Let $M$ be the midpoint of $\overline{AC}$ and $F$ be the common point of the $B$-symmedian with $\odot(ABC)$; the key is that $\overline{FM}$ passes through the point $B'$ on $\odot(ABC)$ with $\overline{BB'} \parallel \overline{AC}$. Since $BM=B'M$, we can easily construct $B'$, and then take the intersections of the line through $M$ parallel to $\overline{BB'}$ with $\odot(BB'F)$ to finish.

Use Centroamerican MO 2016 P2 as a lemma. It follows immediately.