Let $H$ be the orthocenter of $ABC$, and let $D,E,F$ be its projections onto $BC,AC,AB$ respectively. Note that the circle with diameter $AB$ goes through $D,E$, and the circle with diameter $AC$ goes through $D,F$. Now by PoP \[ HY\cdot HX=HD\cdot HA=HK\cdot HL \] So $KLXY$ is cyclic. Additionally, since $AF,AE$ are perpendicular bisectors of $XY,KL$, $A$ is the center of this circle. Now since $KL,XY$ intersect at $H$, by Brokard's theorem it suffices to show that the polar of $H$ is $BC$ with respect to $(KLXY)$, or equivalently that $H,D$ are inverses with respect to this circle. Now since $\angle BXA=\angle BYA=\frac{\pi}{2}$, and $F$ is the midpoint of $XY$, we know $F$ and $B$ are mutually inverse with respect to this circle. Similarly, $E$ and $C$ are mutually inverse. Now since $\angle AFH=\angle ADB=\frac{\pi}{2}$ and $\angle AED=\angle ADC=\frac{\pi}{2}$, $D$ is mapped to $H$ in this inversion, so the problem is solved.

Cross ratio let us show that XK and LY also intersect on BC... Let's denote by H the orthocenter of ABC.Then CXHY and BKHL are both harmonic. Since they've got a common point H, the results follow immediately.

An alternative finish after proving that $KLXY$ is cyclic. Let $E, F$ be the projections from $B$ onto $AC$, $C$ onto $AB$. Clearly, $\angle BXA = 90^\circ$, so $\triangle BXA \sim \triangle BFX$, so $BX^2 = BF \cdot BA = BK \cdot BL$, so $BK$ is tangent to the circumcircle of $KLXY$. Similarly, since $BY = BX$, $BY^2 = BK \cdot BL$, so $BY$ is also tangent. Similarly, $CL$ and $CK$ are tangent to the circumcircle of $KLXY$. So using Pascal's Theorem on $XXLLKY$ gives $XX\cap LK$, $LL\cap XY$, and $XL\cap KY$ are collinear, which is the same as $B$, $C$ and the intersection of $XL$ and $KY$ are collinear, which is equivalent to the problem statement.

Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent.

Sorry for posting just a reworded version of the problem, but I wanted the original wording for this.

very nice problem!

Let the two circles with diameters $AC$ and $AB$ cut at another point $E$ and let $H\equiv BL\cap CY$, than $H$ is the orthocenter of $\Delta ABC$ and $A, H, E$ are collinear. As $YH. HX = AH. HE = KH.HL$, quadrilateral $YKXL$ is cyclic, as $AY^{2} = AF. AB = AX^{2} = AH. AE = AD. AC = AL^{2}$, $A$ it the center of $(YKXL)$. Now notice that because $AH. AE = AY^{2}$ and $AE\perp BC$, $BC$ is in fact the polar of $H$ wrt. $(YKXL)$. But the intersection $G\equiv YK\cap XL$ also lies on the polar of $H$ , thus $G$ lies on $BC$.

Notice by power of point at orthocenter that XLKY is cyclic. Then notice that the orthocenter and the foot of the altitude from A to BC are inverse w.r.t this circle. This can be done by inverting lines XY and KL. Therefore polar line of the orthocenter is BC. But XL and KY also must intersect on the polar line of the orthocenter. So we are done.

Dear Mathlinkers,according with the cocyclicity (with the three chords theorem) of vslmat , we can finish with the Pascal's theorem. Sincerely Jean-Louis

