Let $M$ be the midpoint of $BC$ and $E$ the reflection of $D$ around $M$. Then $E$ also lies on the line $AM$ and moreover $\angle CEB = \angle BDC = 180^\circ - \angle BAC$, thus $ABCE$ are concyclic. This implies that $\triangle BEM \sim \triangle ACM$ and therefore $\frac{AC}{BE}=\frac{AM}{BM}$. Analogously we have $\frac{AB}{CE}=\frac{AM}{CM}$. But it is clear that $BM=CM$, $BE=CD$ and $CE=BD$. Thus $\frac{AC}{CD}=\frac{AB}{BD}$, as desired.

Basically, the point $D$ is the intersection of the circumcircle $BHC$ (where $H$ is the orthocenter) with the median. It is well-known that this belongs to the Apollonius circle of A.