Let $S$ be the intersection of $\overline{AO}\cap (BOC)$. I contend that $S$ lies on $(ALK)$ and $S$ satisfies the given angle condition, which determines the point uniquely. Let $P=\overline{BL}\cap \overline{CK}$, $E=\overline{AD}\cap \overline{BL}$ and $F=\overline{AD}\cap \overline{CK}$. Observe that $\overline{BL}$ is tangent to $(ABC)$ as $L\in (ABD)$ and $\measuredangle ABL=\measuredangle ADL=\measuredangle ACB$. In the same manner, we have $\overline{CK}$ tangent to $(ABC)$. Claim: $S$ lies on $(ABE)$ and, similarly, $S$ lies $(ACF)$. Proof. Indeed, \begin{align*} \measuredangle ASB&= \measuredangle OSB=\measuredangle OCB=90^\circ+\measuredangle CAB\\&=90^\circ+\measuredangle DBE=\measuredangle DEB=\measuredangle AEB. \end{align*}Claim: $P$ is the circumcenter of $\triangle SEF$ and $\triangle ABC\sim\triangle SEF$. Proof. Indeed, \begin{align*} \measuredangle SPE&= \measuredangle SPB=\measuredangle SOB=\measuredangle AOB=2\measuredangle OAB=2\measuredangle SAB=2\measuredangle SEB=2\measuredangle SEP, \end{align*}thus $PS=PE$, similarly $PS=PF$, hence $P$ is the circumcenter of $\triangle SEF$. As $\triangle AOB\sim\triangle SPE$ and $\triangle AOC\sim\triangle SPF$, $\triangle ABC\sim\triangle SEF$. Note that $\measuredangle ASP=\measuredangle OSP=\measuredangle OCP=90^\circ=\measuredangle AKC=\measuredangle AKP=\measuredangle ALP$, which implies that $S$ belongs to the circumcircle of $\triangle ALK$. On top of that, $ \measuredangle SBA=\measuredangle SEA=\measuredangle SEF=\measuredangle ABC$ and $ \measuredangle SCA=\measuredangle SFA=\measuredangle SFE=\measuredangle ACB$. We are done. [asy][asy] import geometry; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,O,D,E,F,P,S,L,K; A=dir(120); B=dir(200); C=dir(340); D=foot(A,B,C);O=(0,0);P=intersectionpoint(perpendicular(B,line(O,B)),perpendicular(C,line(C,O))); E=extension(A,D,B,P);F=extension(A,D,C,P);S=intersectionpoints(circumcircle(B,O,C),circle(P,abs(P-E)))[1]; L= (D-A)/(A-C)*(A-B)+A;K=(D-A)/(A-B)*(A-C)+A; draw(A--B--C--cycle, red+1);draw(circumcircle(A,B,C),royalblue);draw(circumcircle(B,O,C),royalblue); draw(C--F,orange);draw(B--P,orange);draw(A--F,orange);draw(A--S,red+0.3);draw(circumcircle(A,B,E),royalblue+dotted); draw(circumcircle(A,K,L),royalblue+dashed);draw(B--L,orange);draw(C--K,orange);draw(S--P,orange); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$P$",P,dir(P)); dot("$S$",S,dir(S)); dot("$O$",O,dir(90)); dot("$K$",K,dir(K)); dot("$L$",L,dir(L)); [/asy][/asy]

Perform inversion with centre $A$ and radius $\sqrt{AB.AC}$ followed by reflection in angle bisector of $\angle BAC$. By simple angle chasing we get that $K$ goes to point $K'$ s.t. $\angle ABK'=90^{\circ}$ and $\angle AK'B=\angle ABC$ and $L$ goes to point $L'$ s.t. $\angle ACL'=90^{\circ}$ and $\angle AL'C=\angle ACB$. Also $S$ goes to $H$ where $H$ is orthocentre of $\triangle ABC$. Let $H_M$ be $A-\text{Humpty Point}$ of $\triangle ABC$. Note that given problem is equivalent to proving that $H,K'$ and $L'$ are collinear. But now note that $\angle AH_MB=180^{\circ}-\angle ABC$ and $\angle AK'B=\angle ABC$. Hence $AH_MBK'$ is cyclic. So $\angle AH_MK'=\angle ABK'=90^{\circ}$. Similarly $\angle AH_ML'=90^{\circ}$. Also it is well known that $H_M$ is feet of $\perp$ from $H$ to $A-$median. Hence $\angle AH_MH=90^{\circ}$. Thus $H,K',L'$ all lie on $\perp$ to $AH_M$ at point $H_M$ and hence are collinear as desired.

