1. Given Setup:
- Points $A$ and $B$ lie on the circle $S$.
- Tangent lines to $S$ at $A$ and $B$ intersect at $C$.
- $M$ is the midpoint of $AB$.
- Circle $S_1$ passes through $M$ and $C$, intersects $AB$ at $D$, and intersects $S$ at $K$ and $L$.

2. Objective:
- Prove that the tangent lines to $S$ at $K$ and $L$ intersect at a point on the segment $CD$.

3. Lemma Application:
- Consider the lemma provided in the solution. The lemma states that in an acute-angled triangle $\Delta ABC$ with $AB < AC$, circumcenter $O$, and incenter $I$, certain geometric properties hold. Specifically, it involves the intersection of tangents and cyclic quadrilaterals.

4. Proof Strategy:
- We need to show that the tangents at $K$ and $L$ intersect on $CD$. To do this, we will use properties of cyclic quadrilaterals and radical axes.

5. Cyclic Quadrilateral:
- Since $K$ and $L$ lie on both $S$ and $S_1$, the quadrilateral $AKBL$ is cyclic with respect to $S$, and $CKDL$ is cyclic with respect to $S_1$.

6. Radical Axis Theorem:
- The radical axis of two circles is the locus of points that have equal power with respect to both circles. For circles $S$ and $S_1$, the radical axis is the line $CD$.

7. Intersection of Tangents:
- The tangents to $S$ at $K$ and $L$ intersect at a point $P$. By the properties of the radical axis, $P$ must lie on the radical axis of $S$ and $S_1$, which is $CD$.

8. Conclusion:
- Therefore, the tangents to $S$ at $K$ and $L$ intersect at a point on the segment $CD$.

$\blacksquare$