Define $P = XY \cap AB$ so that $P$ is the midpoint of $\overline{AB}$. The tangents imply that $P$ lies on the polar of $K$ wrt $\omega_1$, and $P$ lies on the polar of $L$ wrt $\omega_2$. So La Hire gives $AN$ is the polar of $P$ wrt $\omega_1$, and $BM$ is the polar of $P$ wrt $\omega_2$. Then $PA=PM=PN=PB$. Radical center on $\omega_1, \omega_2, (AB)$ gives a point $Z = AN \cap BM \cap XY$. Define $Z' = XY \cap O_1O_2 = (ANO_1) \cap (BMO_2)$. Observe that inversion about $(AB)$ maps $Z \mapsto Z'$, so $O_1O_2$ is the polar of $Z$ wrt $(AB)$. By Brokard, $H=AM \cap BN$ also lies on the polar of $Z$, so we are done.

Dadgarnia wrote: Circles $\omega_1$ and $\omega_2$ have centres $O_1$ and $O_2$, respectively. These two circles intersect at points $X$ and $Y$. $AB$ is common tangent line of these two circles such that $A$ lies on $\omega_1$ and $B$ lies on $\omega_2$. Let tangents to $\omega_1$ and $\omega_2$ at $X$ intersect $O_1O_2$ at points $K$ and $L$, respectively. Suppose that line $BL$ intersects $\omega_2$ for the second time at $M$ and line $AK$ intersects $\omega_1$ for the second time at $N$. Prove that lines $AM, BN$ and $O_1O_2$ concur. Proposed by Dominik Burek - Poland Invert about $A$ with arbitrary radius. We get that $K^*$ is the $A-$ HM point of $\triangle AX^*Y^*$, so $N^*$ is the midpoint of $X^*Y^*$ this gives $\angle AB^*N^* = 90^{\circ}$ so $\angle ANB = 90^{\circ}$. Similarly, $\angle AMB = 90^{\circ}$. Let $H = AN\cap BM$, so $H$ lies on $XY$ and let $Z = AB\cap O_1O_2$. Let $AM\cap BN = W$. We get $(L,H;M,B) = -1 = (K,H;N,A) = (L,H;NZ\cap LB,B)$, so $M,N,Z$ are collinear. Now center of $ABNM$ is the midpoint of $AB$ call it $O$ which lies on $XY$. From Brokard, $OH\perp WZ$, further $OH\perp O_1O_2$, so $W$ lies on $O_1O_2$ and we are done.

My solution: It's easy to see that $AXNY$ and $BXMY$ are harmonic quadrilaterals. Invert about $X$ with an arbitrary radius to get the following problem- Inverted problem wrote: Let two lines $AN$ and $BM$ meet at $Y$ such that $YA=YB$, and $Y$ is the midpoint of segments $AN$ and $BM$. Let $X$ be some point in the plane, and denote $\odot (XAM) \cap \odot (XBN)=Z$. Then prove that $YX=YZ$. Redefine $Z$ as the point such that $AXZN$ is an isosceles trapezoid. Since $AN=BM$, so we get that $BXZM$ is also an isosceles trapezium (i.e. our definition is correct). Thus, $Y$ lies on the perpendicular bisector of $XZ$, as desired. $\blacksquare$

Let M the intersection of the radical axes of W1 W2 and o1o2 J is the midpoint of AB, X the intersection of AE and W2 and Y is the intersection of BL and W1 : Claim 1: D is on tje polar of J respect to w2: let D* be the inverse of D with respect to inversion around w2 (D*O2)(DO2)=(O2K)^2 and we have by the similar triangles (MO2)(MD)=(O2K)^2 so M=D* and we have O2O1 is perpendicular to KQ so KQ is the polar of D respect to w2 and J is on KQ so J is on the polar of D and by hires D is on the polar of J. Claim 2: DB is the polar of J: we know that the polar of B respect to w2 is the tangent at B so J lies on the polar of J and by hires B is on the polar of J respect to W2 and by claim 1 we have D is on the polar of J respect to w2 so DB is the polar of J respect to W2. Claim 3: AELB is Cyclic quaditeral and AB is a a diameter of his circle: we have that E is on W2 and B E D are collinear so from claim 2 we have that EB is the polar of J respect to W2 so JE is a tangent to W2 so JE=JB and JB=JA so ABE is a right angled triangle. by symmetry we have that ALB is a right angled so AELB is Cyclic quaditeral and AB is a a diameter of his circle. Claim4: The intersection of AE and BL lies on O1O2: from claim 3 we have that AY and BX are diameters of w1,w2: so consired the homothety centered at the intersection of AE and BL, since BX is parallel to AY it's clear that this homothety takes B to Y and X to A, so it takes the midpoint of BX to the midpoint of AY so o1,o2 and the intersection of BY and AX are collinear.

