Let $AB$ be a line segment with midpoint $I$. A circle, centred at $I$ has diameter less than the length of the segment. A triangle $ABC$ is tangent to the circle on sides $AC$ and $BC$. On $AC$ a point $X$ is given, and on $BC$ a point $Y$ is given such that $XY$ is also tangent to the circle (in particular $X$ lies between the point of tangency of the circle with $AC$ and $C$, and similarly $Y$ lies between the point of tangency of the circle with $BC$ and $C$. Prove that $AX \cdot BY = AI \cdot BI$.