Denote by $M$ and $N$ the tangent points of the incircle of $ABCD$ with $AB$ and $CD$, respectively. It follows from $\angle BAD + \angle ADC > 180^{\circ}$ that $AB \parallel CD$ and $\angle MIN < 180^{\circ}$. Also, the equalities $IM = IN$, $\angle IMX = \angle INY$ and $IX = IY$ show that $\triangle IMX \cong \triangle INY$, implying $\angle IYN = \angle IXM$. If $X \in BM$ and $Y \in DN$ (or $X \in AM$ and $Y \in CN$) then the equality $\angle IYN = \angle IXM$ implies $AB \parallel CD$, a contradiction. Therefore $X \in XB$ and $Y \in NC$. It follows from $AXYD$ that
$$\angle AXI = \angle DYI = 180^{\circ} - \frac{\angle A}{2} - \frac{\angle D}{2},$$
giving $\angle AIX = \frac{\angle D}{2}$ and $\angle DIY = \frac{\angle A}{2}$. Therefore $\triangle AIX \sim \triangle IDY$, which implies that $AX \cdot DY = IY \cdot IX$. Analogously, $\triangle BIX \sim \triangle ICY$, i.e. $BX \cdot CY = IY \cdot IX$. Hence $AX \cdot DY = BX \cdot CY$.