This problem is a particular case of the problem Perpendicular line .

Dear Mathlinkers, 1. E the foot of the B-altitude of ABC B' the circumtrace of BE, A'', B'' the antipoles of A, B wrt (O), (1) the circle with diameter BP A' the second point of intersection of (1) with (O). 2. by considering two time a converse of the Reim theorem, B''P goes through A', then AH goes through A' 3. By the Pascal theorem, we are done... Sincerely Jean-Louis

Let $P$ be the point on $AC$ such that $OP\parallel BC$, we will prove that $\angle PKB=90$. Let $M$ be the foot of the perpendicular line from $P$ to $BC$. Suppose that $N$ is the midpoint of $BC$. Then we have $PM=ON=AK=KH$. $AH\parallel PM$ so $APMK$ and $KPMH$ are parallelograms. So $MK\parallel AC$ and $MH\parallel PK$. So we have to say that $MH$ is perpendicular to $BK$. $MK\parallel AC$, so $BH$ is perpendicular to $MK$. $KH$ is also perpendicular to $BM$, So $H$ is the orthocenter of the triangle $\triangle BKM$. And we are done.

My solution: Let $AD,BE$ are the altitudes of $\triangle ABC.$ Easy to see that $BKEP$ is cyclic. $\Longrightarrow \angle KBP =\angle AEK =\angle KAE =\angle EBC \Longrightarrow \angle KBE =\angle PBC .$ We have $\angle OBP =\angle OBC -\angle PBC = 90^{o} - \angle BAC -\angle KBE =\angle ABE - \angle KBE =\angle ABK (1).$ In the other hand, $\angle BAK = \angle PAO.(2)$ (well-known). From $(1),(2)$ we deduce $ O, K$ is isogonal conjugate WRT $\triangle ABP.$ $\Longrightarrow \angle BPO = \angle APK = \angle KBE = \angle PBC \Longrightarrow OP \parallel BC.$ Done

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Solution:Let $B'$ be the symmetric point of $B$ wrt point $K$.We shall start with the following lemma: Lemma: $m(\measuredangle{KBP})=90^0-m(\measuredangle{ACB})$ Proof of the lemma:Quadrilateral $ABHB'$ is parallelogram,so $m(\measuredangle{HAB'})=m(\measuredangle{BHA})=180^0-m(\measuredangle{ACB})$ and since $m(\measuredangle{HAC})=90^0-m(\measuredangle{ACB})$,it results that $B'A\perp AC$.Hence,$m(\measuredangle{B'AP})=90^0=m(\measuredangle{B'KP})$,resulting that quadrilateral $AKPB'$ is concyclic. Therefore $m(\measuredangle{KBP})=m(\measuredangle{KB'P})=m(\measuredangle{KAP})=90^0-m(\measuredangle{ACB})$,which ends the proof.$\blacksquare$ Back to the main problem,using the above lemma we obtain $m(\measuredangle{KBP})=90^0-m(\measuredangle{ACB})=m(\measuredangle{ABO})$,which implies $\measuredangle{ABB'}\equiv\measuredangle{OBP}(\bigstar)$. But,let's observe that $\triangle{AOB}\sim\triangle{B'PB}(\text{case A.A.})$.This helps us see that $\frac{AB}{BO}=\frac{BB'}{BP}(\bigstar\bigstar)$. Finally,from $(\bigstar)$ and $(\bigstar\bigstar)$ we get $\triangle{BAB'}\sim\triangle{BOP}$.So,$m(\measuredangle{BOP})=m(\measuredangle{BAB'})=90^0+m(\measuredangle{BAC})=180^0-m(\measuredangle{OBC})$,following that $OP\parallel BC$,which is what we wanted to prove.

This is nice problem but not too hard to complex bash:$k=a+\frac {b+c}{2}$,because $P$ is on chord $AB$ ,$\overline{p}=\frac {a+c-p}{ac}$, now because $BK$ perpendicular to $KP$ so $\frac{b-k}{\overline{b}-\overline{k}}= - \frac{p-k}{\overline{p}-\overline{k}}$ so after some calculations (about 15 minutes) we get $p=\frac{b^2c+abc+b^2a-c^2b}{b^2+bc-ac-ab}$ and finaly aply $p$ in formula for $OP || BC$ : $\frac{p}{\overline{p}}=\frac{b-c}{\overline{b}-\overline{c}}$ and result follows (I only needed 30 minutes for this bash ).

