$\triangle ABC$ is an acute angled triangle. Perpendiculars drawn from its vertices on the opposite sides are $AD$, $BE$ and $CF$. The line parallel to $ DF$ through $E$ meets $BC$ at $Y$ and $BA$ at $X$. $DF$ and $CA$ meet at $Z$. Circumcircle of $XYZ$ meets $AC$ at $S$. Given, $\angle B=33 ^\circ.$ find the angle $\angle FSD $ with proof. 
1. **Identify the given information and draw the diagram:**
   - $\triangle ABC$ is an acute-angled triangle.
   - Perpendiculars from vertices to opposite sides are $AD$, $BE$, and $CF$.
   - Line parallel to $DF$ through $E$ meets $BC$ at $Y$ and $BA$ at $X$.
   - $DF$ and $CA$ meet at $Z$.
   - Circumcircle of $\triangle XYZ$ meets $AC$ at $S$.
   - $\angle B = 33^\circ$.

2. **Establish the relationships and properties:**
   - Since $AD$, $BE$, and $CF$ are altitudes, $D$, $E$, and $F$ are the feet of the perpendiculars from $A$, $B$, and $C$ respectively.
   - $E$ is the midpoint of $BC$ because $BE$ is an altitude in an acute-angled triangle.
   - $X$ and $Y$ are points such that $EX \parallel DF$ and $EY \parallel DF$.

3. **Use parallel lines properties:**
   - Since $EX \parallel DF$ and $EY \parallel DF$, quadrilateral $EXDF$ is a parallelogram.
   - Therefore, $EX = DF$ and $EY = DF$.

4. **Consider the circumcircle of $\triangle XYZ$:**
   - The circumcircle of $\triangle XYZ$ intersects $AC$ at $S$.
   - We need to find $\angle FSD$.

5. **Use cyclic quadrilateral properties:**
   - Since $S$ lies on the circumcircle of $\triangle XYZ$, $\angle XSY = \angle XZY$ (opposite angles in a cyclic quadrilateral are supplementary).

6. **Calculate $\angle XZY$:**
   - $\angle XZY = \angle DFC$ because $DF \parallel EX$ and $DF \parallel EY$.
   - $\angle DFC = 90^\circ$ because $DF$ is an altitude.

7. **Determine $\angle FSD$:**
   - Since $\angle XZY = 90^\circ$, $\angle XSY = 90^\circ$.
   - $\angle FSD = \angle XSY = 90^\circ$.

The final answer is $\boxed{90^\circ}$