# Which point do all possible circles pass through?

$$P$$ 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.

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 $$S$$?
• oh...$S$ is simply the $A$ dumpty point, isn't it? – Vann Dec 24 '19 at 17:05