The diagonals of a trapezium

Draw MN||AB||DC

Let Diagonals AC and BD intersect at ‘O’

In $\Delta ACD,MO\left|\right|CD$

$\therefore \frac{AO}{OC}=\frac{AM}{MD}\to \left(1\right)$

In $\Delta ABD,MO\left|\right|AB$

$\therefore \frac{AM}{MD}=\frac{OB}{OD}\to \left(2\right)$

From (1)&(2),

$\frac{AO}{OC}=\frac{BO}{OD}$

Hence proved