I never imagined my geo skills could become so bad and rusty that even this problem troubled me . Anyways here's my solution. Sharygin 2022 P1 wrote: Let $O$ and $H$ be the circumcenter and the orthocenter respectively of triangle $ABC$. It is known that $BH$ is the bisector of angle $ABO$. The line passing through $O$ and parallel to $AB$ meets $AC$ at $K$. Prove that $AH = AK$. Redefine $K$ to be the intersection of the line parallel to $OB$ passing through $H$ with $AC$. Let $H'$ be the reflection of $H$ over $AC$. It is well known that $\measuredangle ABH = \measuredangle OBC$. So, firstly, $$\measuredangle BH'O = \measuredangle OBH' = \measuredangle KHH' = \measuredangle HH'K = \measuredangle BH'K \implies \overline{O - K - H'}\text{ are collinear.}$$ Now, $$\measuredangle OH'B = \measuredangle KH'H = \measuredangle H'HK = \measuredangle H'BO = \measuredangle ABH' \implies OH' \parallel AB.$$ Finally, $$\measuredangle HAK = \measuredangle BAO = \measuredangle OBA = 2 \measuredangle OBH = 2 \measuredangle OBH' = 2 \measuredangle BH'O = 2 \measuredangle HH'K \implies \text{ the center of }\odot HH'K \text{ lies on } \odot HAK.$$ So, from the fact that $A$ lies on $\odot HAK$ and that $AH = AH'$, we conclude that $A$ is the center of $\odot HH'K \implies AK = AH$ and we are done . [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = 3.661465847376298, xmax = 18.084874144065477, ymin = -1.7376901539993266, ymax = 6.969180910646938; /* image dimensions */ draw(arc((6.8247962819744785,0.6986593162544235),0.37732919023385686,-0.12981688231938696,24.048175520138585)--(6.8247962819744785,0.6986593162544235)--cycle, linewidth(0.8) + blue); draw(arc((6.8247962819744785,0.6986593162544235),0.47166148779232103,24.048175520138585,48.22616792259648)--(6.8247962819744785,0.6986593162544235)--cycle, linewidth(0.8) + blue); draw(arc((6.8247962819744785,0.6986593162544235),0.5659937853507853,48.22616792259653,72.40416032505446)--(6.8247962819744785,0.6986593162544235)--cycle, linewidth(0.8) + blue); draw(arc((10.335519066080694,4.628800029407053),0.47166148779232103,-131.7738320774035,-107.5958396749456)--(10.335519066080694,4.628800029407053)--cycle, linewidth(0.8) + blue); draw(arc((7.979827467310177,1.9916794051659592),0.47166148779232103,24.048175520138404,48.22616792259648)--(7.979827467310177,1.9916794051659592)--cycle, linewidth(0.8) + blue); /* draw figures */ draw((9.462388221756079,1.875649267877402)--(6.8247962819744785,0.6986593162544235), linewidth(0.4)); draw(circle((9.462388221756079,1.875649267877402), 2.8882860985405046), linewidth(0.4)); draw((7.985188017030506,4.357599336307677)--(6.8247962819744785,0.6986593162544235), linewidth(0.4)); draw((6.8247962819744785,0.6986593162544235)--(12.09461961181735,0.686719288358659), linewidth(0.4)); draw((12.09461961181735,0.686719288358659)--(7.985188017030506,4.357599336307677), linewidth(0.4)); draw(circle((7.985188017030506,4.357599336307677), 2.365926003928893), linewidth(0.4)); draw((6.8247962819744785,0.6986593162544235)--(10.335519066080694,4.628800029407053), linewidth(0.4)); draw((10.335519066080694,4.628800029407053)--(9.462388221756079,1.875649267877402), linewidth(0.4)); draw((7.979827467310177,1.9916794051659592)--(9.749649112487496,2.781438526400997), linewidth(0.4)); /* dots and labels */ dot((9.462388221756079,1.875649267877402),linewidth(1pt) + dotstyle); label("$O$", (9.594967363803697,1.8752368424898602), NE * labelscalefactor); dot((6.8247962819744785,0.6986593162544235),linewidth(1pt) + dotstyle); label("$B$", (6.353701855106839,0.8092818800792123), NE * labelscalefactor); dot((12.09461961181735,0.686719288358659),linewidth(1pt) + dotstyle); label("$C$", (12.348336912022546,0.7526825015441337), NE * labelscalefactor); dot((10.335519066080694,4.628800029407053),linewidth(1pt) + dotstyle); label("$H'$", (10.434524812074029,4.742938688267178), NE * labelscalefactor); dot((7.985188017030506,4.357599336307677),linewidth(1pt) + dotstyle); label("$A$", (7.83358538628642,4.55597440388271), NE * labelscalefactor); dot((7.979827467310177,1.9916794051659592),linewidth(1pt) + dotstyle); label("$H$", (7.791318305309806,2.1865334244327927), NE * labelscalefactor); dot((9.749649112487496,2.781438526400997),linewidth(1pt) + dotstyle); label("$K$", (9.981729783793401,2.809126588318658), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Here's a sol that doesn't add any point a.k.a. T R I G B A S H. Since $BH$ bisects $\angle ABO$, we have $$\angle ABH=\frac{1}{2}\angle ABO\implies 90^{\circ}-A=\frac{1}{2}(90^{\circ}-C)\implies 90^{\circ}+C=2A\quad (\dagger)$$Now that we know the property of triangle $ABC$, let's focus on $K$, more specifically triangle $AOK$. Since $AB\parallel OK$, we have $\angle OKA=180^{\circ}-A$. Also, $\angle OAK=\angle OAC=90^{\circ}-B$. Thus, $$\angle AOK=180^{\circ}-((180^{\circ}-A)+(90^{\circ}-B))=90^{\circ}-C.$$We can finally use sine law on triangle $AOK$. $$\frac{AK}{AO}=\frac{\sin\angle AOK}{\sin \angle OKA}=\frac{\sin(90^{\circ}-C)}{\sin(180^{\circ}-A)}\stackrel{(\dagger)}{=}\frac{\sin 2A}{\sin A}\implies AK=2R\cos A=AH$$We are now done!!!

Let $K$ be the point on $AC$ such that $AH=AK$. We will show $KO || AB$. By angle chase, we get $B=270^\circ -3A$ and $C=2A-90^\circ$. So $\angle HAK = 180^\circ -2A \implies \angle AKH = A$. So if we let $KH$ intersect $AB$ at $P$, $\triangle PAK$ is isosceles with $PA=PK$. Now in $\triangle APH$, we have $\angle AHP = \angle HAK + \angle HKA = 180^\circ - A$ and $\angle APH = 180 ^\circ -2A$. We also know $AH=2R \cos A$ (well known). Applying law of sines, we get $\frac{AH}{\sin 2A}=\frac{AP}{\sin A} \implies \frac{2R \cos A}{\sin 2A}=\frac{AP}{\sin A} \implies \frac{2R \cos A}{2 \cos A \sin A}=\frac{AP}{\sin A} \implies AP=R$. Therefore $KP=AP=R$. Since $\angle APK = 180^\circ - 2A = 2(90^\circ -A) = \angle ABO$, $PK||BO$. But also $PK=BO=R$, so $PBOK$ is a parallelogram $\implies KO||BP \implies KO||AB$, so $K$ is the point on $AC$ such that $OK||AB$.