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 $\frac{O A}{O C}=\frac{O B}{O D}$.
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)
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)