Let $KB\cap LC=T.$ Clearly $S\notin BC,$ so easy to conclude that $A$ is the $S-\text{excenter}$ in $\triangle BCS.$ Next $$\measuredangle ABC=\measuredangle ADL=\measuredangle ACL,$$so $CL$ and analogously $KB$ tangents to $\odot (ABC),$ which yields that $T$ is the midpoint of arc $BSC.$ Hence $$\angle AST=90^{\circ}=\angle ALT=\angle AKT,$$done.

Solved with sriraamster. First, observe that \[\measuredangle LAB = 90^\circ - \measuredangle ABL = 90^\circ - \measuredangle ADL = 90^\circ - \measuredangle ACB,\]and by symmetry this means that $\measuredangle CAK = 90^\circ - \measuredangle CBA\implies \measuredangle LAK = 2\measuredangle BAC.$ Next, let $T$ be the intersection of $BL$ and $CK.$ Then since \[\measuredangle TBC = 180^\circ - \measuredangle CBA - \measuredangle ABL = 180^\circ - \measuredangle CBA - \measuredangle ACB = \measuredangle BAC\]and similarly $\measuredangle BCT = \measuredangle BAC,$ it follows that $\measuredangle CTB = 180^\circ - 2\measuredangle BAC.$ On the other hand, we also have \[\measuredangle CSB = 180^\circ - \measuredangle SBC - \measuredangle BCS = 180^\circ - (180^\circ - 2\measuredangle CBA) - (180^\circ - 2\measuredangle ACB) = 180^\circ - 2\measuredangle BAC,\]so in fact $BCST$ and $ALPK$ are cyclic. The final critical observation is that $A$ is the $S$-excenter of $\triangle BCS.$ This implies that $SA$ is the angle bisector of $\angle BSC,$ so it passes through $M,$ the midpoint of arc $\widehat{BC}$ on $(BCS)$ not containing $S.$ However, $T$ must be the midpoint of arc $\widehat{BSC}$ on $(BCS)$ since $\triangle BCT$ is isosceles, and thus $MT$ is a diameter of $(BCS),$ which means that $\measuredangle AST = \measuredangle MST = 90^\circ = \measuredangle AKT=\measuredangle ALT,$ so $A,K,L,S$ are concyclic as desired. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -15.39606385487529, xmax = 23.694893061224473, ymin = -14.044471292516992, ymax = 9.231428934240359; /* image dimensions */ /* draw figures */ draw(circle((-3.284703129251704,2.1109185487528337), 3.982197386644101), linewidth(0.7)); draw(circle((1.7652968707482957,2.1109185487528337), 5.613367762015716), linewidth(0.7)); draw(circle((-0.7597031292517039,-2.392248745974918), 8.506439430568166), linewidth(0.7)); draw((-7.087510946913467,3.292675176881353)--(-2.3094062585034085,-1.75), linewidth(0.7)); draw((-2.3094062585034085,-1.75)--(7.184765563303621,0.6482987341564392), linewidth(0.7)); draw((5.84,-1.75)--(-2.3094062585034085,5.971837097505667), linewidth(0.7)); draw((-2.3094062585034085,5.971837097505667)--(-4.26,-1.75), linewidth(0.7)); draw((-4.26,-1.75)--(5.84,-1.75), linewidth(0.7)); draw((-7.087510946913467,3.292675176881353)--(0.79,-10.756334589455502), linewidth(0.7)); draw((0.79,-10.756334589455502)--(7.184765563303621,0.6482987341564392), linewidth(0.7)); draw((-4.26,-1.75)--(6.142690534832854,-7.363815248791734), linewidth(0.7)); draw((6.142690534832854,-7.363815248791734)--(5.84,-1.75), linewidth(0.7)); draw((-7.087510946913467,3.292675176881353)--(-2.3094062585034085,5.971837097505667), linewidth(0.7)); draw((-2.3094062585034085,5.971837097505667)--(7.184765563303621,0.6482987341564392), linewidth(0.7)); draw(circle((0.79,-4.837358246846166), 5.918976342609339), linewidth(0.7)); draw((-2.3094062585034085,-1.75)--(-2.3094062585034085,5.971837097505667), linewidth(0.7)); draw((-2.3094062585034085,5.971837097505667)--(6.142690534832854,-7.363815248791734), linewidth(0.7)); draw((0.79,1.0816180957631731)--(0.79,-10.756334589455502), linewidth(0.7)); draw((0.79,-10.756334589455502)--(6.142690534832854,-7.363815248791734), linewidth(0.7)); /* dots and labels */ dot((-4.26,-1.75),dotstyle); label("$B$", (-5.186095600907038,-2.337505306122444), NE * labelscalefactor); dot((5.84,-1.75),dotstyle); label("$C$", (6.189677641723344,-1.9477638095238048), NE * labelscalefactor); dot((-2.3094062585034085,5.971837097505667),dotstyle); label("$A$", (-2.74401850340137,6.382304217687073), NE * labelscalefactor); dot((-2.3094062585034085,-1.75),linewidth(4pt) + dotstyle); label("$D$", (-2.2128257596371976,-1.4614418140589525), NE * labelscalefactor); dot((7.184765563303621,0.6482987341564392),linewidth(4pt) + dotstyle); label("$K$", (7.421079002267562,0.6599096598639481), NE * labelscalefactor); dot((-7.087510946913467,3.292675176881353),linewidth(4pt) + dotstyle); label("$L$", (-7.717914195011346,3.509034376417234), NE * labelscalefactor); dot((6.142690534832854,-7.363815248791734),linewidth(4pt) + dotstyle); label("$S$", (6.382838639455771,-7.766738866213142), NE * labelscalefactor); dot((0.79,-10.756334589455502),linewidth(4pt) + dotstyle); label("$T$", (0.6845892063491962,-11.602394195011326), NE * labelscalefactor); dot((0.79,1.0816180957631731),linewidth(4pt) + dotstyle); label("$M$", (0.8777502040816224,1.2635377777777799), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

