RagvaloD wrote: tangent lines to $C$ at $K$ and $L$ Is this a typo? If $C$ is replaced by $S$, then here's my solution. Since the inverse of $C$ in $S$ is the midpoint of $AB$, so $(CD)$ is orthogonal to $S$, and thus we have the conclusion, as the tangents meet at the midpoint of $CD$.

Yes, it was typo

Let $O$ be the center of circle $S$, $P$ the intersection of the line $KL$ and $AB$, and $Q$ the intersection of tangent lines to $S$ at $K$ and $L$. Since $PA \cdot PB=PK \cdot PL=PD \cdot PM$ and M is the midpoint of $AB$, it can be concluded that $(A,B;P,D)=-1$ or $P$ is on the polar of $D$ wrt $S$. And from $P$ is on the polar of $C$ wrt $S$, $CD$ is the polar of $P$ wrt $S$. Polar of $Q$ wrt $S$ passes through $P$; therefore, polar of $P$ wrt $S$ also passes through $Q$, i.e., $Q$ lies on $CD$.

St. Petersburg 2018 Grade 11 P7 wrote: Points $A,B$ lies on the circle $S$. Tangent lines to $S$ at $A$ and $B$ intersects at $C$. $M$ -midpoint of $AB$. Circle $S_1$ goes through $M,C$ and intersects $AB$ at $D$ and $S$ at $K$ and $L$. Prove, that tangent lines to $S$ at $K$ and $L$ intersects at point on the segment $CD$. Solution: This problem is based on the Lemma: Lemma: In an acute angled $\Delta ABC$ with $AB<AC$ with circumcenter $O$ and incenter $I$. Let $AI$ $\cap$ $BC$ $=$ $D$ and $AI$ $\cap$ $\odot (ABC)$ $=$ $M_A$. Let $OM_A$ $\cap$ $\odot (ABC)$ $=$ $M_{BC}$. Let $M$ be the midpoint of $BC$. Let $KM_A,$ $LM_A$ be the tangents from $M_A$ to $\odot (DMM_{BC})$. Then, $M_A$ is the center of $\odot (BKLC)$ Proof: Firstly, observe, $\angle DMM_A=90^{\circ}=\angle DAM_{BC}$ $\implies$ $A$ $\in$ $\odot (DMM_{BC})$. Now we want to show $BKCL$ cyclic with center $M_A$. Let $DM_{BC}$ $\cap$ $\odot (ABC)$ $=$ $E$. Then, $EDMM_A$ is cyclic. Hence, By Radical Axes Theorem, $AM_{BC}$, $BC$ and $EM_A$ concur, say at $X$. Let $F$ be the center of $\odot (ADMM_{BC})$ $\implies$ $FKEM_AL$ is cyclic. Hence, again By Radical Axes Theorem, $X$ lies on $KL$ $\implies$ $BKCL$ is cyclic. Now just notice, $$\angle FKL= \angle KBL=\frac{1}{2} \angle KM_AL$$$\implies$ $M_A$ is the center of $\odot (BKCL)$ $\qquad \blacksquare$ Apply this Lemma, and the Original Problem Follows!