Diagonals of trapezium ABCD with AB
Question.

Diagonals of trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at the point $\mathrm{O}$. If $\mathrm{AB}=2 \mathrm{CD}$, find the ratio of the areas of triangles $\mathrm{AOB}$ and COD.

Solution:

In $\Delta \mathrm{AOB}$ and $\Delta \mathrm{COD}$,

$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate interior angles)

$\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate interior angles)

$\therefore$ By AA, similarity

$\Delta \mathrm{AOB} \sim \Delta \mathrm{COD}$

So, $\frac{\operatorname{ar} . \Delta \mathrm{AOB}}{\operatorname{ar} . \Delta \mathrm{COD}}=\left(\frac{\mathrm{AB}}{\mathrm{CD}}\right)^{2}$

$=\left(\frac{2}{1}\right)^{2} \quad\{\because \mathrm{AB}=2 \mathrm{CD}\}$

$=4: 1$ [/question]
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