A parallelogram ABCD is given. The excircle of triangle △ABC touches the sides AB at L and the extension of BC at K. The line DK meets the diagonal AC at point X; the line BX meets the median CC1 of trianlge △ABC at Y. Prove that the line YL, median BB1 of triangle △ABC and its bisector CC′ have a common point.(A. Golovanov) geometryparallelogramincenteranalytic geometrygeometry unsolvedTuymaada