1. Identify Key Points and Definitions:
- Let $ABCD$ be a cyclic quadrilateral with diagonals $AC$ and $BD$ intersecting at point $E$.
- Let $F$ be the midpoint of side $AB$.
- Let $P$, $Q$, and $R$ be the feet of the perpendiculars from $E$ to the lines $DA$, $AB$, and $BC$, respectively.

2. Extend Lines and Consider Triangle:
- Extend lines $AD$ and $BC$ to meet at point $K$.
- Consider $\triangle KAB$.

3. Pedal Circle Concept:
- The points $P$, $Q$, and $R$ form the pedal triangle of point $E$ with respect to $\triangle KAB$.
- The pedal circle (or the nine-point circle) of a point with respect to a triangle passes through the feet of the perpendiculars from the point to the sides of the triangle.

4. Isogonal Conjugate and Angle Properties:
- Since $ABCD$ is cyclic, $\angle KAE = \angle KBE$.
- This implies that $\angle BAE^* = \angle ABE^*$, where $E^*$ is the isogonal conjugate of $E$ with respect to $\triangle KAB$.
- The isogonal conjugate of a point with respect to a triangle has the property that the feet of the perpendiculars from the point and its isogonal conjugate to the sides of the triangle lie on the same circle.

5. Midpoint and Perpendicular Foot:
- Since $F$ is the midpoint of $AB$, it is also the foot of the perpendicular from $E$ onto $\overline{AB}$ in the context of the pedal circle of $E$ and $E^*$.

6. Conclusion:
- Therefore, the points $P$, $Q$, $R$, and $F$ lie on the common pedal circle of $E$ and $E^*$.

$$\boxed{\text{The points } P, Q, R, \text{ and } F \text{ are concyclic.}}$$