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Diagonals of a trapezium ABCD with $AB \| DC$ intersect each other at the point O. If $AB = 2 CD$, find the ratio of the areas of triangles AOB and COD.
Given:
Diagonals of a trapezium ABCD with $AB \| DC$ intersect each other at the point O.
$AB = 2 CD$
To do:
We have to find the ratio of the areas of triangles AOB and COD.
Solution:
In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{COD}$,
$\angle \mathrm{AOB} =\angle \mathrm{COD}$ (vertically opposite angles)
$\angle 1=\angle 2$ (Alternate angles)
Therefore, by AA similarity,
$\Delta \mathrm{AOB} \sim \Delta \mathrm{COD}$
This implies,
$\frac{\operatorname{ar} \Delta \mathrm{AOB}}{\operatorname{ar} \Delta \mathrm{COD}}=\frac{\mathrm{AB}^{2}}{\mathrm{CD}^{2}}$
$=\frac{(2 \mathrm{CD})^{2}}{\mathrm{CD}^{2}}$
$=\frac{4 \mathrm{CD}^{2}}{\mathrm{CD}^{2}}$
$=\frac{4}{1}$
The ratio of the areas of triangles AOB and COD is $4:1$.
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