I can't find a solution without analytical method with which I solved it in the Test. Could someone find a geometric methgod?

I found a solution without analytical method in the exam.Let L be a point on AC satisfying LN=HL=HN,and let K be a point on AB satisfying KM=KH=MH .We can prove △DHC∽△MNL,△DHB∽△MNK,so DHC=OHC.So H,O,D are collinear.

This is my solution , we know that for a point $ P $ inside ( not necessary ) an equilateral triangle $ \Delta XYZ $ , a triangle with angles $ \angle YPZ - 60^o ~,~ \angle ZPX - 60^o ~,~ \angle XPY - 60^o $ has side ratio $ PX : PY : PZ $ , which can be proved by rotation . Now apply this result on $ H $ and equilateral triangle $ BDC $ . Since $ \angle MHD = 120^o - \angle A = \angle BHC - 60^o $ and $ HM:HN= BH:CH $ , we find that $ \Delta MHN $ is similar to the triangle with sides $ CH ~,~ DH ~,~ BH $ . Therefore , $ \angle MHO = 90^o - \angle MNH = 90^o - ( \angle CHD - 60^o ) $ $ = 150^o - \angle CHD = \angle MHD $ .

Construct $F$ on the same side of $HC$ as $A$ such that $\triangle FCH$ is equilateral. Then a rotation around $C$ maps $\triangle DCB$ to $\triangle FCH$, so $FD = HB$. Furthermore, $\angle DFH = \angle CHB - 60 = 120 - \angle CAB$. Furthermore, $\angle MHN = 120 - \angle CAB$ also. In addition, $\triangle CMH \sim \triangle BNH$, so $\dfrac {HM} {HN} = \dfrac {CH} {HB} = \dfrac {FH} {FD}$. Therefore, $\triangle DFH \sim \triangle NHM$. Then we have $\angle MHD$ $= 60 + \angle FHD - (\angle CAB + 30)$ $= \angle FHD + 120 - \angle CAB - 90$ $=\angle HMN + \angle MHN - 90$ $= 90 - \angle MNH$ $= \angle MHO$, so $H$, $O$, and $D$ are collinear.

Hint: let HM intersects AC at point E. and HN intersects AB at F. it’s easy to see that points M,N,F,E lie on a same circle, so it’s enough to say that HD is perpendicular to EF. That means we want to show that DF²₋DE²=HF²₋HE². now use cosine law in triangles BDF and CDE and BHF and CHE.the rest is not difficult.

I generalize this problem at http://www.artofproblemsolving.com/community/c6h610521

My solution:Ok,after we look at the picture we see that we need to make some connection beetwen the angles in triangles $CHD$ and $HMN$ and now we get the intution to construct $G$ such that $GHC$ is equilateral(because of spiral similarity we will have $BGC$ is similar to $CHD$ and $BHG$ will be similar to $HMN$).Now,from this we have an easy angle chase.Let $<HMN=x$ and by similarity of $HMN$ and $HBG$ we have $<HBG=<HMN=x$ and since $<CBH=90-<BCA$ we get $<CBG=<<CDH=x+<ACB-90$ and from this we get $<HDB=150-x-<BCA$ and from this we get $<BHD=60+x$ and also $<BHO=<MHO+<MHB=30+x-<BAC+30+<BAC=60+x$,so we get $<BHO=<BHD$ and from this we are finished.

Say $<BHM = <CHN = x+30$. Then $<MON = 240-2x$ Let $S_1$ be spiral similarity around $H$ with angle $x+30$ and $k_1=HM/HB$, Let $R_O$ be rotation around $O$ with angle $240-2x$, Let $S_2$ be spiral similarity around $H$ with angle $x+30$ and $k_2=HC/HN$. Let $R_D$ be rotation around $D$ with angle $60$, Then composition $\mathcal{J} = \mathcal{R}_D\circ \mathcal{S}_2 \circ \mathcal{R}_O \circ \mathcal{S}_1 $ is isometry since $k_1 *k_2=1$. Since it's sum of rotational angles is $360$ and $B$ it is fixed point: \[ \mathcal{J}: B\stackrel{\mathcal{S}_1}{\longmapsto}M\stackrel{\mathcal{R}_O}{\longmapsto}N\stackrel{\mathcal{S}_2}{\longmapsto}C \stackrel{\mathcal{R}_D}{\longmapsto}B\] $\mathcal{J} $ must be identity map. So if $ \mathcal{R}_O: H \longmapsto H'$ and $\mathcal{S}_2: H'\longmapsto H''$ we have: \[ \mathcal{J}: H\stackrel{\mathcal{S}_1}{\longmapsto}H\stackrel{\mathcal{R}_O}{\longmapsto}H'\stackrel{\mathcal{S}_2}{\longmapsto}H'' \stackrel{\mathcal{R}_D}{\longmapsto}H\] Now since $<H'HO = x-30$ and $<H'HH'' = x+30$ we have $<OHH'' = 60$. But since $\mathcal{R}_C $ maps $H''$ to $H$ we have also $<DHH'' = 60$ so $O,D,H$ are collinear.