Let BH cuts (O) at E, midpoint of AD is F. We observe that BDK+KDP=HAC+BCA=90 so BDP=90. Because HGP=HFP=90 then FHPG is cylic therefore AG.AP=AF.AH=AE.AK. Note that BDP=BEP=90 so BEPD is cyclic too and K lies on (DPE) then K lies on (BEP). So BKP=BEP=90, q.e.d

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andria wrote: Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $K$ be the midpoint of $AH$. point $P$ lies on $AC$ such that $\angle BKP=90^{\circ}$. Prove that $OP\parallel BC$. Solution: Let $M$ be the midpoint of $BC$, $X$ be the feet from $P$ on $BC$ and $E$ be the feet from $B$ on $AC$. We have $B,K,P,E,X$ are concyclic with $BP$ as diameter. Therefore \[\angle PXK = \angle PBK = \angle AEK = \angle KAE.\]Now as $AK||PX$, we have $AKXP$ is parallelogram or $PX = AK = OM$ and therefore $POMX$ is a rectangle and we are done. $\square$

We make use of complex numbers.Assume without loss of generality that $ABC$ is the unit circle and let the coordinates of $X$ be $x$ respectivly.It is well-known that $h=a+b+c$ and so $k=\frac{2a+b+c}{2}$.Assume that $P$ is the intersection of line passing throught $O$ and parallel to $BC$ with $AC$ then we have: $P$ lies on a line parallel to $BS$ passing throught $O$ so $\frac{p}{\overline{p}}=-bc$. $P$ lies on $AC$ so $\frac{a-p}{\overline{a}-\overline{p}}=-ac$ Then we have $P=\frac{b(a+c)}{a+b}$ Then it is straightforward calculation to show $PK$ is perpendicular to $BK$.

Iran Round 3 2015 G2 wrote: Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $K$ be the midpoint of $AH$. point $P$ lies on $AC$ such that $\angle BKP=90^{\circ}$. Prove that $OP\parallel BC$. Solution: Let $H',H''$ be reflection of $H$ over $\overline{BC}$, $\overline{AC}$. Hence, $KE||AH''$. By Converse of Reim's Theorem, $BH'PEK$ is cyclic, $$\angle PAH'=\angle KEA=\angle AH'P \implies AP=PH' \implies OP ||BC \qquad \blacksquare$$

a cute problem let $P$ the point on $AC$ such that $OP \parallel BC$ we will prove that $PK \perp KB$ let $M,N$ be the projections from $O,P$ onto $BC$ since $\vec{PN}=\vec{OM}=\frac{1}{2}\vec{AH}=\vec{AK}$ we have $AKNP$ is parallelogram so $KN \parallel AC \implies BH \perp KN $ so $H$ is the orthocenter of $\triangle BKN $ so $NH \perp BK$ and we win

Rename $K$ as $P$ and let a line through $O$ parallel to $BC$ meet $AC$ at $F$. We will show that $F=P$ by proving $\angle CMF=90^\circ$. We use complex numbers with $(ABC)$ as the unit circle and $OF$ as the real line. Observe that $B$ and $C$ are reflections across the imaginary line so $bc=-1$. Clearly we have $M=\frac{A+H}2=\frac{a+a+b+c}2=\frac{2a+b+c}2$. By Lemma 8, we have $F=\frac{a+b}{ab+1}$ and clearly $C=c$. Their conjugates are $\overline{M}=\frac{\frac2a+\frac1b+\frac1b}{2},\overline{F}=F=\frac{a+b}{ab+1},\overline{C}=\frac1c$, respectively. It remains to check that $\frac{m-f}{m-c}+\overline{\left(\frac{m-c}{m-f}\right)}=0$ which can be rewritten as $\frac{m-f}{\overline{m}-\overline{f}}+\frac{m-c}{\overline{m}-\overline{c}}=0$. Upon plugging in the values and substituting $c=-\frac1b$, we now wish to show $$\frac{\frac{2a+b-1/b}2-\frac{a+b}{ab+1}}{\frac{2/a+1/b-b}2-\frac{a+b}{ab+1}}+\frac{\frac{2a+b-1/b}2-\left(-\frac1b\right)}{\frac{2/a+1/b-b}2-(-b)}=0$$which happens to be true upon expansion. $\blacksquare$

Let $PS$ be perpendicular to $BC$. It's well known that $PS = AM/2$ and $PS || AH$ so $PAKS$ and $SPKH$ are parallelogram. we know $BH$ is perpendicular to $AC$ so $BH$ is also perpendicular to $KS$. $KH$ is perpendicular to $BS$ so $H$ is orthocenter of $BKS$ so $SH$ is perpendicular to $BK$ so $PK$ is also perpendicular to $BK$ as wanted. we're Done.

let circumcircle of $ABC$ is the unit circle and $B,C$ are relative to the width axis so : $A=a,B=b,C=-\bar{b},H=a+b-\frac{1}{b},K=\frac{1}{2}(2a+b+\frac{1}{b})$ let $X$ be a point on $AB$ such that $OX||BC \Rightarrow x=\bar{x}$ and $\frac{x-a}{x-b}=\frac{x-\frac{1}{a}}{x-\frac{1}{b}}$ so $x=\frac{a+b}{ab+1}$ now we show that $XK \perp KC \Longleftrightarrow \frac{k-x}{k+\frac{1}{b}} + \frac{\bar{k}-x}{\bar{k}+b} = 0$