In the given figure, ABC and ADBC are on the same base BC. If AD intersects BC at O, prove that ar( ABC)ar( DBC) =…

Draw $AL \perp BC$ and $DM \perp BC\\$In $\Delta AOL$ and $$\begin{aligned}& \Delta DOM, \\ & \angle AOL = \angle DOM\end{aligned}$$ (vertically opposite angles)
$\angle ALO = \angle DMO$ (each $90^\circ$)
$\Delta AOL \sim \Delta DOM$ (AA similarity)
$\Rightarrow \frac{AL}{DM} = \frac{AO}{DO}$
dots (i)
$$\begin{aligned}& \frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DBC)} = \frac{\frac{1}{2} \times BC \times AL}{\frac{1}{2} \times BC \times DM} \\ & = \frac{AL}{DM} = \frac{AO}{DO}\end{aligned}$$ [using (i)]