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

Given:

Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \| DC$ intersect each other at the point $O$.

To do:

We have to show that $\frac{OA}{OC}\ =\ \frac{OB}{OD}$.

Solution:

In $\vartriangle AOB$ and $\vartriangle COD$,

$\angle AOB = \angle COD$ (Vertically opposite angles are equal)

$\angle OAB = \angle OCD$    ($AB \| DC$, Alternate angles)

Therefore,

$\vartriangle AOB ∼ \vartriangle COD$

This implies,

$\frac{OA}{OC} = \frac{OB}{OD}$   (Corresponding sides are proportional)

Hence proved.

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Updated on: 10-Oct-2022

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