Problem 1393

1. Using Newton's Theorem: Newton's theorem states that in a circumscribed quadrilateral, the two diagonals and the line segment joining the points of tangency of opposite sides are concurrent. Therefore, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $S$.

2. **Cyclic Quadrilateral $SKBL$**: Given that quadrilateral $SKBL$ is cyclic, we know that the opposite angles of a cyclic quadrilateral sum to $180^\circ$. This implies:
$$\angle KSL + \angle KBL = 180^\circ$$

3. Tangency Points and Equal Segments: Since $ABCD$ is circumscribed about circle $\Gamma$, the tangency points $K, L, M, N$ satisfy:
$$BK = BL \quad \text{and} \quad DM = DN$$

4. Angle Bisector: Since $BK = BL$, point $B$ is the midpoint of the arc $KL$ in $\odot(SKL)$. This implies that line $BD$ bisects $\angle KSL$. Therefore:
$$\angle KSB = \angle LSB$$

5. Angle Relationship: Since $BD$ bisects $\angle KSL$, we have:
$$\angle KSB = \angle LSB \quad \text{and} \quad \angle KSL = 2\angle KSB$$

6. **Cyclic Condition for $SNDM$**: To show that $SNDM$ is cyclic, we need to show that $\angle MSN + \angle MDN = 180^\circ$. Since $DM = DN$, point $D$ is the midpoint of the arc $MN$ in $\odot(SMN)$. This implies:
$$\angle MSD = \angle NSD$$

7. Conclusion: Since $D$ is the midpoint of the arc $MN$ in $\odot(SMN)$, it follows that:
$$\angle MSN + \angle MDN = 180^\circ$$
Therefore, quadrilateral $SNDM$ is cyclic.

$\blacksquare$