Let $P, Q, R$ be the feet of the perpendiculars from D to the sides $AB$, $BC$, $CA$ respectively. Then $$\angle{QPR} = \angle{QPD} + \angle{DPR} = \angle{QBD}+ \angle{RAD} = \angle{ADC} - \angle{ABC} = 60^o$$Similarly $\angle{PQR} = 60^o$ and $\angle{PRQ} = 60^o$. Thus triangle $PQR$ is equilateral. Let $PQ=QR=RS=k$. Then applying the sine theorem in the circumcircle of inscribed quadrilateral $CQDR$ we have: $$ QR = CD \cdot sin(\angle{C})$$Thus $ CD = \frac{QR}{sin(\angle{C})} = \frac{k}{sin(\angle{C})}$ So: $ AB \cdot CD = 2R \cdot sin(\angle{C}) \frac{k}{sin(\angle{C})} = 2 R \cdot k$ where R is the radius of the circumcircle of $ABC$ Similarly we may obtain: $$ BC \cdot AD = 2R \cdot sin(\angle{A}) \frac{k}{sin(\angle{A})} = 2 R \cdot k$$and $$ CA \cdot CD = 2R \cdot sin(\angle{B}) \frac{k}{sin(\angle{B})} = 2 R \cdot k$$Thus we can conclude: $$AB \cdot CD = BC \cdot AD = CA \cdot BD$$