[asy][asy] size(9cm); defaultpen(fontsize(8.5pt)); defaultpen(linewidth(0.35)); dotfactor *= 1.5; pair O = origin, P1 = (4,1), O1 = (1,0), O2 = (1.35,0), A = foot(O1,O,P1), B = foot(O2,O,P1); path p1 = circle(O1,abs(O1-A)), p2 = circle(O2,abs(O2-B)); pair X = intersectionpoints(p1,p2)[0], Y = intersectionpoints(p1,p2)[1], K = extension(O1,O2,X,X+dir(A--foot(A,O1,X))), L = extension(O1,O2,X,X+dir(B--foot(B,O2,X))), M = L+dir(L--B)*abs(L-X)*abs(L-X)/abs(L-B), N = K+dir(K--A)*abs(K-X)*abs(K-X)/abs(K-A), P = extension(A,M,B,N), R = (A+B)/2, Q = extension(A,N,B,M); draw(p1^^p2, orange); draw(K--A--B--L); draw(L--K, dashed); draw(A--P--B, blue); draw(L--X--K, purple); draw(circumcircle(A,B,M), heavygreen); draw(X--Y, heavyred); dot("$A$", A, dir(135)); dot("$B$", B, dir(60)); dot("$X$", X, dir(90)); dot("$Y$", Y, dir(270)); dot("$L$", L, dir(210)); dot("$K$", K, dir(330)); dot("$M$", M, dir(265)); dot("$N$", N, dir(290)); dot("$P$", P, dir(270)); dot("$O_1$", O1, dir(270)); dot("$O_2$", O2, dir(270)); dot("$Q$", Q, dir(270)); [/asy][/asy] Let $\Gamma$ denote the circle with diameter $\overline{AB}$; all poles and polars are taken with respect to $\Gamma$ unless mentioned otherwise. Redefine $M = \Gamma \cap \omega_2$ and $N = \Gamma \cap \omega_1$, with $L = \overline{BM} \cap \overline{XX}$ and $K = \overline{AN} \cap \overline{XX}$. Let $P = \overline{AM} \cap \overline{BN}$ and $Q = \overline{AN} \cap \overline{BM}$. Then $P$ lies on the polar of $Q$ by Brocard's. Observe that $\overline{O_1A}$ and $\overline{O_1N}$ are tangents to $\Gamma$, so $Q$ lies on the polar of $O_1$. Thus $O_1$ lies on the polar of $Q$ by La Hire's and similar for $O_2$. Note that $Q \in \overline{XY}$ by radical axis and $AXNY$ is harmonic, so $$-1 = (A,N;X,Y) \overset{X}{=} (A,N;K,Q),$$implying $K$ lies on the polar of $Q$ and similar for $L$. Therefore $P$, $O_1$, $O_2$, $K$, and $L$ all lie on the polar of $Q$ and we may conclude. $\blacksquare$

Wanna kill myself for not solving such an easy problem last year in 3 hours time Solve this now in 10 min: Let $C$ be the midpoint of $AB$, then it is well known that $C,X,Y$ are collinear. Claim 01. $AN,BM,XY$ concur and $AMNB$ cyclic. Proof. Notice that $KX$ and $KY$ are tangents to $\omega_1$ since $K \in O_1 O_2$. Thus, $C$ lies on polar of $K$. By La Hire, $K$ lies on polar of $C$< which means $CN$ tangent to $\omega_1$. Similarly, $CM$ tangent to $\omega_2$. This gives us $CA = CM = CN = CB$, which means $AMNB$ cyclic. Now, $AXNY$ is harmonic. Therefore, taking pencil on $A$, \[ (C,X;AN \cap XY, Y) = -1 \]Similarly, $(C,X;BM \cap XY, Y) = -1$. So we are done. Claim 02. $AB, MN, O_1 O_2$ concur. Proof Let the other tangent of $\omega_1, \omega_2$ be $A_1, B_1$. We'll prove that $AB,MN, A_1 B_1$ concur instead. This is an easy application of radical axis of $(ABMN), (ABA_1 B_1), (MNA_1B_1)$. SubClaim. $A_1 B_1 MN$ cyclic. Proof. Notice that \[ \measuredangle MB_1 A_1 = \measuredangle MBB_1 = \measuredangle ABB_1 - \measuredangle ABM = \measuredangle AA_1 B_1 - \measuredangle ANM = \measuredangle ANA_1 - \measuredangle ANM = \measuredangle MNA_1\] We are essentially finished since by Brocard Theorem on cyclic quadrilateral $AMNB$, the circumcenter of $(AMNB)$, which is $C$ is the orthocenter of the triangle formed by $AB \cap MN$, $AN \cap BM$ and $AM \cap BN$. Line passes through $AN \cap BM$ and $C$is $XY$, which is perpendicular to $O_1 O_2$. Since $AB \cap MN \in O_1 O_2$, we are done.

