aops, 'A great perpendicular problem', https://artofproblemsolving.com/community/c6t48f6h2611905_a_great_perpendicular_problem

 Reworded text:
Let CBVE be a cyclic kite, CV being a diameter of its circumcircle. Take F, G onto VE, BE so that FG||BV, call K the midpoint of BG. Prove that KF_|_CK.

Proof:
FG=FB; construct the parallelogram GCBL, we need to prove FL=FC. We have GL=BC=CE, FG||BV, GL||BC, so FG_|_GL. Triangles FGL and FEC are congruent, thus FL=FC, so the median FK is perpendicular bisector of CL as well, done.