Let $A',I$ the foot of $A$ and the midpoint of $BH$ ; we have $\angle AKA'= \angle BKA=\pi - \angle C \implies \angle BKA'=2\angle C$ but $I $ is the circumcenter of $BHA' \implies \angle BIA'= 2\angle C$ thus $BKA'I$ cyclic so $\angle BKI = \angle BA'I =\angle IBC = \frac{\pi}{2}- \angle C$ in the other hand $\angle BKC=2\pi -\angle BKA -\angle AKC=2\pi -(\pi -\angle C)-(\frac{\pi}{2} ) =\frac{\pi}{2} +\angle C$ hence $C,K,I $ are collinear $\square $

Let $F$ denote the foot of the altitude from $C$ and $D$ denote the foot of the altitude from $A$. Then $AFKDC$ and $ABKH$ are cyclic. From these cyclic figures we get $\angle HBK = \angle HAK = \angle DAK = \angle DCK$ and $\angle BHK = \angle BAK = \angle FAK = \angle FCK$. This means that the circumcircle of $HKC$ is tangent to $BH$ at $H$ and that the circumcircle of $BKC$ is tangent to $BH$ at $B$. In these two circles $KC$ is a common chord and $BH$ is a common tangent. The result follows, since the radical axis of two circles (the common chord) bisects a common tangent. [asy][asy] import olympiad; size(8cm); pair A=(3,6), B=(0,0), C=(9,0), D, E, F, H, K,M; H=orthocenter(A,B,C); D=foot(A,B,C); F=foot(C,A,B); M=(B+H)/2; K=intersectionpoint(C--M,circumcircle(A,C,D),0); draw(A--B--C--cycle); draw(circumcircle(A,D,C),dotted); draw(circumcircle(A,B,H),dotted); draw(C--F); draw(C--M); draw(B--H); draw(A--D); draw(B--K--A); draw(K--H); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,NE); label("$K$",K,S); label("$H$",H,NE); label("$D$",D,S); label("$F$",F,W); [/asy][/asy]

Invert around $\odot (BH) $ and note that the midpoint of $BH $ is the intersection of the tangents at $AH\cap BC $ and $CH\cap AB $ to the circle with diameter $AC $.

tchebytchev wrote: Let $ABC$ be a triangle with $H$ its orthocenter. The circle with diameter $[AC]$ cuts the circumcircle of triangle $ABH$ at $K$. Prove that the point of intersection of the lines $CK$ and $BH$ is the midpoint of the segment $[BH]$ [asy][asy] import olympiad; size(8cm); pair A=(3,6), B=(0,0), C=(9,0), D, E, F, H, K,M; H=orthocenter(A,B,C); D=foot(A,B,C); F=foot(C,A,B); M=(B+H)/2; K=intersectionpoint(C--M,circumcircle(A,C,D),0); draw(A--B--C--cycle); draw(circumcircle(A,D,C),dotted); draw(circumcircle(A,B,H),dotted); draw(C--F); draw(C--M); draw(B--H); draw(A--D); draw(B--K--A); draw(K--H); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,NE); label("$K$",K,S); label("$H$",H,NE); label("$D$",D,S); label("$F$",F,W); [/asy][/asy] \(\angle BHK = \angle BAK = \angle HCK ; \angle HBK = \angle DAK = \angle BCK \Rightarrow K=C\) vertexed HM point of \(\Delta BCH,\) which clearly \(\in C-\) median of the \(\Delta BCH.\)

Let $M$ be the midpoint of $BH$ and $AD$ be the $A-$altitude.Notice that $\angle BKD=360^{\circ}-\angle AKB-\angle AKD=360^{\circ}-2(180^{\circ}-\angle C)=2\angle C$.It is well known that $\angle BMD=180^{\circ}-2(90^{\circ}-\angle C)=2\angle C$.Hence,$B,M,K,D$ are concyclic $\implies \angle AKM=\angle AKB-\angle BKM=\angle AKB-\angle BDM=180^{\circ}-\angle C-(90^{\circ}-\angle C)=90^{\circ}$.So,we conclude that $C,K,M$ are collinear.

Let $E=AH\cap BC$,$F=CH\cap AB$. It is clear that $ACKE$ and $AFKC$ are cyclic so, $\angle HBK=\angle HAK=\angle EAK=\angle ECK=\angle BCK$ So $(BKC)$ is tangent with $BH$. $\angle BHK=\angle BAK=\angle FAK=\angle HCK$ So $(HKC)$ is also tangent to $BH$ and we get the desired result.

Let $KC \cap HB=S$ By cyclics we obtain: Let: $\angle BHK=\angle BAK=\alpha ; \angle KAH=\angle KCD=\beta \implies \angle KCH=\alpha$ $\angle BCK=\angle KAD=\angle HBK=\beta$ By PoP: $SB^2=SH^2 \implies SB=SH$.$\blacksquare$

Let $AK \cap BC = T$,$AH\cap BC = A'$ and $CH \cap AB = C'$ $\angle BKT = \pi - \angle BKA = \pi - \angle BHA = \angle C$ $\angle TKA' = \angle ACA' = \angle C$ and $\angle TKC = \frac{\pi}{2} \Rightarrow (B,A';T,C)=-1$ $-1= (B,A';T,C)=A(C',A';K,C) = C(H,B;CK\cap BH,\infty) \Rightarrow CK\cap BH$ is the midpoint of $BH$

Good Problem Let $I$ b the intersection of lines $CK$ and $BH$, $$\boxed{\textbf{Claim:} FIKH \text{ is cyclic quad}}$$Proof: $$\measuredangle HIC=\measuredangle HBC+\measuredangle ICB $$$$ \measuredangle KFH=\measuredangle DAC+\measuredangle KAD$$ $$\boxed{\textbf{Claim:} AK \text{bisect} \measuredangle FKH}$$proof: $$\measuredangle FKA= \measuredangle FCA=\measuredangle AKH=\measuredangle FBH$$$$\measuredangle FIH=\measuredangle FKH=2\times\measuredangle FBH$$ we know that $\triangle HFB$ is right angle triangle and $$\measuredangle BFI =\measuredangle IBF$$so $$FI=IB=IH$$hence we are done $\blacksquare$

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