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In a trapezium ABCD, $AB || DC$ and its diagonals intersect at O. Prove that $\frac{OA}{OC} = \frac{OB}{OD}$.
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Proving $\triangle OAB \sim \triangle OCD$ (By AA similarity criterion) ($1\frac{1}{2}$ marks)
$\therefore \frac{OA}{OC} = \frac{OB}{OD}$ (corresponding sides of similar triangles are proportional) ($\frac{1}{2}$ mark)
$\therefore \frac{OA}{OC} = \frac{OB}{OD}$ (corresponding sides of similar triangles are proportional) ($\frac{1}{2}$ mark)