In $\triangle ABC$, let $AK$ be the symmedian through $A$, and let the tangents to the circumcircle at $B$ and $C$ meet at $X$. I want to prove that $N$, the midpoint of $AK$, is concyclic with $B$, $C$, and $X$.
The book I am reading says that this can be done by angle chasing, but I am missing how to utilize that $N$ is the midpoint.
It is practical to consider a couple of extra points: $M$, the midpoint of $BC$; $O$, the circumcenter of $ABC$; $Y$, the intersection of the $OX$ segment with the circumcircle of $ABC$.
$OX$ is clearly a diameter of the circle through $B,C,X$ and $\widehat{XNO}=\widehat{KNO}=90^\circ$ (since $N$ is the midpoint of $AK$). In particular $\widehat{XNO}=\widehat{XBO}=90^\circ$ and $N$ lies on the circumcircle of $BCX$.