Let's phrase the converse of the Angle Bisector Theorem as follows:

In $\triangle ABC$, let $D$ be a point on side $\overline{BC}$ such that $\dfrac{BD}{DC}=\dfrac{AB}{AC}$.  Prove that $\overline{AD}$ bisects $\angle BAC$.

To prove this result, you may consider the "phantom point" method.  Construct the angle bisector $\overline{AD'}$ of $\angle BAC$, where $D'$ is on $\overline{BC}$.  Then try to prove that $D'$ and $D$ are in fact the same point.