Let us construct the equilateral triangle $BCE$ in the half-plane determined by line $BC$ and point $D$.
Then $\frac{AB}{AD} = \frac{BC}{CD} = \frac{CE}{CD}$ and $\angle DAB = \angle DCE$, so $\Delta DAB \sim \Delta DCE$.
Therefore $\frac{AD}{DC} = \frac{DB}{DE}$ and $\angle ADB = \angle CDE$, hence $\angle ADC = \angle BDE$.
It follows that $\Delta ADC \sim \Delta BDE$, so $\frac{AD}{BD} = \frac{AC}{BE} = \frac{AC}{BC}$, which leads to $AC \cdot BD = BC \cdot AD = AB \cdot CD$.