IMO 2018 P6

Discussion on International Mathematical Olympiad (IMO)
M Ahsan Al Mahir
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Joined: Wed Aug 10, 2016 1:29 am

IMO 2018 P6

Unread post by M Ahsan Al Mahir » 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 $\angle{BXA} + \angle{DXC} = 180^\circ$.

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