Let $ ABC$ be an isosceles triangle with $ AB=AC$, and $ D$ be midpoint of $ BC$, and $ E$ be foot of altitude from $ C$. Let $ H$ be orthocenter of $ ABC$ and $ N$ be midpoint of $ CE$. $ AN$ intersects with circumcircle of triangle $ ABC$ at $ K$. The tangent from $ C$ to circumcircle of $ ABC$ intersects with $ AD$ at $ F$.
Suppose that radical axis of circumcircles of $ CHA$ and $ CKF$ is $ BC$. Find $ \angle BAC$.
1. suppose$ \angle A \leq 90$
line $ l$ parallel to $ CE$ from$ A$ intersect circumcircle at $ S$ we have $ A(ECNS)=-1$so $ (BCKS)=-1$
now assume that $ FK$ intersect circumcircle at $ T$.we have $ (BCKT)=-1$ so $ S \equiv T$
we have$ DB.DC=DH.DA=DC^2$ means $ DC$ is tangent to circumcircle of $ AHC$ so it should be tangent to circumcircle $ FKC$
let$ \angle KFC=\angle BAK=\alpha$ but $ \angle ASK=\alpha +\angle B$ .and $ \angle AFK=90-\alpha-\angle A$
but$ \angle AFK+\angle SAK+\angle FSA=180$
so $ 90-\alpha-\angle A+\angle B+\alpha +\angle B=180$ means $ \angle A=45$
2.if $ \angle A> 90$ we have $ \angle A=135$