Problem 1595

1. Identify Key Points and Projections:
- Let $F$ be a point on the circumscribed circle of $\Delta ABC$.
- Let $P$ and $Q$ be the projections of $F$ onto $AB$ and $AC$, respectively.
- Let $M$ be the midpoint of $BC$.
- Let $N$ be the midpoint of $PQ$.

2. Simson Line:
- By the Simson line theorem, the points $P$, $Q$, and the foot of the perpendicular from $F$ to $BC$ (let's call this point $K$) are collinear.

3. Cyclic Quadrilateral:
- Since $P$ and $Q$ are projections of $F$ onto $AB$ and $AC$, respectively, the quadrilateral $PAQF$ is cyclic.
- This implies that $\angle FPQ = \angle FAQ$ and $\angle PFQ = \angle PAQ$.

4. Angle Relationships:
- Since $F$ lies on the circumcircle of $\Delta ABC$, $\angle FBC = \angle FAC$ and $\angle BFC = \angle BAC$.
- Therefore, $\angle FPQ = \angle FBC$ and $\angle PFQ = \angle BFC$.

5. Similarity of Triangles:
- From the above angle relationships, we have $\triangle PFQ \sim \triangle BFC$ by AA similarity criterion.

6. Medians:
- $FN$ is the median of $\triangle PFQ$.
- $FM$ is the median of $\triangle BFC$.

7. **Cyclic Quadrilateral $FNMK$:**
- Since $FN$ and $FM$ are medians, and $\angle FNK = \angle FMK$, the quadrilateral $FNMK$ is cyclic.
- This implies that $\angle FNM = \angle FKM = 90^\circ$.

8. Conclusion:
- Since $\angle FNM = 90^\circ$, $MN$ is perpendicular to $FN$.

$$\boxed{\text{MN is perpendicular to FN}}$$