Another solution to this interesting problem, which just uses elementary knowledge (angle chasing and similar triangles): Assume $AB \le AC$. Let $H$ is the orthocenter. $D$ is the pedal of $H$ on $BC$. $(AB), (AC)$ are the circle of diameter $AB$ and $AC$. Observe that $(AB)$ and $(AC)$ meet at $D$. $H$ is a point in $(AB)$, so $HA.HD=HX.HY$. Similarly, $HA.HD=HK.HL$. Therefore, $X, Y, L, K$ are on the same circle. Also, $A$ is the intersection of the bisectors of $KL$ and $XY$. So, $A$ is the center of this circle. We need to prove: $\angle KYD = \angle XLD$. Having, $\angle KYD = \angle HYD - \angle HYK$, $\angle XLD = \angle HLX - \angle HLD$. Now we need to show that: $\angle HYK + \angle HLK = \angle HYD + \angle HLD$. It's equivalent to: $\angle KAX = \angle HYD + \angle HLD$, which is true because $\angle KAX = \angle KAD + \angle XAD$. We need orther equal angles: $\angle DYK = \angle DLX$. Or $\angle XYD - \angle XYK = \angle XLK - \angle DLK$. It's equivalent to: $\angle KAX = \angle XLK + \angle XYK$. That's true because A is the center of $(XYLK)$. From two pairs of equal angles, we've shown that $\Delta DYK$ is similar to $\Delta DLX$. As a result, $\Delta DYL$ is similar to $\Delta DKX$. Finishing the proof: Let $KY$ and $XL$ meet at $N$. It's easy to see that $N$ is the intersection of $(DKX)$ and $(DYL)$. Assume $(DKX)$ meets $BC$ at $N'$. We have $L, X, N'$ are collinear. So $\angle LN'C = \angle DKX = \angle DYL$. This means $DYLN'$ is a cyclic quadrilateral. Thus, $N \equiv N'$. We are done.

Nice,but a bit too trivial for TST.Of course you have $KH \cdot HM=CH \cdot HD=NH \cdot HL$ so $KLMN$ is cyclic.$C$ lies on the perpendicular bisector of $KM$ as well as $FN$ so it is the center of $KLMN$.$CH \cdot CD=CE \cdot CA=CN^2$ so $D$ is the inverse point of $H$ w.r.t this circle and consequently $AB$ is the polar line of $H$.Now the result is just a consequence of Brokard's theorem.

My solution: Let $ H $ be the orthocenter of $ \triangle ABC $ . Since $ B $ lie on the polar of $ H $ WRT $ (AC) $ , so we get $ Y(B,H;K,L)=-1 $ . ... $ (1) $ Since $ C $ lie on the polar of $ H $ WRT $ (AB) $ , so we get $ L(H,C;X,Y)=-1 $ . ... $ (2) $ From $ (1) $ and $ (2) $ we get $ LX \cap KY \in BC $ . Q.E.D

Define $H\equiv BK\cap CX$, $F\equiv AH\cap BC$; it is easy to see that an inversion centered at $A$ with power $r^2=AH\cdot AF$ fixes each of $\{Y, K, X, L\}$, so it follows that $AY=AK=AX=AL$ and thus quadrilateral $YKXL$ is cyclic. Then $BY, BX, CL, CK$ are tangent to $\odot(YKXL)$, and so $B$ and $C$ lie on the polar of $H$ and we're done by Brokard's theorem.

Note that $\{B, H, L, N\}=-1$ due to $AH$ being the bisector of $\angle LFN$. Analogously $\{A, H, K, M\}=-1$, define $KN \cap LM=P$ then $\{PB, PH, PL, PN\}=\{PA, PH, PK, PM\}=-1$. So $PA$ and $PB$ must be the same line.

v_Enhance wrote: Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent. Let $H$ be the orthocenter of $\triangle ABC$, and $CH \cap AB=D$. Inversion around $C$ with radius $\sqrt{CH \cdot CD}$ fixes $M,N,K,L$, thus quadrilateral $MNKL$ is cyclic with center $C$. Now, since $D$ is the inverted image of $H$, and $CH \perp DB$, $AB$ is the polar of $H$. But by Brocard's theorem, $ML \cap NK$ lies on the polar of $H$, so we are done. $\square$

v_Enhance wrote: Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent.