We just need to prove there's a spiral similarity $\phi$ centered at $S$ taking $LC \to KB$ (i.e. $S$ be the miquel point of $KLCB$). Easy angle chase to get $\angle LCS=\angle KBS$, and thus we just need $KB/BS=LC/CS \implies KB/LC=BS/CS$. Now, again by angle chase, if $\angle B=y,\angle C=x$, then by sine law on $BCS$, we get $BS/CS=sin 2y/sin 2x=( sin y/sin x )(cos y/cos x)=AB/AC.[DC/AC]/[DB/AB]=(AB/AC)^2.DC/DB$. Then, notice that by easy angle chase and because of the tangents, we get $\triangle AKB\sim \triangle ADC, \triangle ALC\sim \triangle ADB$. Thus $KB/DC=AB/AC$ and $LC/BD=AC/AB$ which implies $KB/LC=(AB/AC)^2.DC/DB$, and thus the ratios are equal, proving what we wanted. $\blacksquare$

Lemma$1 :$ Let $F$ be center of $A$-excircle and $M$ be midpoint of arc $BAC$. $\angle FAM = \angle 90$. Proof $:$ Let perpendicular bisector of $BC$ meet $ABC$ at $M,N$. we have $\angle NAM = \angle 90$ so we need to prove $A,N,F$ are collinear. $AF$ is angle bisector and $N$ is midpoint of arc $BC$ so $N$ lies on $AF$. Let $\omega$ be circumcircle of $ABC$. $\angle ACD = \angle ADK = \angle ABK$ and $\angle ABD = \angle ADL = \angle ACL$ so $BK$ and $CL$ are tangent to $\omega$. Let $BK$ and $CL$ meet at $E$. note that $AKEL$ is cyclic. $\angle 180 - \angle ABS = \angle ABC$ and $\angle 180 - \angle ACS = \angle ACB$ so $BA$ and $CA$ are exterior angle bisectors of $\angle BSC$ so $A$ is center of $S$-excircle so $ \angle BSC = \angle 180 - 2\angle BAC = \angle BEC$ so $BCSE$ is cyclic and note that $T$ is midpoint of arc $BSC$ so by the lemma $\angle ASE = \angle 90$ so $ALSEK$ is cyclic. we're Done.

