Easy problem for sharygin 9.5. Here's my solution. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -143.87782622508433, xmax = 159.6131305556715, ymin = -83.60386922556246, ymax = 116.81840547965157; /* image dimensions */ /* draw figures */ draw((-5.519536595786453,69.48830105976538)--(38.09219842126832,17.46462707747975), linewidth(0.4)); draw((-24.9803790621171,16.097529312963665)--(38.09219842126832,17.46462707747975), linewidth(0.4)); draw((-5.519536595786453,69.48830105976538)--(-24.9803790621171,16.097529312963665), linewidth(0.4)); draw(circle((-54.5115111772032,33.673480958004), 60.68702424989806), linewidth(0.4) + linetype("4 4")); draw(circle((6.161262548399664,34.98856298947976), 36.42352255843927), linewidth(0.4)); draw((-5.519536595786453,69.48830105976538)--(-4.01370537897934,0.01510129004338931), linewidth(0.4)); draw((-5.519536595786453,69.48830105976538)--(3.7594952816383698,16.720465928772192), linewidth(0.4)); draw((-5.519536595786453,69.48830105976538)--(6.161262548399663,34.98856298947975), linewidth(0.4)); draw(circle((6.950556810751496,-1.426406599036249), 36.42352255843917), linewidth(0.4)); draw(circle((-53.722216914851316,-2.741488630512079), 60.68702424989807), linewidth(0.4) + linetype("4 4")); draw((-4.730242333434549,33.07333147124929)--(3.7594952816383698,16.720465928772192), linewidth(0.4)); draw((-4.01370537897934,0.01510129004338931)--(-3.2244111166274556,-36.39986829847271), linewidth(0.4) + linetype("4 4")); draw((6.95055681075156,-1.4264065990363333)--(-3.2244111166274556,-36.39986829847271), linewidth(0.4)); draw((3.7594952816383698,16.720465928772192)--(6.95055681075156,-1.4264065990363333), linewidth(0.4)); draw((6.95055681075156,-1.4264065990363333)--(-4.730242333434549,33.07333147124929), linewidth(0.4)); draw((6.161262548399663,34.98856298947975)--(6.95055681075156,-1.4264065990363333), linewidth(0.4)); draw((-4.730242333434549,33.07333147124929)--(6.161262548399663,34.98856298947975), linewidth(0.4)); /* dots and labels */ dot((38.09219842126832,17.46462707747975),dotstyle); label("A", (36.246027344792815,20.450173099268422), NE * labelscalefactor); dot((-5.519536595786453,69.48830105976538),linewidth(4pt) + dotstyle); label("B", (-4.744959025594984,71.09768991268042), NE * labelscalefactor); dot((-24.9803790621171,16.097529312963665),dotstyle); label("C", (-24.255091961500717,18.08530850097681), NE * labelscalefactor); dot((6.161262548399663,34.98856298947975),linewidth(4pt) + dotstyle); label("$O$", (6.882291916005403,36.61008118759443), NE * labelscalefactor); dot((3.7594952816383698,16.720465928772192),linewidth(4pt) + dotstyle); label("$L$", (4.517427317713798,18.282380550834446), NE * labelscalefactor); dot((-4.01370537897934,0.01510129004338931),linewidth(4pt) + dotstyle); label("D", (-3.1683826267339144,1.531256312935536), NE * labelscalefactor); dot((-4.730242333434549,33.07333147124929),linewidth(4pt) + dotstyle); label("$H$", (-3.9566708261644488,34.63936068901809), NE * labelscalefactor); dot((-3.2244111166274556,-36.39986829847271),linewidth(4pt) + dotstyle); label("$B'$", (-2.3800944273033795,-34.73000086086916), NE * labelscalefactor); dot((6.95055681075156,-1.4264065990363333),linewidth(4pt) + dotstyle); label("$O'$", (7.6705801154359365,0.15175196393209633), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Firstly notice that clearly $AHOC$ is cyclic since $\angle COA=120^\circ=\angle CHA$. Now reflect everything about $\overline{BC}$. Denote by $B',O'$ reflection of $B,O$ respectively. Now notice that it just suffices to show $HLO'B'$ cyclic. But notice that since $\angle ABC=60^\circ \implies \overline{BH}=\overline{BO} \implies \overline{LH}=\overline{LO}=\overline{LO'} \implies L$ is the circumcentre of $\Delta HOO' \implies \angle HLB=\angle HOO'=\angle HBO=\angle HB'O'$ . Done $\blacksquare$.

parmenides51 wrote: In triangle $ABC$ $\angle B = 60^o, O$ is the circumcenter, and $L$ is the foot of an angle bisector of angle $B$. The circumcirle of triangle $BOL$ meets the circumcircle of $ABC$ at point $D \ne B$. Prove that $BD \perp AC$. (D. Shvetsov) Redefine $D$ as the intersection of $B-\text{altitude}$ and $\odot(ABC)$. So, the reflection of $D$ on $BC$ is the Orthocentre $H$ of $\triangle ABC$ and it's well known that $AH=AO$ also $\{H,O\}$ are pairs of Isogonal Conjugates. So, $\angle DBL=\angle OBL$. So, $LO=LH=LD\implies$ BOLD is a cyclic quadrilateral. So, $D\in\odot(BOL)$. $\blacksquare$