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$P$ is a variable point on side $BC$ of triangle $ABC$. $M, N$ are respectively on $AB, AC$ such that $PM||AC$,$PN||AB$. Prove that as $P$ varies on $BC$, then the circumcircle of $AMN$ passes through a fixed point.

I predict that the fixed point would be the "$A$ dumpty point", i.e. the intersection point of the circumcircle of $BOC$ and the $A$ symmedian. But I couldn't prove it. Can anyone help proving this, or otherwise, if the fixed point I mentioned is incorrect, then what is it, and can you prove it? Thanks for helping.

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Hint:

Notice that $PMAN$ is a paralelogram and that $BM:MA = AN:NC$, which means there is a spiral similarity which takes $B-M-A$ to $A-N-C$ (wherever $P$ is) and let $S$ be a center of this spiral similarity. What can you deduce for $S$?

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  • $\begingroup$ oh...$S$ is simply the $A$ dumpty point, isn't it? $\endgroup$ – Vann Dec 24 '19 at 17:05
  • $\begingroup$ Could be.... :) $\endgroup$ – Aqua Dec 24 '19 at 17:06
  • $\begingroup$ Well, I think I get it already...thanks for the useful hints anyway :) $\endgroup$ – Vann Dec 24 '19 at 17:12

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