Lemma: $(AH;KM)=-1$, where $H$ is the orthocenter. Proof of Lemma: Let $P$ be the midpoint of $AH$ and $O$ be the midpoint of $BC$. Since $\triangle EAH\sim \triangle EBC$, there exists a spiral similarity (with angle $90^\circ$) sending $AH\to BC$. In fact, $P \to O$, so $\angle PEO = 90^\circ$. Hence, $PE$ is tangent to $(BC)$, so $PE^2=PK\cdot PF$. This implies that $K,F$ are inverses in $(AH)$, which means $(AH;KF)=-1$. $\square$ By symmetry, $(BH;LN)=(AH;KM)=-1$. Let $X=MN\cap AB$. \begin{align*} -1&=(BH;LN) \stackrel{M}{=} (BA;ML\cap AB,X) \\ -1&= (AH;KM) \stackrel{N}{=} (AB;NK\cap AB,X) = (BA;NK\cap AB,X). \end{align*}Therefore, $ML\cap AB = NK\cap AB$.

USA TST 2013 P2 wrote: Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent. Solution:- Claim 1:- $NKLM$ is a cyclic quadrilateral. $NH.HL=HX.HC=HK.HM\implies NKLM$ is a cyclic quadrilatreral where $X$ is the feet of perpendicular from $C$ to $AB$ which is actually $\omega_1\cap\omega_2\neq C$. Claim 2:- $C$ is the center of $\odot(NKLM)$. $CK^2=CF.CB$. Now notice that $\angle FAC=\angle FLC=\angle LBC\implies CL^2=CF.CB=CK^2$. So, $CK=CL=CL=CN\implies C$ is the circumcenter of $\odot(NKLM)$. Claim 3:-$AB$ is the Polar of $H$ WRT $\odot(NKLM)$. Consider $\Psi$ as the $\sqrt{AH\cdot AD}$ Inversion around $A$. Then $-1=(K,M;F,\infty_{\overline{KM}})\overset{\Psi}{=}(A,H;K,M)\implies A\in$ Polar of $H$. Similarly $B$ lies on Polar of $H$ which implies $\overline{AB}$ is the Polar of $H$ WRT $\odot(NKLM)$. So as $NK\cap LM$ lies on Polar of $H$ so we can directly conclude that $\overline{AB},\overline{NK},\overline{LM}$ are concurrent. $\blacksquare$

Let $H$ be the orthocenter of $\triangle ABC$. Because $(C, A; N, L)$ and $(K, M; C B)$ are harmonic, so are $(N, L; H, B)$ and $(A, H; K, M)$. Projecting $(N, L; H, B)$ through $K$ and $(A, H; K, M)$ through $L$, we are done.

We use the actual wording that Evan wrote Evidently, $E$ and $F$ are the feet of the obvious altitudes, so add $D$, the foot of the $C$-altitude, so $ADFC$ and $BDEC$ are cyclic. By radical center on these two circles, it follows that $KLMN$ is cyclic as well. Furthermore, because $C$ lies on the perpendicular bisectors of $\overline{LN}$ and $\overline{KM}$, $C$ is the center of $(KLMN)$. Now, note that since $\angle ALC=\angle ANC=90^\circ$, $A$ is the intersection of the tangents at $L$ and $N$ to $(KLMN)$, hence $A$ and $E$ are polar and $CE\cdot CA=r^2$ where $r$ is the radius of $(KLMN)$. But since $AEHD$ is also cyclic, we have $CH \cdot CD=r^2$ as well, hence $\overline{AB}$ is the polar of $H$. By Brocard, it thus follows that we should have $\overline{KN} \cap \overline{LM} \in \overline{AB}$, as desired. $\blacksquare$

