Question

Diagonals of a trapezium $ABCD$ with $AB||DC$, intersect each other at the point $O$. If $AB=2CD$, find the ratio of the areas of triangles $AOB$ and $COD.$

Answer 1

Construction:
Draw $OD$ perpendicular to $DC$ and $OP$ perpendicular to $AB.$
In $△AOB$ and $△DOC$
$\angle CDO=\angle OBA$ (Alternate Angles)
$\angle DCO=\angle OAB$ (Alternate Angles)
$\angle DOC=\angle AOB$ (Vertically opposite angles)
$\therefore$ By AAA Criterion of Similarity $△AOB$ $\sim$ $△DOC$
$\therefore$ $\frac{AB}{DC}=\frac{QO}{PO}.....\left(1\right)$
$\therefore$ $\frac{A\left(△AOB\right)}{A\left(△DOC\right)}$ $=\frac{(2DC{)}^2}{(DC{)}^2}=\frac{4}{1}$