Question

In the figure, ABCD is a trapezium in which AB || DC. Prove that $ar(△AOD=ar(△BOC)$

Answer 1

In trapezium ABCD, diagonals AC and BD intersect each other at O.
$\therefore △ADBand△ACB$ are on he same base AB and between the same parallels
$\therefore ar(△ADB=ar(△ACD)$
Subtracting, ar $\left(△AOB\right)$ from both sides, $ar\left(△ADB\right)-ar\left(△AOB\right)=ar\left(△ACD\right)-ar\left(△AOB\right)\Rightarrow ar\left(△AOD\right)=ar\left(△BOC\right)$