Nice question

We can find that sin(EHO)/sin(FHO) = sin(EHD)/sin(FHD) where E and F are the intersects of altitudes with AC and AB , the problem is solved .

projective >>> synthetic [asy][asy] import graph; size(10cm); real labelscalefactor = 1.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(11); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.868953582044828, xmax = 19.051346935130493, ymin = -1.424030804570298, ymax = 10.436451535195745; /* image dimensions */ draw((4.1011591118959165,8.443093158764478)--(0,0)--(10,0)--cycle, linewidth(1.5)); /* draw figures */ draw((0,0)--(5,8.660254037844386), linewidth(0.75)); draw((5,8.660254037844386)--(10,0), linewidth(0.75)); draw((0,0)--(6.719869915485527,4.694895895949962), linewidth(1)); draw((1.9090225772392457,3.930121953830859)--(10,0), linewidth(1)); draw((2.523791648867181,5.195752572215299)--(4.1011591118959165,2.8653106868493228), linewidth(0.75)); draw((4.1011591118959165,2.8653106868493228)--(5.663558402505913,6.206809283332205), linewidth(0.75)); draw(circle((4.395606767138283,4.763652767795307), 1.9210419250937896), linewidth(1) + dotted); draw((5,8.660254037844386)--(4.1011591118959165,2.8653106868493228), linewidth(0.75) + linetype("4 4")); draw(shift((5,0))*xscale(5)*yscale(5)*arc((0,0),1,0,180), linewidth(1) + linetype("4 4") + heavygrey); draw((1.9090225772392457,3.930121953830859)--(4.542830594759458,5.712823808268292), linewidth(0.75) + heavygrey); draw((4.542830594759458,5.712823808268292)--(7.5,4.330127018922194), linewidth(0.75) + heavygrey); draw(circle((4.32199485332769,4.289067247558802), 1.4407814438203435), linewidth(0.75) + dotted); /* dots and labels */ dot((4.1011591118959165,8.443093158764478),dotstyle); label("$A$", (4.1011591118959165,8.443093158764478), dir(90) * labelscalefactor); dot((0,0),dotstyle); label("$B$", (0,0), dir(225) * labelscalefactor); dot((10,0),dotstyle); label("$C$", (10,0), dir(315) * labelscalefactor); dot((5,8.660254037844386),linewidth(4pt) + dotstyle); label("$D$", (5,8.660254037844386), dir(90) * labelscalefactor); dot((6.719869915485527,4.694895895949962),linewidth(4pt) + dotstyle); label("$E$", (6.719869915485527,4.694895895949962), dir(45) * labelscalefactor); dot((1.9090225772392457,3.930121953830859),linewidth(4pt) + dotstyle); label("$F$", (1.9090225772392457,3.930121953830859), dir(135) * labelscalefactor); dot((4.1011591118959165,2.8653106868493228),linewidth(4pt) + dotstyle); label("$H$", (4.1011591118959165,2.8653106868493228), dir(60) * labelscalefactor); dot((2.523791648867181,5.195752572215299),linewidth(4pt) + dotstyle); label("$M$", (2.523791648867181,5.195752572215299), dir(135) * labelscalefactor); dot((5.663558402505913,6.206809283332205),linewidth(4pt) + dotstyle); label("$N$", (5.663558402505913,6.206809283332205), dir(45) * labelscalefactor); dot((4.395606767138283,4.763652767795307),linewidth(4pt) + dotstyle); label("$O$", (4.395606767138283,4.763652767795307), dir(45) * labelscalefactor); dot((2.5,4.330127018922194),linewidth(4pt) + dotstyle); label("$K$", (2.5,4.330127018922194), dir(90) * labelscalefactor); dot((7.5,4.330127018922194),linewidth(4pt) + dotstyle); label("$L$", (7.5,4.330127018922194), dir(45) * labelscalefactor); dot((4.542830594759458,5.712823808268292),linewidth(4pt) + dotstyle); label("$P$", (4.542830594759458,5.712823808268292), dir(45) * labelscalefactor); dot((2.918133514624364,4.613142100873805),linewidth(4pt) + dotstyle); label("$Q$", (2.918133514624364,4.613142100873805), dir(90) * labelscalefactor); dot((5.2729585798534115,5.371434634211479),linewidth(4pt) + dotstyle); label("$R$", (5.2729585798534115,5.371434634211479), dir(30) * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $E=BH\cap AC$, $F=CH\cap AB$, and $\omega = (BCEF)$. Let $K=BD\cap\omega$ and $L=CD\cap\omega$. Let $P=EL\cap FK$, $Q=FK\cap HM$, and $R=EL\cap HN$. By Pascal's theorem, $D$, $P$, and $H$ are collinear, so it suffices to show that $P$, $O$, and $H$ are collinear. Note that $\measuredangle PEH = \measuredangle LCB = 60^{\circ}$ and $\measuredangle HFP = \measuredangle CBK = 60^{\circ}$, thus $EP\perp HN$ and $FP\perp HM$. Thus the center of $(PQHR)$ lies on $HP$ and it suffices to show that $(HQR)$ and $(HMN)$ are tangent at $H$. Note that $\tfrac{HQ}{HM}=\tfrac{HR}{HN}=\tfrac{3}{4}$, which implies a homothety centered at $H$, as desired. $\blacksquare$