Question

ABCD is a trapezium having AB $\parallel$ DC. Prove that O, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that $\frac{ar\left(△OCD\right)}{ar\left(△OAB\right)}=\frac{1}{9}$ , if AB = 3 CD.

Answer 1

In $△OCD$ and $△OAB$

$\angle CDO=\angle OBA$ (Alternate angles)

$\angle DCO=\angle OAB$ (Alternate angles)

$△OCD\sim △OAB$ (AA similarity criterion)

=> $\frac{OD}{OB}=\frac{OC}{OA}$ (CPST)

Hence, proved that point $O$ divides the two diagnols in the same ratio.

$AB=3CD$ (Given)

$\frac{AB}{CD}=\frac{3}{1}$ or $\frac{CD}{AB}=\frac{1}{3}$

$\frac{ar.\left(△OCD\right)}{ar.\left(△OAB\right)}={\left(\frac{CD}{AB}\right)}^2$ (Using theorem of areas of similar ${△}^{\prime }s$)

$\frac{ar.\left(△OCD\right)}{ar.\left(△OAB\right)}={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$

Hence, proved.