Question 76
A quadrilateral $ABCD$ with $\angle BAD + \angle ADC \gt 180^{\circ}$ is circumscribed around a circle of center $I$. A line through $I$ meets $AB$ and $CD$ at points $X$ and $Y$, respectively. Prove that if $IX = IY$ then $AX \cdot DY = BX \cdot CY$.