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<BKC = <CDB if MA x MC + MA x CD = MB xMD

Source: 2001 Cuba MO 1.2

September 15, 2024
geometryequal angles

Problem Statement

Let MM be the point of intersection of the diagonals ACAC and BDBD of the convex quadrilateral ABCDABCD. LetK K be the intersection point of the extension of side ABAB (from AA) with the bisector of the ∠ACD\angle ACD. If MA⋅MC+MA⋅CD=MB⋅MDMA \cdot MC + MA \cdot CD = MB\cdot MD , prove that ∠BKC=∠CDB\angle BKC = \angle CDB.
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