In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (\(\triangle\) ABC)/area (\(\triangle\) DBC) = AO/DO.

Given, ABC and DBC are two triangles on the same base BC. ADintersects BC at O.

We have to prove: Area (\(\triangle\) ABC)/Area (\(\triangle\) DBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.



We know that area of a triangle = \(\frac{1}{2} \) × Base × Height
\({{ar(\triangle ABC) } \over {ar(\triangle DEF}} = {{{{1} \over {2}}BC(AP)}\over {{{1} \over {2}}BC(DM)}} ={{AP} \over {DM}}\)

In \(\triangle\) APO and \(\triangle\) DMO,
\(\angle\) APO = \(\angle\) DMO (Each 90°)
\(\angle\) AOP = \(\angle\) DOM
(Vertically opposite angles)
\(\triangle\) APO ~ \(\triangle\) DMO (AA similarity criterion)
\(\frac{AP}{DM} = \frac{AO}{DO} \)
\(\frac{Area (\triangle ABC)}{Area ( \triangle DBC)} = \frac{AO}{DO} \) .