Diagonals AC and BD of a trapezium ABCD with AB
Question.

Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{O A}{O C}=\frac{O B}{O D}$.

Solution:

In figure, $\mathrm{AB} \| \mathrm{DC}$

$\Rightarrow \angle 1=\angle 3, \angle 2=\angle 4$

(Alternate interior angles)

Also $\angle \mathrm{DOC}=\angle \mathrm{BOA}$

(Vertically opposite angles)

$\Rightarrow \Delta \mathrm{OCD} \sim \Delta \mathrm{OAB} \quad \Rightarrow \frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{OD}}{\mathrm{OB}}$

(Ratios of the corresponding sides of the similar triangle)

$\Rightarrow \frac{O A}{O C}=\frac{O B}{O D}$ (Taking reciprocals)
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