Problem 1389

1. Identify the given elements and their properties:
- $ABCD$ is a cyclic quadrilateral.
- Circles with diameters $AB$ and $CD$ touch at $S$.
- $M$ and $N$ are the midpoints of $AB$ and $CD$, respectively.
- We need to prove that the perpendicular through $M$ to $MN$ meets $CS$ on the circumcircle of $ABCD$.

2. Define the intersections and homothety:
- Let line $CS$ and $DS$ intersect the line through $M$ at $X$ and $Y$, respectively.
- Let $I$ be the intersection of the radical axes of the three circles.
- Since the circles are tangent at $S$, they are homothetic with respect to $S$.

3. Consider the intersections on the circumcircle:
- Let $CS$ and $DS$ intersect the circumcircle $(ASB)$ at points $J$ and $K$, respectively.
- Since $JK \parallel DC$, we have a homothety centered at $S$ that maps $J$ to $K$.

4. Establish the homothety relationships:
- Let $SK \cap AB = P$.
- By the power of a point theorem, we have $JP \cdot PS = AP \cdot PB$.
- We need to show that $JP \cdot PS = PX \cdot PC$.

5. **Use the homothety centered at $P$:**
- Since $XY \parallel AS$, there is a homothety centered at $P$ that maps $X$ to $S$ and $M$ to $I$.
- Therefore, $\frac{MP}{PI} = \frac{PX}{PS}$.

6. Relate the segments using parallel lines:
- Similarly, since $JM \parallel ID$, we have $\frac{MP}{PI} = \frac{JP}{PC}$.

7. Combine the ratios to prove the desired equality:
- From the above ratios, we get $\frac{PX}{PS} = \frac{JP}{PC}$.
- This implies $JP \cdot PS = PX \cdot PC$, as desired.

8. Conclude the proof:
- Since $JP \cdot PS = PX \cdot PC$, the perpendicular through $M$ to $MN$ meets $CS$ on the circumcircle of $ABCD$.

$\blacksquare$