Solved with SnowPanda and jelena_ivanchic. Let $W = XY\cap AB$, or the midpoint of $AB$. Since $W$ lies on the polar of $L$ wrt $\omega_2$, $LMB$ is the polar of $W$ wrt $\omega_2$ and $WM = WB$. Similarly $WN = WA$, so $\angle AMB = \angle ANB = 90^{\circ}$. Now let $P = AM\cap \omega_2$ and $Q = BN\cap \omega_1$; the right angles imply that $O_1$ is the midpoint of $AQ$ and $O_2$ is the midpoint of $BP$. Since $ABPQ$ is a trapezoid, $AM, BN,$ and $O_1O_2$ concur.

Since $(MB;XY)=-1$, Tangent to $\omega_2$ at $M$, $XY$ and $AB$ concur at a point $C$. Similarly, tanget to $\omega_1$ at $N$,$XY$ and $AB$ concur at the same point $C$. By radical axis, $CA=CB$ and $CM=CM$, $CN=CA$ by tangents. So, $AMNB$ is cyclic with center at $C$. Moreover, $D=AN\cap MB$ lies on $XY$ (by radical axis). It is easy to see that $(AN;DK)=-1$ and $(BM;DL)=-1$. Hence, $KL$ is the polar of $D$ wrt $(AMNB)$. By Brocards theorem, $AM\cap BN$ lies on the polar of $D$, aka line $KL$

First $BXMY$ is harmonic because $L$ is the intersection of tangents at $X$ and $Y$. Now denote by $Z$ the midpoint of $\overline{AB}$. Observe that $\overline{ZM}$ and $\overline{ZN}$ are tangent to $\omega_2$ and $\omega_1$ respectively, so $ZA=ZM=ZN=ZB$ r$AMNB$ cyclic with center $Z$. Then $\overline{AM} \parallel \overline{ZO_2}$ and $\overline{BN} \parallel \overline{ZO_1}$. Denote $Q = \overline{AM} \cap \overline{O_1O_2}$ and $R = \overline{AB} \cap \overline{O_1O_2}$. Then $$\frac{RA}{RQ} = \frac{RQ}{RO_2} \implies \frac{RA}{RZ} \cdot \frac{RB}{RA} = \frac{RB}{RZ} = \frac{RQ}{RO_2} \cdot \frac{RO_2}{RO_2} = \frac{RQ}{RO_1},$$so the line through $B$ parallel to $ZO_1$, i.e. $BN$ also passes through $Q$. It follows $Q$ is the desired concurrency point.

Let $Z$ be the intersection of $AB$ and $XY$. Note that $ZA=ZB$ by radical axis. Now if we look at the polar of $Z$ wrt $\omega_1$, note that $A$ lies on it. But since $XY$ is the polar of $K$, and $Z,X,Y$ are collinear, by La Hire's, $K$ also lies on it, so $AK$ is the polar, and so $ZN=ZA$ as they are both tangents. Similarly, $ZM=ZB$. Thus, we have $Z$ is the center of the circle that contains the 4 points $A,B,M,N$. All polars/poles following will be wrt the circle with diameter $AB$. Let $P=AM \cap BN$, we will show it lies on $O_1O_2$. By radical axis, we have $Q=BM \cap AN \cap XY$. Now by Brocard's, we have $P$ lies on the polar of $Q$. Now note that $(X,Y;N,A)=-1$ and projecting through $X$ gives $(A,N;Q,K)=-1$ so $K$ also lies on the polar of $Q$. Similarly, $L$ does too, so $P \in LK=O_1O_2$ and so we are done.