Make a $\sqrt{bc}$ inversion, let $A'$ the A-antipode in $(ABC)$, let $O$ the circumcenter and let $H$ the ortocenter and let $BH,CH$ meet $(ABC)$ again at $H_B,H_C$ respectivily. Now we have that the circles with diameters $AB,AC$ are send to $CA',BA'$ respectivily and now note that $\angle LAB=\angle LDB=\angle DAC$ and $\angle KAC=\angle KDC=\angle BAD$ so we have that $\angle LAD=\angle BAC=\angle KAD$ which means that $AH_C \cap BO'=K'$ and $AH_B \cap CO'=L'$ and now by pascal in $(ABC)$ we get that $K',H,L'$ are colinear but guess what since the reflection of $A$ over $BC$ inverts into $O$ in this inversion one can do a similar logic to end with that $S'$ is $H$ so inverting back we get $AKSL$ cyclic as desired thus we are done

Let $H$ and $O$ be the orthocenter and circumcenter of $ABC$ respectively, $E = BH \cap AC$, $F = CH \cap AB$, the medial triangle of $ABC$ be $MNP$, the midpoints of $BH$ and $CH$ be $X$ and $Y$ respectively, and $T = BL \cap CK$. In addition, we redefine $S$ as the second intersection between $AO$ and $(BOC)$. The existence of the Nine-Point Circle implies $DFPY$ is cyclic. Thus, $$\angle PDY = \angle PFY = 90^{\circ} = \angle PDK$$which means $D, K, Y$ are collinear. Now, since $\angle CDH = 90^{\circ}$, we have $YC = YD$. Combining this result with $F \in (AC)$ yields $CDFK$ as an isosceles trapezoid. This gives $$\angle KCA = \angle KFA = \angle CBA$$so $\overline{TCK}$ is tangent to $(ABC)$. Similarly, we find that $\overline{TBL}$ is tangent to $(ABC)$. At this point, it's easy to see $ALTK$ is cyclic with diameter $AT$. Because $T = BB \cap CC$, we know $BOCT$ is cyclic with diameter $OT$. This implies $$\angle AST = \angle OST = 90^{\circ}$$so $ALTSK$ is a cyclic pentagon. Now, observe that $$\angle BCS = \angle BOS = 180^{\circ} - \angle BOA = 180^{\circ} - 2 \angle C$$yielding $$\angle ACB + \angle ACS = \angle ACB + (\angle ACB + \angle BCS) = 180^{\circ}.$$An analogous process gives $\angle ABC + \angle ABS = 180^{\circ}$, which means we've recovered the original definition of $S$, as required. $\blacksquare$ Remark: My solution is very overcomplicated. Merely noticing $$\angle KCA = \angle KDA = \angle DBA = \angle CBA$$is enough to show that $\overline{TCK}$ is tangent to $(ABC)$. By the way, solving ELMO 2016/2 and Taiwan TST Round 2 Quiz 2 2016/1 helped me finish off this question.

Let $T=KB\cap LC$. It's easy to see that $AKTL$ is cyclic. We get $$\angle ADK=\angle C\text{ and }\angle AKD=\angle B\implies\angle KAD=\angle A,$$and similarly $\angle LAD=\angle A$ implying $\angle BTC=180^{\circ}-2\angle A$. Also $$\angle BSC= 180^{\circ}-(\angle CBS+\angle BCS)=180^{\circ}-2\angle A=\angle BTC,$$implying $BCST$ cyclic and therefore $\angle KBS= \angle LCS$. Applying law of sines in $\triangle BCS$ we get $$\frac{BS}{CS}=\frac{\sin(2\angle C)}{\sin(2\angle B)},$$and we have $KB=AB\cdot\sin(\angle KAB)$ and $LC=AC\cdot\sin(\angle CAL)$. Thus we get $$\frac{BS}{CS}=\frac{KB}{LC}\iff\frac{\sin(\angle C)\cos(\angle C)}{\sin(\angle B)\cos(\angle B)}=\frac{AB\cdot\sin(\angle KAB)}{AC\cdot\sin(\angle CAL)}\iff$$$$\iff\frac{\cos(\angle C)}{\cos(\angle B)}=\frac{\sin(\angle A+\angle B-90^{\circ})}{\sin(\angle A+\angle C-90^{\circ})},$$which is true, and thus we get $\triangle SBK\sim\triangle SCL$ $$\implies\angle TKS=\angle TLS\implies ATKLS\text{ is cyclic.}\blacksquare$$

