Let $ABC$ be an acute triangle and $AD$ a height. The angle bissector of $\angle DAC$ intersects $DC$ at $E$. Let $F$ be a point on $AE$ such that $BF$ is perpendicular to $AE$. If $\angle BAE=45º$, find $\angle BFC$.
1. **Identify the given elements and relationships:**
   - $ABC$ is an acute triangle.
   - $AD$ is the height from $A$ to $BC$, so $\angle BDA = 90^\circ$.
   - The angle bisector of $\angle DAC$ intersects $DC$ at $E$.
   - $F$ is a point on $AE$ such that $BF \perp AE$.
   - $\angle BAE = 45^\circ$.

2. **Establish cyclic quadrilateral $ABDF$:**
   - Since $\angle BDA = 90^\circ$ and $\angle BFA = 90^\circ$, quadrilateral $ABDF$ is cyclic (opposite angles sum to $180^\circ$).

3. **Determine $\angle CDF$:**
   - Since $ABDF$ is cyclic, $\angle BFA = \angle BDA = 90^\circ$.
   - $\angle BAE = 45^\circ$.
   - $\angle CDF = \angle BAE = 45^\circ$ because $E$ lies on the angle bisector of $\angle DAC$.

4. **Identify $F$ as the incenter of $\triangle CDA$:**
   - Since $\angle CDF = 45^\circ$ and $E$ is on the angle bisector of $\angle DAC$, $F$ is the incenter of $\triangle CDA$.

5. **Calculate $\angle BFC$:**
   - $\angle BFC = \angle BFE + \angle EFC$.
   - $\angle BFE = 90^\circ$ (since $BF \perp AE$).
   - $\angle EFC = 90^\circ - \frac{\angle CDA}{2}$ (since $F$ is the incenter of $\triangle CDA$ and $\angle CDA = 90^\circ$).
   - Therefore, $\angle BFC = 90^\circ + (90^\circ - \frac{90^\circ}{2}) = 90^\circ + 45^\circ = 135^\circ$.

\[
\boxed{135^\circ}
\]