Extremely easy, yet very cute. Let $D=\overline{XY}\cap \overline{AB}$. Note that by PoP, $DA=DB$. Due to symmetry, $LY$ and $KY$ are tangent to $\omega_2,\omega_1$, respectively. Thus, $BMXY$ and $ANXY$ are harmonic quadrilaterals. This means that $DN$ and $DM$ are tangent to $\omega_1,\omega_2$, respectively. Hence, $D$ is the center of $(AMNB)$. As $\omega_1,\omega_2$ are orthogonal wrt $(AMNB)$, we obtain that $O_1O_2$ is polar of $AN\cap BM$. However, by Brokard, $AM\cap BN$ lies on the polar of $AN\cap BM$. We are done.

Obviously, the tangents from $X$ and $Y$ to $\omega_1$ and $\omega_2$ intersect at $\overline{O_1O_2}$ by symmetry. By radical axis, $\overline{XY}$ intersects $\overline{AB}$ at a point $O$ with $OA=OB$. Let $M' \neq B$ be the point on $\omega_2$ such that $\overline{OM}$ is tangent to $\omega_2$, and define $N'$ similarly. By Lemma 2.3, both $XMYB$ and $XM'YB$ are harmonic, so $M'=M$. Similarly, $N'=N$. Thus, $OA=OB=OM=ON$, so $AMNB$ is cyclic. Thus, lines $AN$, $BM$, and $XY$ are concurrent at a point which we call $S$. Let $T$ be the exsimilicenter of $\omega_1$ and $\omega_2$. By symmetry, $T$ passes through $\overline{O_1O_2}$. Claim: $M$, $N$, and $T$ are collinear. Proof: Let $\overline{MN}$ intersect $\omega_1$ at a point $R$. We know that $$\widehat{AR}=2\angle ANM=2\angle ABM=\widehat{BM}.$$By homothety, $N$, $R$, and $T$ are collinear. Thus, it is sufficient to show that the polar of $S$ with respect to the circumcircle of $AMNB$ is $O_1O_2$ by Brocard's. Let $Z$ be the intersection point of $\overline{O_1O_2}$ and $\overline{XY}$. Claim: $\overline{OM}$ is tangent to the circumcircle of $\triangle MSZ$. Proof: Since $\angle OBO_2=\angle OMO_2=\angle OZO_2=90^\circ$, $OBO_2ZM$ is cyclic. Thus, we have $$\angle MZS=\angle MBO=\angle OMB,$$so our claim is proven. Thus, we have $OS \cdot OZ=OM^2$, so $\overline{O_1O_2}$ is the polar of $S$ and we are done.

Solved with MrOreoJuice, SatisfiedMagma, Executioner230607. Here is a godly diagram, enjoy! Let $Z= AB \cap XY$. Now since $Z$ lies on the radical axis, $ZA=ZB$. Also $ZM$ and $ZN$ must be tangents to the respective circles.$$ZN=ZA=ZB=ZM$$Therefore $A, B, M$ and $N$ lie on a circle with diameter $AB$ and center $Z$. Now, by theradical axis theorem, $AN$, $BM$ and $XY$ are concurrent. Let that concurrency point be $O$. Let $AM \cap BN =F$. Clearly $F$ lies on the polar of $O$ by Brokard. Now, $O_1A, O_1N, O_2M, O_2N$ are tangents to $(AMNB)$. Now, since $O$ lies on the polar of $O_1$, $O_1$ lies on the polar of $O$ and similarly $O_2$ lies on the polar of $O$ by La Hire and therefore $F$ must lie on $O_1O_2$. $\square$

Attachments:

Liked this one Note that the problem is symetric taking $A,B$ up or below $O_1O_2$ so W.L.O.G. $A,B$ closer to $X$. Claim 1: $AK \cap BL \cap XY=D$ exists. Proof: Let $XY \cap AB=C$, it is well known that $C$ is midpoint of $AB$ and note that taking polars w.r.t. $\omega_2$ we have $C \in XY=\mathcal P_L$ thus by La'Hire $L\in \mathcal P_C$ meaning that since $M \in BL$ we have $\mathcal P_C=BM$ and by the same way on $\omega_1$ we get $\mathcal P_D=AN$ thus $DA=DM=DN=DB$ thus $(AMNB)$ is cyclic.And by radax on $\omega_1,\omega_2,(AMNB)$ we are done. Claim 2: $KL$ is the polar of $D$ w.r.t. $(AMNB)$ (From here we take polars w.r.t. $(AMNB)$ till the end of the main proof) Proof: Since $BXMY$ is harmonic we have: $$-1=(B, M; X, Y) \overset{X}{=} (B, M; L, D) \implies L \in \mathcal P_D$$And similarily for $AXNY$ harmonic we get $K \in \mathcal P_D$ thus $\mathcal P_D=KL$ as desired. Main proof: We are basicaly done since by brokard we have that $AM \cap BN \in \mathcal P_D$ thus $AM,BN,O_1O_2$ are concurrent as desired!.

