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IMO 2018 P6

Posted: Wed Jan 09, 2019 11:46 pm
A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that $\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.$Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$.