JBMO-SL 2021 G4

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 $AZ$ .