Diagonals $AC$ and $BD$ of trapezium $ABCD$ with $AB \parallel DC$ intersect each other at point $O$. Show that $\frac{OA}{OC} = \frac{OB}{OD}$.

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Question

Diagonals $AC$ and $BD$ of trapezium $ABCD$ with $AB \parallel DC$ intersect each other at point $O$. Show that $\frac{OA}{OC} = \frac{OB}{OD}$.

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Accepted Answer

Step 1: In $	riangle AOB$ and $	riangle COD$: $\angle OAB = \angle OCD$ (alternate angles, $AB \parallel DC$), $\angle OBA = \angle ODC$ (alternate angles). Therefore $	riangle AOB \sim 	riangle COD$ (AA similarity).
Step 2: $	herefore \frac{OA}{OC} = \frac{OB}{OD}$.

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