Question

Question 3
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{AO}{OC}=\frac{OB}{OD}$.

Answer 1

In $\Delta DOC$ and $\Delta BOA$,
$\angle CDO=\angle ABO$ [Alternate interior angles as AB || CD]
$\angle DCO=\angle BAO$ [Alternate interior angles as AB || CD]
$\angle DOC=\angle BOA$ [Vertically opposite angles]
$\therefore \Delta DOC\sim \Delta BOA$ [AAA similarity criterion]
$\therefore \frac{DO}{BO}=\frac{OC}{OA}$ [Corresponding sides are proportional]
$\Rightarrow \frac{OA}{OC}=\frac{OB}{OD}$