Let $\ell=\overline{O_1O_2}.$ It is well-known that $C=\overline{AB}\cap\overline{XY}$ is the midpoint of $\overline{AB}.$ Notice by symmetry, $\overline{KY}$ is tangent to $\omega_1,$ so $C$ lies on the polar of $K$ with respect to $\omega_1.$ Hence, La Hire yields that $K$ is on the polar of $C$ so $\overline{CN}$ is tangent to $\omega_1$ and $CA=CN.$ Therefore, $N$ and similarly $M$ lies on the circle with diameter $\overline{AB},$ $\gamma.$ By Radical Axis on $\omega_1,\omega_2,$ and $\gamma,$ we see $Z=\overline{AK}\cap\overline{BL}\cap\overline{XY}$ exists. Notice $$-1=(AN;KZ)\stackrel{B}=(\overline{AB}\cap\ell,\overline{BN}\cap\ell;K,L)$$and similarly $(\overline{AB}\cap\ell,\overline{AM}\cap\ell;K,L)=-1$ so $\overline{BN}\cap\ell=\overline{AM}\cap\ell.$ $\square$

oops I swapped $M$ and $N$ apparently Let $P = \overline{XY} \cap \overline{AB}$. By symmetry, $\overline{KY}$ is also tangent to $\omega_1$, so $XY$ is the polar of $K$ wrt $\omega_1$. Therefore, by La Hire's, $K$ must lie on the polar of $P$ wrt $\omega_1$; but $A$ evidently lies on this polar, too, so $\overline{PM}$ is tangent to $\omega_1$. Similarly, $\overline{PN}$ is tangent to $\omega_2$. Claim: $AM$, $BN$, and $XY$ concur Proof: Note that quadrilateral $AMXY$ is harmonic, so we have $$-1 = (A, M; X, Y) \overset{A}{=} (P, \overline{AM} \cap \overline{XY}; X, Y),$$and similarly $(P, \overline{BN} \cap \overline{XY}; X, Y) = -1$. Therefore, the three lines indeed concur at some point $Z$. Now, to finish the problem, note that $Z$ lies on the polar of $K$ wrt $\omega_1$, so we can project as follows: $$-1 = (Z, K; A, M) \overset{B}{=} (L, K; \overline{AB} \cap \overline{O_1O_2}, \overline{BM} \cap \overline{O_1O_2}).$$Analogously, we find that $(L, K; \overline{AB} \cap \overline{O_1O_2}, \overline{AN} \cap \overline{O_1O_2}) = -1,$ so we are done.

Let $P$ be the midpoint of $AB$. Note that quadrilaterals $AXNY$ and $BXMY$ are harmonic since $XY$ is the polar of $K$ and $L$ wrt to $\omega_1$ and $\omega_2$ respectively. Claim: Quadrilateral $AMNB$ is cyclic and $P$ is its center. Proof. Wrt to $\omega_2$ we have that $P$ lies on $XY$, so by La Hire's, $L$ lies on the polar of $P$, thus the polar of $P$ is $LB$ and $PM$ is a tangent to $\omega_2$. Similarily, $PN$ is tangent to $\omega_1$. As such, $PM = PB = PA = PN$. $\blacksquare$ Thus, $AN$, $BM$, and $XY$ concur at some point $Q$. By the above, it follows that $(AMNB)$ is orthogonal to $\omega_1$ and $\omega_2$. As such, the polar wrt to $(AMNB)$ of $O_1$ and $O_2$ is $AN$ and $BM$ respectively, so the polar of $O_1O_2$ is $Q$. Then, by Brokard's $\overline{AM} \cap \overline{BN}$ lies on $O_1O_2$, giving the result.

