Let $D, E, F$ be the feet of the altitudes from $A, B, C$ respectively and denote by $T$ the intersection of the tangents to $\Gamma$ at $B$ and $C$. Let $O'$ be the center of $\odot(BHC)$. Define $P = DE \cap HC$ and $Q = DF \cap HB$. First note that a homothety at $A$ takes the nine-point circle to $\odot(BHC)$, so $A, K, O'$ collinear. Next, note that $P$ is the radical center of $\odot(BHC), \odot(HDCE)$, and the nine-point circle, and similar observations with $Q$ imply that $PQ$ is the radical axis of $\odot(K)$ and $\odot(BHC)$, i.e. $PQ \perp O'K$. So it remains to show that $LM \parallel PQ$. It is easy to see that $PD \parallel MT$ and $HD \parallel OT$ and $HP \parallel OM$. So $\triangle HDP \sim OTM$. Similarly $\triangle HDQ \sim \triangle OTL$. Thus (by similar quadrilaterals, or some applications of Desargues) $\triangle HPQ \sim \triangle OML$. But two pairs of sides are parallel, so the third pair of sides is parallel, i.e. $LM \parallel PQ$ as desired.

parmenides51 wrote: Let $ABC$ be a triangle inscribed in circle $\Gamma$ with center $O$. Let $H$ be the orthocenter of triangle $ABC$ and let $K$ be the midpoint of $OH$. Tangent of $\Gamma$ at $B$ intersects the perpendicular bisector of $AC$ at $L$. Tangent of $\Gamma$ at $C$ intersects the perpendicular bisector of $AB$ at $M$. Prove that $AK$ and $LM$ are perpendicular. by Michael Sarantis, Greece Let $A’$ be a point such that $ABCA’$ is a parallelogram. Let $K_c, K_b$ be points on $\Gamma$ such that $K_cC \parallel AB$, $K_bB \parallel AC$. Let $P$ be the reflection of $O$ in $BC$. Note that $A, K, P$ are collinear. Also, $AP \parallel OA’$ as $A, P, O, A’$ form a parallelogram. So, $AK \parallel PA’$. It is easy to see that $BK_b, CK_c$ are the polars of $L, M$ respectively. Since $OP \perp LM$, we have $AK \perp LM$. $\blacksquare$

Complex coordinates: $|a|=|b|=|c|=1$. Then we have $$h=a+b+c\wedge k=\frac{a+b+c}{2}$$$$OM\perp AB\iff \frac{m}{a-b}=-\overline{\left(\frac{m}{a-b}\right)}\iff \overline{m}=\frac{m}{ab}$$$$CM\perp OC\iff \frac{c-m}{c}=-\overline{\left(\frac{c-m}{c}\right)}$$Two equations give $$m=\frac{2abc}{ab+c^2}$$. In the same way we obtain $$l=\frac{2abc}{ac+b^2}$$Finally $$\frac{l-m}{k-a}=\frac{4abc\cdot (ab+c^2-ac-b^2)}{(ab+c^2)(ac+b^2)(b+c-a)}=\frac{4abc\cdot (c-b)}{(ab+c^2)(ac+b^2)}$$Clearly $$\frac{l-m}{k-a}=-\overline{\left(\frac{l-m}{k-a}\right)}\iff LM\perp AK$$

[asy][asy] size(250); pair P=dir(105),Q=dir(-120),R=dir(-60),O=incenter(P,Q,R),A=foot(O,Q,R),B=foot(O,R,P),C=foot(O,P,Q),K=circumcenter( 1/2*(B+C),1/2*(C+A),1/2*(A+B)),L=extension(P,B,Q,O),M=extension(P,C,R,O),IP=2*dir(-90)-O,T=circumcenter(P,Q,R),IR=extension(Q,IP,R,O),IQ=extension(R,IP,P,IR); draw(circumcircle(A,B,C),blue); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$K$",K,dir(230)); dot("$M$",M,dir(180)); dot("$L$",L,dir(90)); dot("$I_P$",IP,dir(IP)); draw(P--Q--R--P,red); draw(M--L,brown); draw(IP--T^^A--K,green); draw(Q--L^^R--M,purple); [/asy][/asy] Let $PQR$ be the tangential triangle of $ABC$. Note $\overline{QL},\overline{RM}$ are angle bisectors. Let $I_P,I_Q,I_R$ be the excenters of $\triangle PQR$. By Theorem 2.7.2 in Muricaaaaaaa (on Page 50), we know line joining $I_P$ and nine-point center of $\triangle I_PI_QI_R$ is perpendicular to $\overline{LM}$. Since $\triangle ABC, \triangle I_PI_QI_R$ are homothetic, thus $\overline{LM} \perp \overline{AK}$. $\blacksquare$

Let $X = ML \cap AK$ just draw Tangent from X to nine-point circle with angle bisector & harmonic

nice bashing exercise