The points A′ and C′ are chosen on the diagonal BD of a parallelogram ABCD so that AA′ || CC′. The point K lies on the segment A′C, and the line AK meets CC′ at L. A line parallel to BC is drawn through K, and a line parallel to BD is drawn through C; these meet at M. Prove that D, M, L are collinear.
Let M′ and M′′ be the intersections of DM with MK and MC, respectively. Since M′K || DA, we have LM′/LD = LK/LA. Since CK || C′A, we have LK/LA = LC/LC′. Finally, since CM || C′D, we have LM′′/LD = LC/LC′. Therefore LM′/LD = LM′′/LD and so M′ = M′′ = M.