Attachments:

By symmetry, $AXNY$ and $BXMY$ are harmonic. Let $P=\overline{AB} \cap \overline{XY}$ be the midpoint of $\overline{AB}$ (by PoP). Then $\overline{PN}$ is tangent to $\omega_1$, so $PA=PN$. Likewise, $\overline{PM}$ is tangent to $\omega_2$, so $PB=PM$. Since $PA=PB$, it follows that $ABMN$ is cyclic with center $P$, i.e. $\angle AMB=\angle ANB=90^\circ$, so the second intersection of $\overline{AM}$ with $\omega_2$ is the $B$-antipode, and likewise the second intersection of $\overline{BN}$ with $\omega_1$ is the $A$-antipode. This allows us to restate the problem as follows: Define $\omega_1,\omega_2,O_1,O_2,A,B$ as before, and let $A'$ and $B'$ be the antipodes of $A$ and $B$ in $\omega_1$ and $\omega_2$ respectively. Prove that $\overline{AB'} \cap \overline{A'B}:=T$ lies on $\overline{O_1O_2}$. But this is clear since $\triangle TAA'$ and $\triangle TB'B$ are homothetic (since $\overline{AA'}$ and $\overline{BB'}$ are both perpendicular to $\overline{AB}$) and $O_1,O_2$ are the respective midpoints of $\overline{AA'}$ and $\overline{BB'}$.

Let $P$ be the midpoint of $AB$, $Q = AN\cap BM$, $R = AM\cap BN$, and $T = XY\cap O_1O_2$. Since $K$ is the pole of $XY$ wrt $\omega_1$ and $A$ is the pole of $AB$ wrt $\omega_1$, $AK$ is the polar of $P$ wrt $\omega_1$. Similarly, $BL$ is the polar of $P$ wrt $\omega_2$. This gives us a lot of information, particularly 1. $Q$ lies on $XY$ (by PoP) 2. $(PQ;XY)$ is harmonic 3. $A$, $B$, $M$, and $N$ all lie on a circle centered at $P$ The latter implies that $R$ is the orthocenter of $\triangle QAB$. The former implies that $$PQ\cdot PT = PX\cdot PY = PA^2 = PB^2$$ so $T$ is the $Q-$ HM point of $\triangle QAB$, which means $RT\perp Y$, as desired.

Let $P=AB\cap XY$; it is well-known that $P$ is the midpoint of $AB$. Let $M'\neq B$ be the unique point such that $PM'$ is tangent to $\omega_2$ at $M'$. Note that $LX$, $LY$, $PB$, and $PM'$ are tangent to $\omega_2$, $B$ lies on $LM$, and $P$ lies on $XY$, implying that $XMYB$ and $XM'YB$ are both harmonic. Hence $M=M'$, so that $PB=PM$. Symmetrical logic implies that $XNYA$ is harmonic with $PA$ and $PN$ being the tangents to its circumcircle $\omega_1$. As a result, we find that $PA=PB=PM=PN$, so that $AMNB$ is cyclic with diameter $AB$ and circumcenter $P$. Applying the Radical Axis Theorem on $\omega_1=(ANX)$, $\omega_2=(BMX)$, and $(AMNB)$, it follows that $AN$, $BM$, and $XY$ concur at some point $Z=AN\cap BM\cap XY$. Recall that $XMYB$ is harmonic, so that $(XY;MB)=-1$. Projecting from $X$, we find that \begin{align*} (LZ;MB)&\overset{X}=(XY;MB)\\ &=-1. \end{align*}Similarly, we also have that $(KZ;NA)=-1$. Now observe that $\triangle AQB$ has orthocenter $Z$ and altitudes $AN$, $BM$, and $QT$, as having $(ABMN)$ be centered at $P$ implies that $\angle AMB=\angle ANB=90^\circ$. Projecting from $Q$, it then follows that \begin{align*} ((QL\cap AB)T;AB)&\overset{Q}=(LZ;MB)\\ &=-1, \end{align*}and we similarly have that $((QK\cap AB)T;AB)=-1$. As a result, we equate lines to get that $(QL\cap AB)T=(QK\cap AB)T$, so that $$QK\cap AB=QL\cap AB,$$forcing $Q$ to lie on $KL$. However, $KL$ is exactly $O_1O_2$, implying that $Q$ lies on $O_1O_2$. Thus $Q=AM\cap BN\cap O_1O_2$, so that $AM$, $BN$, and $O_1O_2$ concur at $Q$, as desired. $\blacksquare$ - Jörg