Please reply if this is incorrect I am new to projective geo: [asy][asy] import graph; size(10.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1., xmax = 9., ymin = -4., ymax = 7.; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); draw((3.3975155288628054,6.263686393496784)--(0.,0.)--(6.,0.0028)--cycle, linewidth(0.) + blue); draw(circle((4.698757764431402,3.133243196748392), 3.3901188716173145), linewidth(0.4) + linetype("4 4") + xdxdff); draw(circle((3.,0.0014), 3.0000003266666493), linewidth(0.4) + linetype("4 4") + xdxdff); draw((1.5333035322697934,2.4256017909047967)--(1.6245103320390752,2.593751365455956)--(1.456360757487916,2.684958165225238)--(1.3651539577186342,2.5168085906740787)--cycle, linewidth(0.) + blue); draw(arc((1.3651539577186342,2.5168085906740787),0.27052896238994345,331.5239453490917,372.6885987258671)--(1.3651539577186342,2.5168085906740787)--cycle, linewidth(0.)); draw(arc((1.3651539577186342,2.5168085906740787),0.27052896238994345,-69.64070802768387,-28.476054650908353)--(1.3651539577186342,2.5168085906740787)--cycle, linewidth(0.)); /* draw figures */ draw((3.3975155288628054,6.263686393496784)--(3.401825314218796,-2.971567940771136), linewidth(0.4)); draw((1.3651539577186342,2.5168085906740787)--(6.,0.0028), linewidth(0.4)); draw((0.,0.)--(8.081338196952771,3.3591980179392906), linewidth(0.4)); draw((1.5303158821608756,2.821302415665919)--(3.401825314218796,-2.971567940771136), linewidth(0.4) + dcrutc); draw((1.5303158821608756,2.821302415665919)--(8.081338196952771,3.3591980179392906), linewidth(0.4) + red); draw((1.3651539577186342,2.5168085906740787)--(3.401825314218796,-2.971567940771136), linewidth(0.4)); draw((1.3651539577186342,2.5168085906740787)--(3.3990503697278167,2.974741682756977), linewidth(0.4)); /* dots and labels */ dot((0.,0.),linewidth(1.pt) + dotstyle); label("$B$", (0.03393876314784184,0.014467362289105373), NE * labelscalefactor); dot((6.,0.0028),linewidth(1.pt) + dotstyle); label("$C$", (6.039681728204586,0.02348499436877013), NE * labelscalefactor); dot((3.3975155288628054,6.263686393496784),linewidth(1.pt) + dotstyle); label("$A$", (2.9376162927999014,6.525197723807063), NE * labelscalefactor); dot((3.400437841973307,0.0015868709929208766),linewidth(1.pt) + dotstyle); label("$F$", (3.4335860571814645,0.02348499436877013), NE * labelscalefactor); dot((5.117022060035043,2.1270103964096974),linewidth(1.pt) + dotstyle); label("$E$", (5.155953784397438,2.142628533089989), NE * labelscalefactor); dot((1.3651539577186342,2.5168085906740787),linewidth(1.pt) + dotstyle); label("$D$", (1.016860659831303,2.512351448356244), NE * labelscalefactor); dot((3.3990503697278167,2.974741682756977),linewidth(1.pt) + dotstyle); label("$K$", (3.4335860571814645,2.9902859485784763), NE * labelscalefactor); dot((3.401825314218796,-2.971567940771136),linewidth(1.pt) + dotstyle); label("$M$", (3.4335860571814645,-2.952333591920601), NE * labelscalefactor); dot((2.1527059231173156,0.8948227748801043),linewidth(1.pt) + dotstyle); label("$L$", (2.1891528301877243,1.0965832118488765), NE * labelscalefactor); dot((8.081338196952771,3.3591980179392906),linewidth(1.pt) + dotstyle); label("$N$", (7.8432081441375425,3.5313438733583618), NE * labelscalefactor); dot((3.399779090101343,1.4131980251991205),linewidth(1.pt) + dotstyle); label("$H$", (3.190109991030515,1.4753237591947965), NE * labelscalefactor); dot((1.5303158821608756,2.821302415665919),linewidth(1.pt) + dotstyle); label("$X$", (1.5669362166908547,2.8369862032241753), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Denote the foot of the perpendicular from $C$ to $\overline{AB}$ as $D$, the orthocenter of $\triangle ABC$ as $H$ and $\overline{ML}\cap\overline{AB} = X $. It is enough to show that $X,K,N$ are collinear. First of all note that $\measuredangle{HDA} = 90^{\circ}$ and since $BC$ is the diameter of $\omega_2$ we deduce that $C$ is the midpoint of $\overset{\huge\frown}{KM} \Longleftrightarrow \measuredangle{HDM} = \measuredangle{KDH}$.Those conditions imply that $$-1 = (M,K;H,A)$$Similarly we get : $-1 = (L,N;H,B)$. We can now project from $X$: $$(L,N;H,B)=-1 = (M,K;H,A)\overset{X}{=}(L,\overline{XK}\cap\overline{BN};H,B)$$from which we derive that $\overline{XK}\cap\overline{BN} = N$ or $X,K,N$ collinear. Sorry for the bad write up. I am still learning

