Question

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

Answer 1

The given figure can be re-drawn as shown in the above figure where $AE$ and $DF$ are heights of $\Delta$ ABC and $\Delta$ BCD respectively.

Now $\Delta AOE$ and $\Delta FOD$ are similar.

As $\angle EOA=\angle FOD$ [Opposite angle], $\angle AEO=\angle DFO$ [right-angles] and $\angle EAO=\angle FDO$ [$AE\parallel FD$ and $AD$ intersector].

So $\frac{AE}{DF}=\frac{AO}{OD}$......(1).

Now

$\frac{ar\left(ABC\right)}{ar\left(DBC\right)}=\frac{\frac{1}{2}\times BC\times AE}{\frac{1}{2}\times BC\times FD}=\frac{AE}{FD}=\frac{AO}{OD}$ [Using (1)].