$KX$ and $KY$ are tangents to $\omega_1$ and $K,N,A$ are collinear; hence, $-1=(X,Y;N,A)$. So, the tangents drawn to $\omega_1$ from $N$ and $A$ intersect on the line $XY$. Let $R$ be the midpoint of $AB$. It is clear that $R$ is on $XY$. Then, $RN$ is tangent to $\omega_1$. Similarly, $RM$ is tangent to $\omega_2$. All these give that $|RN|=|RA|=|RB|=|RM|$. Thus, $M,A,N,B$ lie on a circle centered at $R$. Because $\angle BNA=90 $, $BN$ intersects $\omega_1$ for the second time at the antipode of $A$. Let this point be $A'$. Define $B'$ similarly. Let $P$ be the insimilicenter of $\omega_1$ and $\omega_2$. Define $l_1$ as the line $AB$, and define $l_2$ as the second external tangent of the two circles. Let $\omega_3$ be the degenerate circle $l_2$ with center $O_\infty$ which is on the opposite side of the circles with respect to $l_2$. Let $\omega_4$ be the degenerate circle $l_1$ with center $O_\infty$ which is on the opposite side of the circles with respect to $l_1$. Now, when Monge d'Alembert Theorem is applied to the triple $\omega_1$,$\omega_2$,$\omega_3$, it is seen that $A'$,$P$,$B$ are collinear. If Monge d'Alembert Theorem is applied to the triple $\omega_1$,$\omega_2$,$\omega_4$, it is seen that $B'$,$P$,$A$ are collinear. Hence; each of $AM$,$BN$,$O_1O_2$ contains point $P$.

Let $C = \overline{XY} \cap AB, Q = AM \cap XY, Q' = BN \cap XY.$ Note that $AXMY, BXNY$ are harmonic quadrilaterals. Hence $\overline{AA}, \overline{XY}, \overline{MM}, \overline{NN}, \overline{BB}$ must concur at $C.$ Furthermore, $(C, Q; X, Y) = (C, Q'; X, Y) = -1$ or $Q = Q'.$ Also it is well known that $C$ is the midpoint of $AB$ and $CM = CA, CN = CB$ are tangents to $\omega_1, \omega_2$, so $C$ is the center of $(ABMN) = \omega_3$. Observe that $\angle O_1AC = \angle O_2MC = 90^\circ$ since $C = \overline{AA} \cap \overline{MM}$, so $Q \in AM$ lies on the polar of $O_1$ w.r.t $\omega_3$. Conversely by LaHire, $O_1$ lies on $Q$'s polar. Similarly $O_2$ lies on it's polar, hence $\text{polar}(Q) = \overline{O_1O_2}.$ Define $P = \overline{AM} \cap \overline{BN}.$ By Brokard, $P$ lies on the polar of $Q$, or $O_1O_2.$ Finally $AM, BN, O_1O_2$ concur at $P.$

Let $Q=AB\cap O_1O_2$. Now, \[ -1 = (L,C;M,B) \overset{A}{=} (L,K;AM\cap O_1O_2, Q). \]Similarly $(L,K;BN\cap O_1O_2, Q)=-1$. Thus we get that $AM\cap O_1O_2 \equiv BN \cap O_1O_2$ and we are done.

Let $\overline{AB} \cap \overline{XY} = Z$. Since $XY$ is the radical axis of $\omega_1$ and $\omega_2$, it follows that $ZA = ZB$. Notice that $XY$ is the polar of $K$ wrt $\omega_1$, so $Z$ lies on the polar of $K$. By La Hire's, $K$ lies on the polar of $Z$ and $ZA$ is tangent to $\omega_1$ so it follows that $ZN$ is a tangent to $\omega_1$. Similarly, $ZM$ is a tangent to $\omega_2$. From this, we find that $ZA = ZN = ZB = ZM$, so $Z$ is the center of $(ABMN)$. By Radical Axis, $AN$, $AM$, and $XY$ concur at a point $R$. Then notice that $AXNY$ is a harmonic quadrilateral so $(A, N; R, K) = -1$. Then taking perspectivity at $B$ gives $-1 = (A, N; R, K) \overset{B}= (\overline{BA} \cap \overline{O_1O_2}, \overline{BN} \cap \overline{O_1O_2}; L, K)$. Similarly, we have $-1 = (B, M; R, L) \overset{A}= (\overline{AB} \cap \overline{O_1O_2}, \overline{AM} \cap \overline{O_1O_2}; K, L)$ so we are done.

