In the figure,$D$ is a point on side $BC$ of $\Delta ABC$ such that $\frac{BD}{CD} = \frac{AB}{AC}$. Prove that $AD$ is the bisector of $\angle BAC$.

(N/A) Let us extend $BA$ to $P$ such that $AP = AC$. Join $PC$.
It is given that,
$\frac{BD}{CD} = \frac{AB}{AC}$
Since $AP = AC$,we can substitute $AP$ for $AC$:
$\frac{BD}{CD} = \frac{AB}{AP}$
By using the converse of the Basic Proportionality Theorem $(BPT)$ in $\Delta BPC$,we obtain:
$AD \parallel PC$
Therefore,$\angle BAD = \angle APC$ (Corresponding angles) $\dots(1)$
And,$\angle DAC = \angle ACP$ (Alternate interior angles) $\dots(2)$
By construction,we have $AP = AC$,which implies that in $\Delta APC$,the angles opposite to equal sides are equal:
$\angle APC = \angle ACP \dots(3)$
From equations $(1), (2),$ and $(3)$,we get:
$\angle BAD = \angle DAC$
Thus,$AD$ is the bisector of $\angle BAC$.