In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect at $O$. $\overrightarrow{AO}$ intersects $\overline{BC}$ at $P$. Prove that $\frac{AB}{AC} = \frac{BP}{PC}$.

(N/A) $1$. In $\Delta ABC$,let $BO$ be the bisector of $\angle B$ and $CO$ be the bisector of $\angle C$. Since $O$ is the intersection of these bisectors,$AO$ is the bisector of $\angle A$ by the property of the incenter of a triangle.
$2$. According to the Angle Bisector Theorem,in $\Delta ABC$,if $AP$ is the bisector of $\angle A$ intersecting $BC$ at $P$,then the ratio of the sides forming the angle is equal to the ratio of the segments of the opposite side.
$3$. Therefore,$\frac{AB}{AC} = \frac{BP}{PC}$.
$4$. This confirms the Angle Bisector Theorem for $\Delta ABC$ where $AP$ is the internal angle bisector of $\angle A$.