Let $ ABC$ be an isosceles triangle with $ AB\equal{}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. **Assume $\angle A \leq 90^\circ$**

   - Let $l$ be a line parallel to $CE$ from $A$ intersecting the circumcircle of $\triangle ABC$ at $S$. 
   - Since $A(ECNS) = -1$, it follows that $(BCKS) = -1$.
   - Assume $FK$ intersects the circumcircle at $T$. We have $(BCKT) = -1$, so $S \equiv T$.
   - Given $DB \cdot DC = DH \cdot DA = DC^2$, it implies $DC$ is tangent to the circumcircle of $\triangle AHC$. Therefore, $DC$ should also be tangent to the circumcircle of $\triangle FKC$.
   - Let $\angle KFC = \angle BAK = \alpha$. Since $\angle ASK = \alpha + \angle B$, and $\angle AFK = 90^\circ - \alpha - \angle A$.
   - Using the fact that $\angle AFK + \angle SAK + \angle FSA = 180^\circ$, we get:
     \[
     90^\circ - \alpha - \angle A + \alpha + \angle B + \angle B = 180^\circ
     \]
     Simplifying, we find:
     \[
     90^\circ - \angle A + 2\angle B = 180^\circ \implies \angle A = 45^\circ
     \]

2. **Assume $\angle A > 90^\circ$**

   - If $\angle A > 90^\circ$, then $\angle A = 135^\circ$.

However, the problem constraints and the geometric configuration suggest that $\angle A = 45^\circ$ is the correct solution.

The final answer is $\boxed{45^\circ}$