Let $I_A, I_B$ be the incenters of $\triangle BCD, \triangle ACD$ respectively, and let $X_A, X_B$ be the feet of the $A$-angle bisector of $\triangle ACD$ and $B$-angle bisector of $\triangle BCD$ respectively. Since $AI_A$ and $BI_B$ concur, it follows that $A, B, I_A, I_B$ all lie on a single plane $\mathcal{P}$. Since $X_A\in AI_B$, then $X_A\in\mathcal{P}$. Similarly, $X_B\in\mathcal{P}$. If $\mathcal{P}$ intersects $CD$ more than once, then $CD\in \mathcal{P}$ so $A, B, C, D$ are coplanar. In this case the problem is actually false, so we ignore it. Otherwise, $\mathcal{P}$ intersects $CD$ at most once, so $X_A=X_B$. By angle bisector theorem, $$\dfrac{AC}{AD} = \dfrac{X_AC}{X_AD} = \dfrac{X_BC}{X_BD} = \dfrac{BC}{BD}$$$$\iff AC\cdot BD = AD\cdot BC$$Since this solution works if we consider any two vertices of $ABCD$ instead of just $A$ and $B$, then cyclically, $$AB\cdot CD = AC\cdot BD = AD\cdot BC$$as desired. $\Box$