A problem I came across in EGMO Let $D$ be the second intersection of $\omega_1$ and $\omega_2;$ clearly $D,E,F$ are the feet of the altitudes from $C,B,A$ to $BA,AC,CB,$ respectively. We will now prove that $\measuredangle CHM = \measuredangle DMC.$ But this is easy, we have $$\measuredangle CHM = \measuredangle CHF = 90^\circ - \measuredangle FCH = 90^\circ - (90^\circ - \measuredangle DBC) = \measuredangle DBC = \measuredangle DMC.$$Hence we know that $\triangle CHM \sim \triangle CMD.$ Consequently, $CH \cdot CD = CM^2.$ Similarly, we can deduce that $$CH \cdot CD = CM^2 = CN^2 = CK^2 = CL^2.$$Therefore, not only do we know that $K,L,M,N$ are concyclic, but we also know that the polar of $H$ is just the line through $D$ perpendicular to $DH,$ which is precisely $AB.$ On top of that, if $X = ML \cap NK$ and $Y = KL \cap MN,$ then by Brocard the polar of $H$ is $XY.$ Therefore, the lines $XY$ and $AB$ are equivalent to each other. In particular, $X,$ which is the intersection of $ML$ and $NK,$ lies on $AB,$ as desired.

Clearly, $E$ and $F$ are just the feet of the altitudes. Let $P$ be the feet of the altitude from $C$ to $AB$ (so that $P$ lies on both $\omega_1$ and $\omega_2$). Let $H$ denote the orthocenter of $\triangle ABC$. Claim: $MKNL$ is cyclic. By Power of a Point, $$HN\cdot HL=HP\cdot HC=HK\cdot HM,$$hence shown. Furthermore, note that the perpendicular bisector of $NL$ is side $AC$, and the perpendicular bisector of $KM$ is side $BC$. Hence, since $MKNL$ is cyclic, its circumcenter is the intersection of these two perpendicular bisectors, which is $C$. Note that $\angle ALC=\angle ANC=90$, so $AN$ and $AL$ are tangent to $(MNKL)$. Hence, $MNKL$ is a harmonic quadrilateral. Let $NK$ intersect $AB$ at $T_1$, and let $ML$ intersect $AB$ at $T_2.$ Finally, let $KL$ intersect $AB$ at $X$. Projecting through $K$, we have $$(MK;NL)=^K(AB;T_1X)=-1.$$Similarly, $$(MK;NL)=^L(T_2X;BA)=-1.$$Hence, $T_1=T_2$, as desired.

Relatively easier problem Since H is on the radax of the two circles (it's on CD), HL\cdot HN=HK\cdot HM so KLMN is cyclic; in particular, since the perp. bisectors of MK,NL are BC,AB, respectively (due to them passing through the center and perp.), C is the center of (KLMN). It suffices to prove that AB is the polar of H, which means it suffices to prove CH\cdot CD=CM^2, since it would finish by Brocard's. Indeed, CMB\sim CFM\implies CM^2=CF\cdot CB=CH\cdot CD, as desired.

