Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Let $E$ be the point of intersection of $BC$ with $AD$ and let $M$ be the midpoint of $AE$. On the extension of $CD$, beyond the point $D$, we pick a point $Z$ such that $MZ = \frac{AE}{2}$. Let $U$ and $V$ be the projections of $A$ and $E$ respectively on $BZ$. The circumcircle of the triangle $DUV$ meets again $AE$ at the point $L$. If $I$ is the point of intersection of $BZ$ with $AE$, prove that the lines $BL$ and $CI$ intersect on the line $AZ$.