Question

In $\Delta ABC$, if $AD$ is the bisector of $\angle A$, prove that $\frac{Area\left(\Delta ABD\right)}{Area\left(\Delta ACD\right)}=\frac{AB}{AC}$.

Answer 1

In $\Delta ABC,AD$ is the bisector of $\angle A$.
$\therefore \frac{AB}{AC}=\frac{BD}{DC}.....\left(i\right)$
[By internal bisector theorem]
From $A$ draw $AL\perp BC$
$\therefore \frac{Area\left(\Delta ABD\right)}{Area\left(\Delta ACD\right)}=\frac{\frac{1}{2}BD\cdot AL}{\frac{1}{2}DC\cdot AL}=\frac{BD}{DC}=\frac{AB}{AC}\left[From\right(i\left)\right]$
Hence proved.