Let $XY \cap AB = S.$ We know $(X,Y; A,N)=-1$ (from tangents at $K$), so perspectivity at $A$ gives $(X,Y; S,XY \cap AK)=-1.$ Symmetrically, $(X, Y; S, XY \cap BL)=-1,$ which means $XY \cap AK = XY \cap BL, $ so $AK, BL, XY$ concur. Thus, $$(K, L; BN \cap O_1O_2, AB \cap O_1O_2)\overset{B}{=}(K, BL \cap AK; N, A) = (K, XY \cap AK; N, A) = -1.$$Similarly, $$(K, L; AM \cap O_1O_2, AB \cap O_1O_2) = -1,$$so $AM \cap O_1O_2 = BN \cap O_1O_2$ and $BN, AM, O_1O_2$ concur.

Let $P = \overline{AB} \cap \overline{O_1O_2}$, $Q = \overline{AB} \cap \overline{XY}$, $T_1=\overline{AM} \cap \overline{O_1O_2}$, and $T_2 = \overline{BN} \cap \overline{O_1O_2}$ Notice that $\overline{XY}$ is the polar of $K$ with respect to $\omega_1$. Thus, $Q$ lies on the polar of $K$ and it follows that $K$ lies on the polar of $Q$ due to La Hire. Since $\overline{AQ}$ is a tangent to $\omega_1$ and $\overline{AK}$ intersects $\omega_1$ again at point $N$, we must also have that $\overline{QN}$ is tangent to $\omega_1$. Similarly, $\overline{QM}$ is tangent to $\omega_2$, which gives \[QA=QB=QM=QN.\] Hence, $Q$ is the center of $(ABMN)$. Denote the radical center of $\omega_1$, $\omega_2$, and $(ABMN)$ as $S$. In other words, $S = \overline{AN} \cap \overline{BM} \cap \overline{XY}$. Since $AXNY$ is a harmonic quadrilateral, we have \[-1 = (A,N;S,K) \overset{B}{=} (P,T_2;L,K).\] Analogously, we get $(P,T_1;L,K)=-1$, which means $T_1 \equiv T_2$. $\square$

Note that $(XY;BM)=-1$, so the tangents at $M$ and $B$ intersect on $XY$. Let $Z=XY\cap AB$, then $ZM$ is tangent to $\omega_2$ and $ZN$ is tangent to $\omega_1$. Since $Z$ is on the radical axis of $\omega_1$ and $\omega_2$, $ZA=ZN=ZM=ZB$, so quadrilateral $AMNB$ is cyclic with center $Z$. Since $AB$ is a diameter, we get that $AM\perp MB$ and $AN\perp NB$. Now use radical axis on $(AB),\omega_1,\omega_2$ to get that $AN,BM,XY$ concur, say at $Q$. Also, let $AB$ and $O_1O_2$ intersect at $C$, the exsimilicenter of $\omega_1$ and $\omega_2$. Invert around $C$ with radius $CX$; this swaps $\omega_1$ and $\omega_2$ and fixes $(AB)$. Therefore, it swaps $M$ and $N$, so $C,M,N$ are collinear. Let $P$ be the intersection of the angle bisector of $\angle O_1XO_2$ with $O_1O_2$. I claim that $AM$ and $BN$ meet exactly at $P$, so it suffices to show that $A,M,P$ are collinear. Let the foot of the perpendicular from $P$ to $AB$ be $R$. Note that $(CP;O_1O_2)=-1$, and if we project this onto $AB$, we get $(CR;AB)=-1$. Let $AM$ and $BN$ meet at $P'$. Then note that $Q$ is the orthocenter of triangle $ABP'$, so $QP'\perp AB$. By the cevian config, $QP'\cap AB=R$, so $P,Q,R$ are collinear. But now note that \[(CR;BA)=-1=(XY;BM)\stackrel{X}{=}(LQ;BM)\stackrel{P}{=}(C,R;B,PM\cap AB),\]so we get $P,M,A$ are collinear, as desired!

It is well known that $XY$ intersects $AB$ at its midpoint. If we let $P$ be the midpoint of $AB$ from the tangent condition we get that $ABMN$ is cyclic. Thus from brokards if we let $AN\cap BM$ be $J$ proving $(A,M;J,K)=-1$ and $(B,N;J,L)=-1$ suffices. However as $LX$ and $LY$ are tangent to $W_1$ and from radax $J$ lies on $XY$ we get one of the harmonics and using a similar argument we get the other.