Question

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

Answer 1

$ABCD$ is a trapezium with $AB\parallel CD$ and diagonals $AB$ and $CD$ intersecting at $O$.

$\Rightarrow$ In $△OAB$ and $△OCD$

$\Rightarrow$ $\angle AOB=\angle DOC$ [ Vertically opposite angles ]

$\Rightarrow$ $\angle ABO=\angle CDO$ [ Alternate angles ]

$\Rightarrow$ $\angle BAO=\angle OCD$ [ Alternate angles ]

$\therefore$ $△OAB\sim △OCD$ [ AAA similarity ]

We know that if triangles are similar, their corresponding sides are in proportion.

$\Rightarrow$ $\frac{OA}{OC}=\frac{OB}{OD}$ $\left[henceproved\right]$