IAmTheHazard wrote: We use the actual wording that Evan wrote Evidently, $E$ and $F$ are the feet of the obvious altitudes, so add $D$, the foot of the $C$-altitude, so $ADFC$ and $BDEC$ are cyclic. By radical center on these two circles, it follows that $KLMN$ is cyclic as well. Furthermore, because $C$ lies on the perpendicular bisectors of $\overline{LN}$ and $\overline{KM}$, $C$ is the center of $(KLMN)$. Now, note that since $\angle ALC=\angle ANC=90^\circ$, $A$ is the intersection of the tangents at $L$ and $N$ to $(KLMN)$, hence $A$ and $E$ are polar and $CE\cdot CA=r^2$ where $r$ is the radius of $(KLMN)$. But since $AEHD$ is also cyclic, we have $CH \cdot CD=r^2$ as well, hence $\overline{AB}$ is the polar of $H$. By Brocard, it thus follows that we should have $\overline{KN} \cap \overline{LM} \in \overline{AB}$, as desired. $\blacksquare$ ive done this problem before its actually over We first prove the following lemma. Lemma: Let $\ell$ be a line and $A$ be a point on it. Suppose lines $\ell,\ell_1,\ell_2$ are concurrent at a point $X$. Let $P_1,P_2 \in \ell_1$ and $Q_1,Q_2 \in \ell_2$ such that $\ell$ bisects $\angle P_1AP_2$ and $\angle Q_1AQ_2$. Then $\overline{P_1Q_1} \cap \overline{P_2Q_2}$ lies on the perpendicular to $\ell$ through $A$. Proof: Let the perpendicular to $\ell$ through $A$ intersect $\ell_1$ and $\ell_2$ at $T_1$ and $T_2$ respectively. Then $(P_1,P_2;X,T_1)=(Q_1,Q_2;X,T_2)=-1$, so by Prism Lemma $\overline{P_1Q_1} \cap \overline{P_2Q_2}$ lies on $\overline{T_1T_2}$ as desired. Use Evan's wording, and let $D$ be the foot of the $C$-altitude. Clearly $\overline{CD}$ bisects $\angle LDN$ and $\angle KDM$, so we are done by the lemma. $\blacksquare$

v_Enhance wrote: Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent. We first note $\omega_1$ and $\omega_2$ will pass through the feet of the altitudes in $\triangle ABC$. We see $KLMN$ is cyclic, as \[HK \cdot HM = HB \cdot HE = HC \cdot HG = HL \cdot HN,\] where $G$ is the foot from $C$ to $AB$. In particular, the center of $(KLMN)$ is the intersection of the perpendicular bisectors of $KM$ and $LN$, or $C$. Next we find that the inverse of $H$ with respect to $(KLMN)$ is $G$, as \[CH \cdot CG = CF \cdot CB = CK^2.\] Thus $AB$ is the polar of $H$, and Brocard tells us $ML \cap NK$ lies on $AB$. $\blacksquare$

short. Step 1: Consider $(BC)$; the power of $A$ with respect to this circle is $AX^2=AY^2=AK^2=AL^2$, and now I claim $H=XY\cap KL$ has polar $BC$. Step 2: Previous claim is trivial after noting that inverting $B$ and $C$ around $(YKXL)$ maps them to the feet of the $C$ and $B$-altitudes.

Let $D$ be the second intersection point. Note that $AF$, $BE$, and $CD$ concur at the orthocenter $H$ of $\triangle ABC$. Then from Radical Axis, we have $KLMN$ cyclic since $\overline{KM} \cap \overline{LN} \cap \overline{CD} = H$. Then let $\overline{KL} \cap \overline{MN} = X$ and $\overline{KN} \cap \overline{LM} = Y$. By Brokard's, $H$ is the pole of $XY$ wrt $(KLMN)$. We wish to show that $AB$ is also the pole of $H$ which would imply that $X \in AB$. Note that $\angle MLC = \angle MNC = 90^{\circ} \implies LA, LN$ tangents to $(MLKN)$ so $MLKN$ is a harmonic quadrilateral. We also have $(A, H; K, M) = -1$ and similarly $(B, H; L, N) - 1$. Note that $(A, H; K, M) \overset{X}= (B, H; XK \cap BN, L) = -1$ which implies that $XK \cap BN = N$ which finishes by symmetry.

