Question

In given figure $ABC$ and $DBC$ are two triangles on the same base $BC.$ If $AD$ intersects $BC$ at $O,$ show that:

$\frac{area\left(ABC\right)}{area\left(DBC\right)}=\frac{AO}{DO}$.

Answer 1

Construction:
Draw $AM$ perpendicular to $BC$ and $DN$ perpendicular to $BC.$

Now,
In $△AMO$ and $△DNO$

$\angle AOM=\angle DON$.......opp.angles

$\angle AMO=\angle DNO={90}^o$

$\therefore$ By AA Criterion of Similarity, $△AMO$ $\sim$ $△DNO$

$\therefore$ $\frac{AM}{DN}=\frac{AO}{DO}$ (Corresponding Sides of Similar Triangles) $......\left(1\right)$

Now,

$\frac{A\left(△ABC\right)}{A\left(△BDC\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times BC\times DN}$

$\therefore$ $\frac{A\left(△ABC\right)}{A\left(△BDC\right)}$ $=\frac{AO}{DO}$ (From 1) $\left[henceproved\right]$