Bleh. I'm bad at geo, so this may be really easy, but I'm stuck. Can anyone finish the lemma? [asy][asy] import geometry; import olympiad; size(8cm); defaultpen(fontsize(10pt)); pen pri=green; pen sec=springgreen; pen tri=chartreuse; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A, B, C, M, H, Hb, Ha, Hc; A=(0,0); B=(sqrt(3),sqrt(3)); C=(1+sqrt(3),0); H=orthocenter(A,B,C); M=midpoint(B--C); Hb=foot(B,A,C); Ha=foot(A,B,C); Hc=foot(C,A,B); path circ = circumcircle(A,B,C); draw(A--B--C--cycle,pri); draw(B--Hb,sec); draw(A--Ha,tri+dotted); draw(C--Hc,tri+dotted); draw(circ, tri); line MH = line(M,H); draw(MH, dashed+pri); pair ip_MH_circ[]=intersectionpoints(MH,circ); pair P = ip_MH_circ[1]; pair Mc = midpoint(A--B); pair O = circumcenter(A,B,C); draw(line(P,Hb),dashed+tri); draw(line(O,M),dashed+tri); dot("$O$",O); dot("$P$",P,dir(300)); dot("$M_C$",Mc); dot("$A$",A,W); dot("$B$",B,N); dot("$C$",C,E); dot("$M$",M,dir(230)); dot("$H$",H,dir(30)); dot("$H_B$",Hb,S); [/asy][/asy] Let $P = MH \cap \overarc{\textit{AB}}$. Let $O$ be the circumcenter of $\Delta ABC$, let $H_B$ be the foot of the altitude from $B$ to $\overline{AC}$, and let $M_C$ be the midpoint of $\overline{AB}$. Lemma: We claim that the points $P, M_C, O,$ and $H_B$ are collinear. Proof: $AH_BB$ is an isosceles right triangle, which can be derived from angle chasing. Clearly $O$ lies on $M_CH_B$, since the latter is the perpendicular bisector of triangle $AH_BB$ (medians are also perpendicular bisectors in an isosceles triangle). finish proof of lemma here Thus $P$ lies on $M_CH_B$. This implies that $AM_C = M_CB, \angle PM_CA = \angle PM_CB, $ and $PM_C = PM_C$. Thus $\Delta PM_CA \cong \Delta PM_CB$. Thus $PA = PB$, and clearly $P$ is the midpoint of the desired arc ($\angle PBA = \angle PAB$). $\square$ Also, this is probably easily coord-bashed, but I wanted to find a synthetic solution.

@above. $P \in OM_C$ instantly implies the result, so your lemma is mostly the problem itself. Don't assume that this is "really easy", since it's P5.

I found a simple solution Let $A'$ be a points such that $AA'$ is the diameter of circumcircle of $\triangle ABC$. By a well-known fact, $H$ and $A'$ are symmetric with respect to $M$. We want to prove that $A'H$ is the angle bisector of $\angle AA'B$. By simple angle chasing, one can prove that $H$ is the incenter of triangle $AA'B$. This means, that $A'H$ is passes through the midpoint of arc $AB$. P.S. clearly $A'$ is one the arc $AB$ which does not contain the point $C$

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