We rephrase the problem because $C-$ centered problems are... annoying. Great problem though Problem: Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $\overline{AC}$, intersects side $\overline{AB}$ at $F$ (other than $A$). Circle $\omega_2$, with diameter $\overline{AB}$, intersects side $\overline{AC}$ at $E$ (other than $A$). Ray $CF$ intersects $\omega_2$ at $K$ and $M$ with $CK < CM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $BC, ML, NK$ are concurrent. Solution: We know that $BE,CF$ are altitudes. Let $AD$ be the third altitude and $H$ be the orthocenter. Note that since $\overline{AC}$ is the diameter of $\omega_1$, and $BN\perp AC$. Hence, $ALNC$ is a cyclic kite, which means that $(A,C;L,N)=-1$. Projecting onto line $BN$ with perspective at $D$ gives $(B,H;L,N)=-1$. Similarly, we get $(C,H;K,M)=-1$. Let $T=LK\cap BC$, $X_1=NK\cap BC$ and $X_2=ML\cap BC$. Projecting $(B,H;L,N)$ onto line $BC$ with perspective at $K$ gives $(B,C;T,X_1)=(B,H;L,N)=-1$, while projecting $(C,H;K,M)$ onto line $BC$ with perspective at $L$ gives $(B,C;T,X_2)=(C,H;K,M)=-1$. Hence, $(B,C;T,X_1)=(B,C;T,X_2)\Longrightarrow X_1=X_2$. Hence, $ML, NK, BC$ concur at a point. $\blacksquare$

Swap the definitions of $A$ and $C$ as @above does, and let $H$ be the orthocenter of $\triangle ABC$. By symmetry, we know that $E$ is the midpoint of $\overline{LN}$, and since $$BL \cdot BN = BD \cdot BC = BH \cdot BE,$$by some famous harmonic bundle lemma we have $(B, H; L, N) = -1$. So, $$-1 = (B, H; L, N) \overset{M}{=} (B, C; \overline{ML} \cap \overline{BC}, \overline{MN} \cap \overline{BC}).$$Analogously, we find that $(B, C; \overline{NK} \cap \overline{BC}, \overline{MN} \cap \overline{BC}) = -1$, so $ML$ and $NK$ intersect $BC$ at the same point, and we are done.

v_Enhance wrote: Let $ABC$ be an acute triangle. Circle $\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, $NK$ are concurrent. Notice $KLMN$ is cyclic with center $C$, as \[CK = CM = \sqrt{CF \cdot CB} = \sqrt{CE \cdot CA} = CL = CN.\] By Brocard, we have that $ML \cap NK$ lies on the polar of $MK \cap NL = H$, which is simply $AB$. $\blacksquare$

USA 2013 TST Let the tangent at $A$ intersect $\omega_2$ at $X$ and $Y$, then we have the claim. Claim: $X-H-Y$ are collinear. Pf: Radical axis. Then we have that $(AH;KM)=-1$, and $(B,H;N,L)=(B,H;L,N)=-1$, applying prism Lemma finishes the problem.

Similar as the above posted but doing it anyway. It is obvious that $H$ lies on the radical axis of the mentioned circles which implies $HK.HM=HL.HN$ giving us that $(KLMN)$ is cyclic. Since $AC$ is a diamter in a circle having $KM$ as a chord perpendicular to it, this implies it bisects it. Similarly $C$ also lies on the $\perp$ bisector of $LN$ which implies $C$ is the center of $(KLMN)$. Now it is obvious that $CL \perp AL$ so $AL$ and similarly $AN$ are tangent to $(KLMN)$ so $A$ is pole of $LN$. Similarly $B$ is pole of $KM$, so $AB$ is the polar of $KM \cap LN$ which is $H$, but by brokard's, $KN \cap LM$ must lie on this polar which is what we needed to show.

who else is here from egmo? (Using @v-Enhance’s statement) Observe that a)The two given circles meet in $D$, the foot of perpendicular from $C$ on $AB$. b) $CD,BE,AF$ concur in $H$, the orthocentre of $\triangle ABC$ c)$MLKN$ is cyclic since $NH \cdot HL=DH \cdot HC=KH \cdot HM$ d) $C$ is the centre of $\odot (MLKN)$ since the perpendicular bisectors $LN,KM$ of $AC,BC$ meet in $C$ e)$AB$ is the polar of $H$ since $AB \perp CD$ and $CE \cdot CA=CN^2=CH \cdot CD$, so inversion around $\odot(MLKN)$ sends $H$ to $D$ f)$ML,KN$ meet in the polar of $H$, that is, $AB$ $\blacksquare$