Denote $M$ as the midpoint of $AB$ and $\angle ABC = b, \angle ACB = m$. Then by angle chasing we get \[\angle KBS = \angle LCS, \triangle AKB \sim \triangle ADC, \triangle KBD \sim \triangle DCL, \triangle ABD \sim \triangle ACL\](*) since $AKBD$ and $ADCL$ are cyclic. Claim: $\triangle KBS = \triangle LCS$ Proof:: From (*) and law of sines on $\triangle ABC$ we get \[\dfrac{KB}{DC}=\dfrac{BD}{CL}=\dfrac{AB}{AC}=\dfrac{\sin m}{\sin b}\](1) and also from law of sines on $\triangle KMA$ and $\triangle BMD$ we have \[\dfrac{BD}{\sin 2b}=\dfrac{MB}{\sin b}, \dfrac{KB}{\sin 2m}=\dfrac{MA}{\sin m}=\dfrac{MB}{\sin m}\]$\Rightarrow$ \[\dfrac{KB}{BD} \cdot \dfrac{\sin m}{\sin b} = \dfrac{\sin 2m}{\sin 2b} = \dfrac{BS}{CS}\](2) and so we get from (1) and (2): \[\dfrac{KB}{BD} \cdot \dfrac{\sin m}{\sin b} = \dfrac{KB}{BD} \cdot \dfrac{BD}{CL} = \dfrac{KB}{CL} = \dfrac{BS}{CS}\]and thus since $\angle KBS = \angle LCS$ we have $\triangle KBS = \triangle LCS$. $\triangle KBS = \triangle LCS$ means that $\angle BSC = \angle KSL = 2(b+m)-180^{\circ}$ and since $\angle KAL = 360^{\circ}-2(b+m)$ we get $\angle KAL +\angle KSL = 180^{\circ}$ and thus $A,K,S,L$ are concyclic. $\square$.

Thus, we reduce to the following problem (which is much more aesthetically pleasing compared to the original): Let $A'$ and $H$ be the $A$-antipode and the orthocenter of $\triangle ABC$, respectively. Let $K'$ and $L'$ be the points on lines $A'B$ and $A'C$ such that $\triangle AK'A'\sim \triangle ABC\sim \triangle AA'L'$. Show that $H$ lies on $K'L'$. This can actually be done in a very beautiful way. Consider the midpoints of $BC$, $A'K'$, and $A'L'$, call them $M$, $X$, and $Y$. The spiral similarity centered at $A$ mapping $B$ to $C$ also maps $K'$ to $A'$ and $A'$ to $L'$, so it maps line $A'B$ to line $A'C$. If $P$ is a point moving linearly on line $A'B$, and $Q$ is the image under the spiral similarity, then $Q$ also moves linearly on line $A'C$. The midpoint of $P$ and $Q$ also moves linearly, so $M$, $X$, and $Y$ are collinear. Taking a homothety at $A'$ with a ratio of $2$ gives the desired result.

