Question

$\square ABCD$ is a trapezium in which AB$\parallel$DC and its diagonals intersect each other at the point O. Show that
$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$.

Answer 1

Given : $\square ABCD$ is a trapezium in which AB$\parallel$DC.

Construction :
Draw seg EO $\parallel$DC.


To show :$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$

Proof:
$$\begin{gathered} \text{In}△\mathrm{ADC}, \\ \mathrm{EO}\parallel \mathrm{DC} \\ \mathrm{Therefore},\mathrm{according}\mathrm{to}\mathrm{the}\text{b}\mathrm{asic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}...\left(1\right) \\ \mathrm{AB}\parallel \mathrm{DC}. \\ \mathrm{Hence},\mathrm{EO}\parallel \mathrm{AB}. \\ \text{In}△\mathrm{DAB}, \\ \mathrm{EO}\parallel \mathrm{AB}. \\ \mathrm{Therefore},\mathrm{according}\mathrm{to}\mathrm{the}\text{b}\mathrm{asic}\mathrm{proportionality}\mathrm{theorem}, \\ \frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{OB}}...\left(2\right) \\ \mathrm{By}\mathrm{invertendo}\mathrm{of}\left(2\right),\text{we get}: \\ \frac{\mathrm{EA}}{\mathrm{DE}}=\frac{\mathrm{OB}}{\mathrm{DO}}...\left(3\right) \\ \mathrm{From}\left(1\right)\&\left(3\right),\text{we get:} \\ \frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{DO}} \\ \Rightarrow \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$