Solution:
Let $M'$ be the symmetric point of $M$ with respect to $T$. Observe that $T$ is equidistant from $B$ and $M$, therefore $M$ belongs on $\omega_{2}$ and $M'M$ is a diameter of $\omega_{2}$. It suffices to prove that $M'A$ is perpendicular to $LM$, or equivalently, to $AK$. To see this, let $S$ be the point of intersection of $M'A$ with $LM$. We will then have $\angle M'SM=90^{\circ}$ which shows that $S$ belongs on $\omega_{2}$ as $M'M$ is a diameter of $\omega_{2}$. We also have that $S$ belongs on $\omega_{1}$ as $AL$ is diameter of $\omega_{1}$.

Since $T$ and $C$ are the midpoints of $M'M$ and $KM$ respectively, then $TC$ is parallel to $M'K$ and so $M'K$ is perpendicular to $AB$. Since $KA=KB$, then $KM'$ is the perpendicular bisector of $AB$. But then the triangles $KBM'$ and $KAM'$ are equal, showing that $\angle M'AK=\angle M'BK=\angle M'BM=90^{\circ}$ as required.

Alternative Solution by Proposers.
Since $CA=CL$, then $L$ belongs on $\omega_{1}$. Let $S$ be the other point of intersection of $\omega_{1}$ with the line $LM$. We need to show that $S$ belongs on $\omega_{2}$. Since $TB=TM$ ($T$ is on the perpendicular bisector of $BM$) it is enough to show that $TS=TM$.

Let $N, T'$ be points on the lines $AL$ and $LM$ respectively, such that $MN \perp LM$ and $TT' \perp LM$. It is enough to prove that $T'$ is the midpoint of $SM$. Since $AL$ is diameter of $\omega_{1}$ we have that $AS \perp LS$. Thus, it is enough to show that $T$ is the midpoint of $AN$. We have
$$AT=\frac{AN}{2} \Leftrightarrow AC-CT=\frac{AL-LN}{2} \Leftrightarrow 2AC-2CT=AL-LN \Leftrightarrow LN=2CT$$
as $AL=2AC$. So it suffices to prove that $LN=2CT$.

Let $D$ be the midpoint of $BM$. Since $BK=KC=CM$, then $D$ is also the midpoint of $KC$. The triangles $LMN$ and $CTD$ are similar since they are right-angled with $\angle TCD=\angle CAK=\angle MLN$. ($AK=KC$ and $AK$ is parallel to $LM$.) So we have
$$\frac{LN}{CT}=\frac{LM}{CD}=\frac{AK}{CD}=\frac{CK}{CD}=2$$
as required.