Problem

Source: Iran Third Round MO 1998, Exam 1, P2

Tags: geometry, cyclic quadrilateral, geometry unsolved



Let $ABCD$ be a convex pentagon such that \[\angle DCB = \angle DEA = 90^\circ, \ \text{and} \ DC=DE.\] Let $F$ be a point on AB such that $AF:BF=AE:BC$. Show that \[\angle FEC= \angle BDC, \ \text{and} \ \angle FCE= \angle ADE.\]