Define: $D_1=LM\cap \omega_1$ $E=(ABC)\cap \omega_2$ , $B\neq E$ $(D,E_1)=\omega_1 \cap \omega_2$ $N=$ midpoint of $\overline{BM}$ $P=$ midpoint of $BE$ $\angle ABC=\angle BAK=\alpha$ Claim 1:The circles $(ABC),\omega_1,\omega_2$ have a common point, $E\equiv E_1$ Since $BE$ is the radical axis of $(ABC)$ and $\omega_2$,we have $TK\perp BE$ and we get that $TBPN$ is cyclic. Now: $$\angle CAE=\angle CBE \equiv \angle KBP=\angle NTK=\angle NTC =90^\circ-\angle ACB=90^\circ-(90^\circ-\alpha)=\alpha$$and it follows $AE\perp BC$. If $E_2=(ABC)\cap \omega_1$, we have $AE_2\perp BC$. Now it is easy to conclude $E\equiv E_1\equiv E_2$. $\square$ Claim 2: $D\equiv D_1$ Now since we know that $DE$ is the radical axis of $\omega_1$ and $\omega_2$, it is well-known that $TC$ is the perpendicular bisector of $\overline{DE}$ and also $\triangle LDA\cong \triangle LEA$. $$\angle CDA=\angle DAC \equiv \angle DAL=\angle LAE \equiv \angle CAE=\alpha$$But also since $LD_1 \parallel AK$, $\angle LD_1A=\angle D_1AK=90^\circ$, we have $$\angle AD_1C=\angle D_1AC=90^\circ-\angle CAK=\alpha$$Now $\angle DAC=\angle D_1AC=\alpha$, and we are done.$\blacksquare$

Let $LM$ intersect $(\omega_1)$ at $P$. Then, $CA=CP=CL$, hence $\angle APL=90^\circ$. Let $T$ be the point on $AC$ such that $TB=TM$. We just have to prove that $T$ is the circumcenter of $PBM$. Let $TQ \perp BM$. Then, $BQ=BM/2=3BC/4$, so $CT \cdot CA=CQ \cdot CB=CB^2/4=CM^2$, therefore $\angle TMC=\angle CAM$. We now prove the following Claim: $\vartriangle APB$ and $\vartriangle CAM$ are similar. Proof: Note that $\angle PAB=90^\circ+\angle PAL=180^\circ-\angle C=\angle ACM$. Also, $\dfrac{PA}{BA}=\dfrac{AL \sin \angle C}{BA}=\dfrac{2AC \sin \angle C}{BA}=\dfrac{2AC}{BC}=\dfrac{CA}{CM}$, so indeed $\vartriangle APB$ and $\vartriangle CAM$ are similar $\square$. So, $\angle APB=\angle CAM \Rightarrow \angle BPM=90^\circ-\angle CAM=90^\circ-\angle TMC=\angle QTM=\angle BTM/2$, so $\angle BTM=2\angle BPM$, which together with $TB=TM$ implies the result.

Barycentric bashing $A=(1,0,0), B=(0,1,0), C=(0,0,1).$ Easy to find, that $M=(0,-\frac{1}{2},\frac{3}{2})$ and $L=(-1,0,2)$ If $F$ is midpoint of $BM$, then $F=(0,\frac{1}{4},\frac{3}{4})$. Assume that $T=(t,0,1-t)$, since $A,B,F,T$ lie on a circle, easy to find that $T=(\frac{a^2}{4b^2},0,\frac{4b^2-a^2}{4b^2}).$ Circle with center $C$ with radius $CA$ intersect the line $LM$ at $S\neq L.$ Lets find coordinates of $S$. $S=(x,y,z)$ with $x+y+z=1.$ Since $S,L,M$ lie on a one line. \[ A= \left[{\begin{array}{ccc} -1&0&2\\ 0&-\frac{1}{2}&\frac{3}{2}\\ x&y&z \end{array}}\right]=0 \]From this one can find that $S=(-1-2y,y,y+2).$ On a line $BC$ take a point $A_1$ such that $CA_1=CA$. $A_1=(0,\frac{b}{a},\frac{a-b}{a}.)$ Since $A,L,A_1,S$ lie on a circle, one can find that $S=(-1+\frac{4b^2}{a^2},-\frac{2b^2}{a^2},-\frac{2b^2}{a^2}+2)$ Then by a length equation prove that $TB=TS$ by using $c^2+b^2=a^2$. (It takes for me 15 minutes, but I don’t want to write it because I thought it ugly if you make it beautifully then let me know)

$\angle{TAB}=90=\angle{TEB} \implies TAEB$ cyclic By power of a point $CA \cdot CT=CE \cdot CB=\frac{CK}{2} \cdot 2CK=CK^2=CM^2$, so $\Delta ACM \sim \Delta TCM$ $\angle{LTM}=\angle{CTM}=\angle{AMC}=\angle{AMK}=\angle{LKM} \implies MTKL$ cyclic Let $D=\omega_1 \cap ML$ Let $F$ be the midpoint of $MD$ We want to prove that $TM=TD$ Observe that $\angle{ADL}=90$, so we want to prove $\angle{TFD}=90$ or rather $TF \parallel AD$ $\frac{LD}{DF}=2\frac{LD}{MD}$ $LD=\sin\angle{DAL} \cdot AL$ $\sin\angle{DAL}=\cos\angle{ALD}=\cos\angle{ALM}=\cos\angle{LAK}=\cos\angle{ACB}=\frac{CA}{CB}$ $2\frac{LD}{MD}=2\frac{CA \cdot AL}{CB \cdot MD}=\frac{AL^2}{CB \cdot MD}$ $MD=\cos\angle{AMD} \cdot AM$ $\cos\angle{AMD}=\cos(180-\angle{MLK})=\cos\angle{MTK}=\cos\angle{BTC}=\frac{AT}{TB}$ $\frac{AL^2}{CB \cdot MD}=\frac{AC \cdot AL}{CM \cdot MD}=\frac{AC \cdot AL}{CM \cdot \frac{AT}{TB} \cdot AM}=\frac{AC \cdot AL \cdot TB}{CM \cdot AT \cdot AM}=\frac{AC \cdot AL \cdot TM}{CM \cdot AT \cdot AM}$ We already proved that $\Delta ACM \sim \Delta TCM$, which gets us ratio $\frac{AC}{CM}=\frac{AM}{TM}$ hence $\frac{LD}{DF}=\frac{AC \cdot AL \cdot TM}{CM \cdot AT \cdot AM}=\frac{AL}{AT}$ $\implies TF \parallel AD \implies TM=TD=TB$

Let D be midpoint of CK and let LM touch w1 at P. we will prove T is circumcenter of BMP. CT.CA = CD.CB = CK^2 = CM^2 ---> ∠CMT = ∠MAT first we will prove triangles ABP and CAM are similar. ∠CTM = 180 - ∠TCD = ∠180 - ∠CLM = ∠BAP and BA/AP = BC/AL = BC/2AC = CT/2CD = CT/CM ---> ABP and CAM are similar. ∠MPB = 90 - ∠BPA = 90 - ∠MAC = 90 - ∠CMT = ∠DTM = ∠BTM / 2 ---> ∠BTM = 2∠BPM ---> T is circumcenter of BMP.