Question

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

Answer 1

Given, ABC and DBC are triangles on the same base BC. AD intersects BC at O.
To Prove that $\frac{Areaof\Delta ABC}{Areaof\Delta DBC}=\frac{AO}{DO}$.
Construction: Let us draw two perpendiculars AP and DM on the line BC.
Proof
We know that area of a triangle =$\frac{1}{2}\times Base\times Height$
$\therefore \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta DBC\right)}=\frac{\frac{1}{2}BC\times AP}{\frac{1}{2}BC\times DM}=\frac{AP}{DM}$..........(i)
In $\Delta APOand\Delta DMO$,
$\angle APO=\angle DMO$ (Each angle equal to ${90}^{\circ }$)
$\angle AOP=\angle DOM$ (Vertically opposite angles)
$\therefore \Delta APO\sim \Delta DMO$ (By AAA similarity criterion)
$\therefore \frac{AP}{DM}=\frac{AO}{DO}$
$\Rightarrow \frac{area\left(\Delta ABC\right)}{area\left(\Delta DBC\right)}=\frac{AO}{DO}$ [from (i)]
Therefore, $\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta DBC\right)}=\frac{AO}{DO}$.Hence proved.