$P$ 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.
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 $S$?