aops, equilateral triangle, https://artofproblemsolving.com/community/c4t48f4h2615750_a_geometry_problem_involving_triangles

 D, E are points onto the sides BC, BA of the equilateral triangle ABC so that BE=CD, F is a point onto (BA so that AF=BE and K is the midpoint of DE. Find <CKF.

Proof
Take G reflection of C about K; triangle BEG is equilateral. See that triangles CAF and FEG are congruent, s.a.s., thus GF=CF and FK is perpendicular bisector of CG, i.e. <CKF=90.