In the figure,$ABC$ and $DBC$ are two triangles on the same base $BC$. If $AD$ intersects $BC$ at $O$,show that $\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(DBC)} = \frac{AO}{DO}$.

(N/A) Let us draw two perpendiculars $AP$ and $DM$ on line $BC$.
We know that the area of a triangle $= \frac{1}{2} \times \text{Base} \times \text{Height}$.
Therefore,$\frac{\operatorname{ar}(\Delta ABC)}{\operatorname{ar}(\Delta DBC)} = \frac{\frac{1}{2} \times BC \times AP}{\frac{1}{2} \times BC \times DM} = \frac{AP}{DM}$.
In $\triangle APO$ and $\triangle DMO$:
$\angle APO = \angle DMO = 90^{\circ}$ (By construction).
$\angle AOP = \angle DOM$ (Vertically opposite angles).
Therefore,$\triangle APO \sim \triangle DMO$ (By $AA$ similarity criterion).
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AP}{DM} = \frac{AO}{DO}$.
Substituting this into our area ratio equation:
$\frac{\operatorname{ar}(\Delta ABC)}{\operatorname{ar}(\Delta DBC)} = \frac{AO}{DO}$.