Question

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

Answer 1

$\Delta DAC\&\Delta DBC$ lies on the same base DC and between same parallel line, line $AD\parallel DC$

$\begin{array}{c}\therefore ar\left(\Delta DAC\right)=ar\left(\Delta DBC\right)\\ \Rightarrow ar\left(\Delta DAC\right)-ar\left(\Delta DOC\right)=ar\left(\Delta DBC\right)-ar\left(\Delta DOC\right)\\ \Rightarrow ar\left(\Delta AOD\right)=ar\left(\Delta BOC\right)\end{array}$