Question

In trapezium ABCD, side AB $\parallel$ side DC, diagonals AC and BD intersect in point O. If AB=$20$, DC=$6$, OB=$15$, then find OD.

Answer 1

From the picture we have $\Delta$ COD is similar to $\Delta$ AOB.

Since, $\angle COD=\angle AOB$ (opposite angle) , $\angle CDO=\angle OBA$ (transversal angle) and $\angle DCO=\angle OAB$ (transversal angle).

Then

$\frac{CD}{AB}=\frac{OD}{OA}$

or, $\frac{6}{20}=\frac{OD}{15}$

or, $OD=4.5$