[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -11.70189688515651, xmax = 6.521474337865648, ymin = -6.2840924169047145, ymax = 4.577456417748465; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); /* draw figures */ draw(circle((-2.8502289804179077,-1.305129237870913), 3.9805191019700463), linewidth(0.5)); draw((-4.603809523809526,2.268312744166403)--(-6.2087086896843004,-3.4417495512617484), linewidth(0.5)); draw((-6.2087086896843004,-3.4417495512617484)--(0.464293105268711,-3.5093242529827915), linewidth(0.5)); draw((0.464293105268711,-3.5093242529827915)--(-4.603809523809526,2.268312744166403), linewidth(0.5)); draw((-2.8099219928615637,2.675185783318058)--(-4.603809523809526,2.268312744166403), linewidth(0.5)); draw((xmin, 0.22681078503101412*xmin-0.6586665653041802)--(xmax, 0.22681078503101412*xmax-0.6586665653041802), linewidth(0.5)); /* line */ draw(circle((-3.4448910265952035,-2.8413141946753826), 2.5062085513074983), linewidth(0.5)); draw(circle((-4.049454465188574,-0.17581732021809857), 2.506208551307498), linewidth(0.5)); draw((xmin, 3.5578947368421087*xmin + 18.648182418351886)--(xmax, 3.5578947368421087*xmax + 18.648182418351886), linewidth(0.5)); /* line */ draw((xmin, -1.14*xmin-2.9800301129764612)--(xmax, -1.14*xmax-2.9800301129764612), linewidth(0.5)); /* line */ draw(circle((-6.24871651602167,1.8952300979597636), 1.6866860033595312), linewidth(0.5) + linetype("4 4")); draw((-4.603809523809526,2.268312744166403)--(-2.8905359679742512,-5.285444259059885), linewidth(0.5)); draw((-2.8099219928615637,2.675185783318058)--(-5.039036050745086,0.7198225746483037), linewidth(0.5) + qqccqq); draw((-3.9992460852161557,-0.39718413029088006)--(-5.7959658966742715,-1.9732541403418573), linewidth(0.5) + qqccqq); draw((-4.647767147389301,-2.072502584336311)--(-6.2087086896843004,-3.4417495512617484), linewidth(0.5) + qqccqq); draw((-2.8099219928615637,2.675185783318058)--(-5.664509450350292,3.477510660422877), linewidth(0.5) + red); draw((-3.9992460852161557,-0.39718413029088006)--(-1.6983795951095055,-1.0438773745516232), linewidth(0.5) + red); draw((-4.647767147389301,-2.072502584336311)--(0.464293105268711,-3.5093242529827915), linewidth(0.5) + red); /* dots and labels */ dot((-4.603809523809526,2.268312744166403),dotstyle); label("$A$", (-4.568939441503671,2.3650952854925062), NE * labelscalefactor); dot((-6.2087086896843004,-3.4417495512617484),dotstyle); label("$B$", (-6.170994054516608,-3.3469922930476645), NE * labelscalefactor); dot((0.464293105268711,-3.5093242529827915),dotstyle); label("$C$", (0.504233499703962,-3.4137445685898706), NE * labelscalefactor); dot((-4.647767147389301,-2.072502584336311),linewidth(4pt) + dotstyle); label("$H$", (-4.60708359895636,-1.992874703477207), NE * labelscalefactor); dot((-2.8502289804179077,-1.305129237870913),linewidth(4pt) + dotstyle); label("$O$", (-2.8143081986799787,-1.229991554423428), NE * labelscalefactor); dot((-2.8099219928615637,2.675185783318058),linewidth(4pt) + dotstyle); label("$N$", (-2.7761640412272896,2.756072899382568), NE * labelscalefactor); dot((-2.8905359679742512,-5.285444259059885),linewidth(4pt) + dotstyle); label("$M$", (-2.852452356132668,-5.2065199688662505), NE * labelscalefactor); dot((-5.50041902681272,-4.275150288156296),linewidth(4pt) + dotstyle); label("$D$", (-5.465327141641862,-4.195699796369994), NE * labelscalefactor); dot((-5.7959658966742715,-1.9732541403418573),linewidth(4pt) + dotstyle); label("$E$", (-5.760944361900202,-1.8975143098454845), NE * labelscalefactor); dot((-1.6983795951095055,-1.0438773745516232),linewidth(4pt) + dotstyle); label("$F$", (-1.6604474357361371,-0.9629824522546051), NE * labelscalefactor); dot((-3.9992460852161557,-0.39718413029088006),linewidth(4pt) + dotstyle); label("$G$", (-3.958632922260648,-0.32406781492206515), NE * labelscalefactor); dot((-5.039036050745086,0.7198225746483037),linewidth(4pt) + dotstyle); label("$J$", (-4.998061212846422,0.7916487905690868), NE * labelscalefactor); dot((-5.664509450350292,3.477510660422877),linewidth(4pt) + dotstyle); label("$K$", (-5.62743981081579,3.5571002058890357), NE * labelscalefactor); dot((-3.495099406567622,-2.6199473846026002),linewidth(4pt) + dotstyle); label("$L$", (-3.453222836012519,-2.545964986541197), NE * labelscalefactor); dot((-6.522440499182117,0.2309029501394421),linewidth(4pt) + dotstyle); label("$X$", (-6.676404140764737,0.2671666255946136), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let us define some points. Let $M$ be the midpoint of arc $BC$, $D'=(EFM)\cap (ABC)$, $G=(EFM)\cap AM$, $L=(AEF)\cap AM$, $X=(AEF)\cap (ABC)$. $J$, $K$ are on $AB$, $AC$ respectively such that $JN\perp AC$ and $KN\perp AB$. By symmetry, $AEMF$, $AELF$ and $MFGE$ are kites, and $AL$ and $GM$ are diameters. In particular, $AEMF$ is a rhombus (by letting $Y=EF\cap AM$, $NO=OM\rightarrow AY=YM$) We have $XA\perp XL, XN\perp XM, AN\perp LM$, so $X$ spirals $AN$ to $LM$. Since $KJAN$ is clearly congruent to $FELM$, $X$ also spirals $KJ$ to $FE$, and $XKAJ$ is cyclic. Hence, $X$ spirals $JEB$ to $KFC$, so $\frac{JE}{EB}=\frac{KF}{FC}$. Since $JN\parallel EG\parallel BH$ (all are $\perp AC$) and $KN\parallel GF\parallel HC$ (all $\perp AB$), this means that $NGH$ is collinear. Since $\angle NDM=90^\circ$, $\angle GDM=90^\circ$, so $GEFDM$ cyclic. Then, since $EG=GF$ by symmetry, we have $\angle EDH=\angle HDF$ as angles subtending equal arcs.

My solution: Let $K$ be the midpoint of minor arc $BC$, $A'$ be the antipode of $A$, $CL$ be the altitude of triangle $ABC$. The line passing through $H$ and parallel $AN$ meets $AB, AC$ at $I$ and $J$ respectively. We can easily get $E, F, I, J$ are concyclic. Since, $EF \parallel AN \parallel AK$, $K$ is the reflection of $A$ in $EF$. Then, $\angle EAF = \angle EKF$. We have $\triangle AHL \sim \triangle AA'C$. Then, $$\dfrac{AH}{AO} = 2 \cdot \dfrac{AH}{AA'} = 2 \cdot \dfrac{AL}{AC} = 2 \cdot \cos \angle A$$Also, we have $\triangle AIH \sim \triangle AOF$. Then, $$\dfrac{AI}{AF} = \dfrac{AH}{AO} = 2 \cos \angle A$$Then, $FIA$ is an isosceles triangle. Then, $\angle AIF = \angle IAF = \angle EKF$. Therefore, $E, F, K, I$ are concyclic. According to the Reim's Theory, since $IJ \parallel AN$, we have $I, H, B, D$ are concyclic and $J, H, D, C$ are concyclic. Then, $$\angle IDJ = \angle IDH + \angle JDH = \angle HBA + \angle HCA = 2(90^o-\angle BAC) = \angle IFJ$$Hence, $I, F, J, D$ are concyclic. Since $EI = FJ$ and $DH$ bisects angle $\angle IDJ$, $DH$ bisects angle $\angle EDF$.

Let $a,b,c$ be complex numbers such that $a^2,b^2,c^2$ are vertices of the triangle and $n=bc$. Then $$h=a^2+b^2+c^2,\ d=\frac{h-n}{1-n\cdot\overline{h}}=\frac{a^2bc(a^2+b^2+c^2-bc)}{a^2bc-(a^2b^2+b^2c^2+c^2a^2)}.$$$$E\in AB\iff \overline{e}=\frac{a^2+b^2-e}{a^2b^2}$$Hence $$EO\parallel AN\iff e=\frac{c(a^2+b^2)}{c-b}.$$Let $P$ be reflection of $C$ wrt the circumcircle. Then $p=-c^2$ and $$e+c^2=\frac{c(a^2+b^2+c^2-bc)}{c-b},\ e-d=\frac{c^2(a^2c+b^3)(a^2+bc)}{(c-b)(a^2b^2+b^2c^2+c^2a^2-a^2bc)}.$$Hence $$\frac{e-d}{e+c^2}=\frac{c(a^2c+b^3)(a^2+bc)}{(a^2+b^2+c^2-bc)(a^2b^2+b^2c^2+c^2a^2-a^2bc)}\in\mathbb{R}\implies P\in DE.$$Analogously we can prove that if $Q$ is reflection of $B$ wrt $O$ then $Q\in DF$. Because $BN=BC$ and $PQ\parallel BC$ there holds $NP=NQ$ and $$\measuredangle EDN=\measuredangle PDN=\measuredangle NDQ= \measuredangle NDF$$that is line $DN$ bisects angle $EDF.\square$ #1713