### Part (a): Prove that the lines $AB$ and $FG$ are parallel.

1. Identify the Altitudes and Projections:
- Let $D$ and $E$ be the feet of the altitudes from $A$ and $B$ respectively.
- Let $F$ be the projection of $D$ on line $AC$.
- Let $G$ be the projection of $E$ on line $BC$.

2. Use Similar Triangles:
- In $\triangle CED$, $FG$ is the line segment joining the projections of $D$ and $E$ on $AC$ and $BC$ respectively.
- Since $D$ and $E$ are the feet of the altitudes, $D$ lies on $BC$ and $E$ lies on $AC$.

3. Parallelism Argument:
- Since $D$ and $E$ are the feet of the altitudes, $DE$ is perpendicular to $AB$.
- The line segment $FG$ is the image of $DE$ under the homothety centered at $C$ that maps $\triangle CED$ to $\triangle CAB$.
- Therefore, $FG$ is parallel to $AB$.

Thus, we have shown that $AB \parallel FG$.

### Part (b): Determine the set of midpoints $S$ of segment $FG$ while along all allowable positions of point $C$.

1. **Fixed Length of $ED$:**
- By Ptolemy's theorem in cyclic quadrilateral $AEDB$, we have:
$$ED \cdot AB + AE \cdot BD = AD \cdot EB$$
- Since $A$ and $B$ are fixed points on the circle, the length $ED$ is fixed.

2. **Midpoint of $ED$:**
- Let $M$ be the midpoint of $AB$ and $I$ be the midpoint of $ED$.
- Since $AB$ is fixed, $M$ is a fixed point.
- The length $IM$ is given by:
$$IM = \sqrt{EM^2 - \left(\frac{1}{2} ED\right)^2} = \frac{1}{2} c \cdot \sin C$$

3. **Locus of Midpoint $S$:**
- From Lemma 1 and Lemma 2, the length $SI$ is fixed and given by:
$$SI = \frac{1}{2} c \cdot \cos C \sin C = \frac{1}{4} c \cdot \sin 2C$$
- Since $SI \perp AB$, the locus of $S$ is a circle centered at $O$ on the perpendicular bisector of $AB$ with radius:
$$\frac{1}{2} c \cdot \sin C$$

Combining the above lemmas, we conclude that the point $S$ moves on a circle whose center $O$ lies on the perpendicular bisector of $AB$ and the radius of the circle is $\frac{1}{2} c \cdot \sin C$.

$\blacksquare$