In rectangle $ABCD$, point $E$ is chosen on $AB$ and $F$ is the foot of $E$ onto side $CD$ such that the circumcircle of $\vartriangle ABF$ intersects line segments $AD$ and $BC$ at points $G$ and $H$ respectively. Let $S$ be the intersection of $EF$ and $GH$, and $T$ the intersection of lines $EC$ and $DS$. If $\angle SF T = 15^o$ , compute the measure of $\angle CSD$.
1. **Identify the given elements and their relationships:**
   - Rectangle \(ABCD\) with \(E\) on \(AB\) and \(F\) as the foot of the perpendicular from \(E\) to \(CD\).
   - The circumcircle of \(\triangle ABF\) intersects \(AD\) at \(G\) and \(BC\) at \(H\).
   - \(S\) is the intersection of \(EF\) and \(GH\).
   - \(T\) is the intersection of \(EC\) and \(DS\).
   - Given \(\angle SFT = 15^\circ\), we need to find \(\angle CSD\).

2. **Establish the cyclic nature of quadrilateral \(CTSF\):**
   - Since \(F\) is the foot of the perpendicular from \(E\) to \(CD\), \(\angle EFC = 90^\circ\).
   - Given that \(\angle SFT = 15^\circ\), we can use the properties of cyclic quadrilaterals to find other angles.

3. **Use angle chasing to find \(\angle CST\):**
   - Since \(CTSF\) is cyclic, \(\angle CST = 90^\circ - \angle SCT\).
   - \(\angle SCT = \angle SFT = 15^\circ\) (since \(S\) lies on \(EF\) and \(GH\), and \(T\) lies on \(EC\) and \(DS\)).
   - Therefore, \(\angle CST = 90^\circ - 15^\circ = 75^\circ\).

4. **Calculate \(\angle CSD\):**
   - \(\angle CSD = 180^\circ - \angle CST\) (since \(C, S, D\) are collinear and form a straight line).
   - \(\angle CSD = 180^\circ - 75^\circ = 105^\circ\).

The final answer is \(\boxed{105^\circ}\).