Question

ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, prove that $\frac{area\left(ABC\right)}{area\left(DBC\right)}=\frac{AO}{DO}$.

Answer 1

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