Easy for a P4. Pluto1708 wrote: Let $O, H$ be the orthocenter and circumcenter of of an acute-angled triangke $ABC$ with $AB<AC$.Let $K$ be the midpoint of $AH$.The line through $K$ perpendicular to $OK$ meet $AB$ and the tangent to the circumcircle at $A$ at $X$ and $Y$ respectively. Prove that $\angle XOY=\angle AOB$ Consider a homothety from $A$ with ratio $2$. This maps $K$ to $H$, $O$ to the $A-$antipode call it $A'$. Let $D,E$ be images of $Y,X$ under this homothety. This gives $D,E,H$ are collinear and $A'H\perp DE$. Further, $AA'\perp AD$ as $AD$ is tangent to $\odot(ABC)$ and $A'B\perp EB$ as $A'$ is the $A-$ antipode. This gives us that $H,E,B,A'$ and $H,A',D,A$ are concylic. Therefore \[\angle XOY = \angle EA'D = \angle EBH + 180^{\circ} - \angle HAD = 90^{\circ} - A + 180^{\circ} - (90^{\circ} - C + B) = 2C = \angle AOB\]and we are done. $~\blacksquare$ [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 = -8.176891049225661, xmax = 21.55128779028212, ymin = -4.849096510144992, ymax = 11.014054657493993; /* image dimensions */ pair A = (0.,5.010548932660872), B = (-2.,0.), C = (6.247422153390056,0.), M = (2.123711076695028,0.), H = (0.,2.4937076704976247), O = (2.123711076695028,1.2584206310816235), K = (0.,3.7521283015792486), Y = (4.405830543060755,7.504256603158495), X = (-0.7609973448857101,3.1040417155734454), D = (8.811661086121502,9.997964273656116); draw(A--B--C--cycle, linewidth(2.)); /* draw figures */ draw(A--H, linewidth(1.)); draw(X--Y); draw(circle((2.1237110766950282,1.2584206310816228), 4.311451673020253), linewidth(1.)); draw(H--(4.247422153390056,-2.493707670497625), linewidth(1.)); draw((-1.5219946897714205,1.1975344984860194)--D, linewidth(1.)); draw(D--A, linewidth(1.)); draw((-1.5219946897714205,1.1975344984860194)--(4.247422153390056,-2.493707670497625), linewidth(1.)); draw((4.247422153390056,-2.493707670497625)--D, linewidth(1.)); draw(A--(4.247422153390056,-2.493707670497625), linewidth(1.)); draw(B--H, linewidth(1.)); draw(B--(4.247422153390056,-2.493707670497625), linewidth(1.)); draw(circle((1.362713731809317,-0.6480865860058028), 3.424596306874666), red); draw(circle((6.52954161975578,3.7521283015792446), 6.649701966919629), red); /* dots and labels */ dot(A,dotstyle); label("$A$", (-0.3,5.1), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-2.6,-0.4), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (6.370422694369542,-0.2924002454007665), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); dot(H,linewidth(4.pt) + dotstyle); label("$H$", (-0.23313178555390862,1.8), NE * labelscalefactor); dot(O,linewidth(4.pt) + dotstyle); label("$O$", (2.3010736237156073,1.1940086965746224), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (0.1,3.7), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (4.1529929612587155,7.700093737680012), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (-1.3,3.1), NE * labelscalefactor); dot((4.247422153390056,-2.493707670497625),dotstyle); label("$A'$", (4.201727680667744,-2.9971771726018837), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (8.904628103639057,10.185564427540498), NE * labelscalefactor); dot((-1.5219946897714205,1.1975344984860194),linewidth(4.pt) + dotstyle); label("$E$", (-1.9,1.4620496533242826), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

I'll write the proof in reverse order, as to reflect the motivation. We need to prove that $\angle XOY=\angle AOB$, but $\angle XOY=\angle XOA+\angle AOY$ and $\angle AOB=\angle BOX+\angle XOA$, so we need to prove that $\angle AOY=\angle BOX$. Since $\angle OAY=90^\circ=\angle OKY$, we have $OKAY$ is cyclic and therefore $\angle AOY=\angle AKY\equiv\angle HKX$. So, we need to prove that $\angle HKX=\angle BOX$. For this, it is sufficient to prove that $\triangle XHK\sim\triangle XBO$, which is equivalent to $\triangle XHB\sim\triangle XKO$ (by SAS or by the properties of spiral similarity). We notice that since $\angle XKO=90^\circ$, then $\angle XHB$ must also be $90^\circ$, but since $BH\perp AC$ as an altitude, then it must be that $XH\parallel AC$. But since $AK=KH$, that would mean that $XK=KZ$, where $Z=XK\cap AC$. So, let's prove that.

So, $XH\parallel AC$ and therefore $\angle XHB=90^\circ=\angle XKO$. Now, we only need one more angle for our desired similarity. From $XH\parallel AC$ we also get that $\angle BXH=\angle BAC=\alpha$. From "Proof that XK=KZ" we already found out that $\angle OXK=\alpha$. Therefore, $\angle BXH=\angle OXK$. By AA, this proves our desired similarity and thus finishes the proof.

Meh you don't need so much stuff for this problem lol Let $M$ be the midpoint of $AB$ and let $AH \cap BO = Z$ First of all the condition is equivalent to proving $\angle AOY = \angle BOX$, now because of the perpendiculars we get $Y \in \odot(AKO)$ and $X \in \odot(MKO)$, but by trivial angle chasing we get $Z \in \odot(MKO)$ which means $X \in \odot(MKOZ)$, this gives us $$\angle AOY=\angle AKY = \angle XKZ = \angle XOB \ \blacksquare$$