XXI SILK ROAD MATHEMATICAL COMPETITION

Let N be a point on the line $AK$ so that $MN\parallel AL$. Since $\frac{CM}{DL}=\frac{NM}{AL}=\frac{KM}{KL}$ and $\angle CMN=\angle DLA$, it follows that $K$ is a center of homothety which sends $△CMN$ into similar $△DLA$. On the other hand, $△DLA$ is similar to $△CBA$ because $\angle DAL=\angle BAC$ and $\angle ADL=\angle ACB$. Consequently, $\angle (BN,BC)=\angle (MN,MC)$, and thus points $N$, $B$, $M$, $C$ are cyclic. Therefore, $\angle CBM=\angle CNM=\angle CAB$ from which it follows that $BM$ touches $\omega$.

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Alternative solution.

For this solution we need the following theorem.

Pascal's theorem. Points $A$, $B$, $C$, $D$, $E$, $F$ (not necessarily in this order) lie on some circle. Then the intersections of the lines $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$ lie on a straight line.

Back to the problem. Let $E$ be the second intersection of the line $AL$ and $\omega$. Define $l_b$ as a tangent line to $\omega$ at $B$. Let's apply Pascal's theorem on points $B$, $B_1$, $A$, $E$, $C$, $D$ (here $B_1$ coincides with $B$) and pairs of lines ($B{B}_1$, $EC$), ($B_{1}A$, $CD$), ($AE$, $DB$). These pairs of lines coincide with pairs of lines ($l_b$, $EC$), ($BA$, $CD$), ($AE$, $DB$). Then from Pascal's theorem, the straight line connecting $K=BA\cap CD$ and $L=AE\cap DB$ will also contain $l_b\cap EC$. Since $M=EC\cap KL$, it follows that $BM$ touches $\omega$, as desired.