For $\Delta ABC$, let $AD$ is its internal angle bisector, then we know that $$\frac{AB}{AC}=\frac{BD}{CD}$$ and I know that it can be proved by geometry by drawing a line through $C$ parallel to $AD$ and extend $AB$ so that it cuts new line and then use various angles and concept of similar triangle to prove our result. But I was wondering if we can prove it with the help of vectors as well?
If we take position vectors of $A,B$ and $C$ as $\overrightarrow{0},\overrightarrow{b}$ and $\overrightarrow{c}$ respectively , then $\overrightarrow{AD}$ can be taken as $\lambda(\hat{b}+\hat{c})$. Could someone help me to proceed after that or provide some alternate approach through vectors?
As mentioned, $\vec{AD} \neq \lambda (\hat b+\hat c)$ in general. I think there's still a decent looking proof using dot product.
As usual, it helps if you sketch a diagram.
Let's suppose $\angle BAD=\angle CAD=\theta$, where $\theta$ is acute, and $\lvert \vec{BD} \rvert:\lvert \vec{CD} \rvert=r:s$. Using the section formula, $$\vec{AD} = \frac{r}{r+s}\vec{AC}+\frac{s}{r+s}\vec{AB}.$$
Then by dot product,
$$\vec{AD}\cdot\vec{AC} =\lvert\vec{AD}\rvert\lvert\vec{AC}\rvert \cos\theta =\frac{r}{r+s}\vert\vec{AC}\vert^2+\frac{s}{r+s}\vec{AB}\cdot\vec{AC}$$
$$\implies\lvert\vec{AD}\rvert \cos\theta =\frac{r}{r+s}\vert\vec{AC}\vert+\frac{s}{r+s}\vert\vec{AB}\vert\cos 2\theta. \tag{1}\label{1}$$
Similarly, by considering $\vec{AD}\cdot\vec{AB}$, we have $$\lvert\vec{AD}\rvert \cos\theta =\frac{r}{r+s}\vert\vec{AC}\vert\cos 2\theta+\frac{s}{r+s}\vert\vec{AB}\vert. \tag{2}\label{2}$$
Combining \ref{1} and \ref{2} gives
\begin{align} \frac{r}{r+s}\vert\vec{AC}\vert+\frac{s}{r+s}\vert\vec{AB}\vert\cos 2\theta &=\frac{r}{r+s}\vert\vec{AC}\vert\cos 2\theta+\frac{s}{r+s}\vert\vec{AB}\vert\\ \frac{r}{r+s}\vert\vec{AC}\vert(1-\cos 2\theta) &= \frac{s}{r+s}\vert\vec{AB}\vert(1-\cos 2\theta)\\ \frac{\vert\vec{AB}\vert}{\vert\vec{AC}\vert}&=\frac{r}{s}\\ &=\frac{\vert\vec{BD}\vert}{\vert\vec{CD}\vert}. \end{align}