that $area\left(△AOD\right)=area\left(△BOC\right)$.

Answer 1

Given ABCD is a trapezium the diagonal AC and BD with $AB\parallel CD$ intersect each other at O.

$AB\parallel CD$

Then $\Delta ADB$ and $\Delta ACB$ lie on same base AB and between two parallel line AB and CD

$\therefore area\Delta ADB=area\Delta ACB$

Subtract the area of triangle AOB on both sides we get

$\Rightarrow area\Delta ADB-area\Delta AOB=area\Delta ACB-area\Delta AOB$

As per figure

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