Write triangle angle bisector as a linear combination of side vectors

Question

Let $AO$ be the bisector of $\angle A$ in a triangle $ABC$. How to write the vector $\overrightarrow{AO}$ as a linear combination of $\overrightarrow{AB}$ and $\overrightarrow{AC}$?

Answer 1

The center $I$ of the inscribed circle of $\triangle ABC$ is expressed as a linear combination of the vertices as \begin{align} I&=\frac{aA+bB+cC}{a+b+c} \tag{1}\label{1} , \end{align}
where

\begin{align} a&=\|\overrightarrow{BC}\|=\|\overrightarrow{AC}-\overrightarrow{AB}\| ,\\ b&=\|\overrightarrow{AC}\|,\ c=\|\overrightarrow{AB}\| , \end{align}

So, if we move the origin to the point $A$, then from \eqref{1} the bisector of $\angle A$ is

\begin{align} \overrightarrow{AI} &= \frac{ \|\overrightarrow{AC}\|\cdot\overrightarrow{AB} +\|\overrightarrow{AB}\|\cdot\overrightarrow{AC} }{ \|\overrightarrow{AC}-\overrightarrow{AB}\| +\|\overrightarrow{AC}\|+\|\overrightarrow{AB}\| } \tag{2}\label{2} . \end{align}

Edit

As @John Hughes pointed out, since $\overrightarrow{AO}$ is expected to be just the direction vector, the expression \eqref{2} can be simplified:

\begin{align} \overrightarrow{AO} &= \|\overrightarrow{AC}\|\cdot\overrightarrow{AB} +\|\overrightarrow{AB}\|\cdot\overrightarrow{AC} . \end{align}

Answer 2

Hint: You might want to let $$ v = \frac{1}{\|\overrightarrow{AB}\|} \overrightarrow{AB} $$ and work from there.