Let ABCD be a convex quadrilateral with ∠B=∠D=90∘. 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=2AE. 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. JuniorBalkanshortlist2021geometryconcurrency