Let $KB \cap CL=T$. Claim 1:$T$ is the point the tangents to $(ABC)$ at $B$ and $C$ intersects. Proof 1: $\angle KBA=\angle C \implies \angle CBT=\angle A$ $\angle ACL=\angle B \implies \angle TCB=\angle A$ So $T$ is the point the tangents to $(ABC)$ at $B$ and $C$ intersects. Also $TK \perp AK$ and $TL \perp AL$ which means $A,K,T,L$ lie on the circle with diameter $AT$. Let $TS$ intersects $AB,AC,KL$ at $P,Q,X$. Claim 2:: $B,C,S,T$ lie on a circle. Proof 2: $\angle CBS=180-2\angle B$ so $\angle KBS=\angle B+\angle C+180-2\angle B=180-\angle B+\angle C$ $\implies \angle SBT=\angle B-\angle C$ $\angle SCB=180-2\angle C$ and $\angle TCB=\angle A$ so $\angle SCT=\angle B-\angle C$ $\implies \angle SBT=\angle B-\angle C=\angle SCT$ $\implies B,C,S,T$ are cyclic. Claim 3::$A,B,S,Q$ and $B,C,P,Q$ and $A,C,S,P$ are cyclic. Proof 3: $\angle A=\angle TCB=\angle TSB \implies A,B,S,Q$ are cyclic. $\angle BPT=180-\angle PSB-\angle SBP=180-\angle A-\angle B=\angle C \implies B,P,Q,C$ are cyclic. $\angle CSQ=180-\angle PSC=180-\angle BSC-\angle A=180-\angle BTC-\angle A=\angle A \implies A,C,S,P$ are cyclic. Claim 4:$T$ is the center of $(BPQC)$. Proof 4: $\angle QCT=\angle B=\angle PQC \implies TQ=TC$ $\angle TBP=\angle C=\angle TPB \implies TP=TB$ Also we know that $TB=TC$ $\implies TP=TB=TC=TQ$ $\implies T$ is the center of $(BPQC)$. $QB \perp AP$ and $PC \perp AQ$ and $T$ is the midpoint of $PQ$ so $(BCT)$ is the nine-point circle of $APQ$. $S \in (BCT)$ and $S \in PQ \implies S$ is the altitude from $A$ to $PQ$. $\implies \angle TSA=90=\angle AKT=\angle TLA$ $\implies S \in (AKLT)$ as desired.

jrsbr wrote: and thus we get $\triangle SBK\sim\triangle SCL$ $$\implies\angle TKS=\angle TLS\implies ATKLS\text{ is cyclic.}\blacksquare$$ Would you please explain this part? Why $\triangle SBK\sim\triangle SCL$ $\implies\angle TKS=\angle TLS$

Claim: $SBK \sim SCL$ Proof: We have $\angle SBK = \angle SCL = 180^\circ + \angle C - \angle B$ Since $\angle CBS = 180^\circ - \angle 2B$ and $\angle BCS = 180^\circ - \angle 2C$, it results that $\frac{BS}{CS} = \frac{\sin(\angle 2C)}{\sin(\angle 2B)} = \frac{\sin(\angle C)\cos(\angle C)}{\sin(\angle B)\cos(\angle B)}$ Now because $\frac{AB}{2}$ is the radius of $(ABD)$ and similarly $\frac{AC}{2}$ is the radius of $(ACD)$, by applying the law of sines we get: $BK = AB\sin(90 - \angle C) = AB\cos(\angle C)$ and $CL = AC\sin(90 - \angle B) = AC\cos(\angle B)$ So $\frac{BK}{CL} = \frac{AB\cos(\angle C)}{AC\cos(\angle B)} = \frac{\sin(\angle C)\cos(\angle C)}{\sin(\angle B)\cos(\angle B)}$ Finally $\frac{BK}{CL} = \frac{BS}{CS} , \angle SBK = \angle SCK \implies SBK\sim SCL$ From which results that $\angle BSC = \angle KSL$ Hence $\angle KAL + \angle KSL = 180^\circ$ Which means $ALSK$ is concyclic

After inversion from $A$, let $H_b,H_c$ be the intresection of $AK, AL$ and $(ABC)$ then pascal from $AH_c CDB H_b$ implies $K,S,L$ linear.

$\sqrt{bc}$ invert, then $S$ goes to the orthocenter $H$ of $\triangle ABC$ because of the angle conditions. $D$ goes to the antipode $A'$ of $A$ on $(ABC)$. Then $K$ goes to point $K'$ on $CA'$ with $BA'$ tangent to $(AK'A')$ and similarly for $L$. Now the problem becomes the following: Triangle $ABC$. $\omega_1$, $\omega_2$ are circles through $A, C$ and $A, B$ and both tangent to $BC$. $E, F$ are antipodes of $A$ on $\omega_1, \omega_2$. Prove that $EF$ passes through orthocenter of $\triangle ABC$. This is trivial by Humpty point and we are done.