Notice that, as $N$ is the midpoint of the arc $AOF$, $NO||AC$, and similarly $MO||AB$. Invert trough $O$, with radius of the sircumcicle $R$, so that the inversion fixes $A, B, C$. From $OE+OF=R$, we get $\frac{1}{OE}+\frac{1}{OF}=\frac{1}{R}$ , and $M, N$ are points on $AE$, $AF$, such that $OM||AB$ and $ON||AC$. We want to prove $EN||FM$. Now, by Thales, we have $\frac{NA}{AF}=\frac{OC}{CF}$, and $\frac{EA}{AM}=\frac{EB}{BO}$. To prove the similarity of $EAN$ and $MAF$(and therefore finish the problem), we have to show that $EB*CF=OB*OC$,and that's a straightforward calculation:$(OE-R)(OF-R)=OE*OF-R(OE+OF)+R^2=R^2 <=> OE*OF=R(OE+OF) <=> \frac{1}{OE}+\frac{1}{OF}=\frac{1}{R}$.

See 2007 Iran TST - Day 3 - P3

Collaborative solution with several (Anant Mudgal, Aditya Khurmi, Anushka Aggarwal, Shuborno Das, Robu Vlad, Misheel, and about 5 others): Since $\overline{NO}$ is the external angle bisector of $\angle AOC$ it follows that $\overline{NO} \parallel \overline{AC}$. Similarly, $\overline{MO} \parallel \overline{AB}$. We perform an inversion around the circumcircle now, letting $\bullet^\ast$ denote the inverse of a point. For example, $N^\ast = \overline{AF^\ast} \cap \overline{NO}$ while $M^\ast = \overline{AE^\ast} \cap \overline{OM}$. [asy][asy] size(12cm); pair O = origin; pair A = dir(100); pair B = dir(200); pair C = dir(340); pair E = 0.6*B; pair F = 0.4*C; pair Es = 1/conj(E); pair Fs = 1/conj(F); pair Ns = extension(A, Fs, O, A-C); pair Ms = extension(A, Es, O, A-B); pair N = 1/conj(Ns); pair M = 1/conj(Ms); filldraw(circumcircle(O, N, E), invisible, deepcyan); filldraw(circumcircle(O, M, F), invisible, deepcyan); filldraw(A--B--C--cycle, invisible, red); filldraw(Ns--A--Es--cycle, invisible, deepgreen); filldraw(Ms--A--Fs--cycle, invisible, deepgreen); draw(Es--O--Fs, blue); draw(Ns--O--Ms, mediumblue); draw(unitcircle, red); draw(circumcircle(A, N, F), red+dotted); draw(circumcircle(A, M, E), red+dotted); dot("$O$", O, dir(270)); dot("$A$", A, dir(A)); dot("$B$", B, dir(260)); dot("$C$", C, dir(280)); dot("$E$", E, dir(270)); dot("$F$", F, dir(270)); dot("$E^\ast$", Es, dir(Es)); dot("$F^\ast$", Fs, dir(Fs)); dot("$N^\ast$", Ns, dir(Ns)); dot("$M^\ast$", Ms, dir(Ms)); dot("$N$", N, dir(80)); dot("$M$", M, dir(120)); /* TSQ Source: !size(12cm); O = origin R270 A = dir 100 B = dir 200 R260 C = dir 340 R280 E = 0.6*B R270 F = 0.4*C R270 E* = 1/conj(E) F* = 1/conj(F) N* = extension A Fs O A-C M* = extension A Es O A-B N = 1/conj(Ns) R80 M = 1/conj(Ms) R120 circumcircle O N E 0.1 lightcyan / deepcyan circumcircle O M F 0.1 lightcyan / deepcyan A--B--C--cycle 0.1 lightred / red Ns--A--Es--cycle 0.1 lightgreen / deepgreen Ms--A--Fs--cycle 0.1 lightgreen / deepgreen Es--O--Fs blue Ns--O--Ms mediumblue unitcircle red circumcircle A N F red dotted circumcircle A M E red dotted */ [/asy][/asy] Claim: We have $\overline{E^\ast N^\ast} \parallel \overline{M^\ast F^\ast}$. Proof. Let $x = BE = OF$. Then $BE^\ast = \frac{R^2}{x}-R = \frac{R(R-x)}{x}$ while $CF^\ast = \frac{R^2}{R-x} - R = \frac{Rx}{R-x}$. Consequently, we conclude that \[ 1 = \frac{BE^\ast}{BO} \cdot \frac{CF^\ast}{CO} = \frac{E^\ast A}{AM^\ast} \cdot \frac{F^\ast A}{AN^\ast}. \]This implies $\triangle AE^\ast N^\ast \sim \triangle AM^\ast F^\ast$ which implies the parallel lines we wanted. $\blacksquare$ Thus in the original diagram, $\triangle ENO$ and $\triangle MOF$ have tangent circumcircles, so $\angle ENO + \angle OMF = \angle EOF = \angle BOC = 2 \angle BAC$.

Solution without inversion: danepale wrote: Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$. Firstly, $ON'$ is external bisector of $\angle OAF$, so $ON||AC$ and similarly $OM||AB$. Let $\odot(AOF)$ meet $AC$ again at $D$, then $\angle OAD = 90^{\circ} - B = \angle NAF$, so $NO = FD$. This gives, \[NO:AO=FD:AO = CF:CA = OE:AC,\]so $NO:OE$ is fixed which gives $\angle ENO$ is fixed and similarly $\angle OMF$ is fixed. We'll fix $E = O$ to get $M$ (and $F = C$) and fix $F = O$ to get $N$ (and $E = B$) and now we'll solve the problem. Now $MN$ is perpendicular bisector of $AO$ and $\angle NAO = \angle OAC$. This gives $\triangle ANM\sim \triangle ABC$. Therefore from spiral similarity,\[\triangle ANB\sim \triangle AMC\implies \angle ANB = \angle AMC.\]This gives $\angle ONB+\angle OMC = 360^{\circ} - 2B - \angle ANB + \angle AMC - 2C = 2A$ and we are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8 cm); 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 = -5.071390865857864, xmax = 11.431848097005963, ymin = -1.2958389640101082, ymax = 7.456288584328322; /* image dimensions */ pair A = (0.,6.455272450515271), B = (-2.,0.), C = (7.,0.), O = (2.5,2.143251320722546), M = (3.587945310591788,5.654743016287771); draw(A--B--C--cycle, linewidth(2.)); /* draw figures */ draw(B--O, linewidth(1.)); draw(O--C, linewidth(1.)); draw(O--A, linewidth(1.)); draw(O--(0.5820156255452035,3.9119815625706624), linewidth(1.)); draw(O--M, linewidth(1.)); draw((0.5820156255452035,3.9119815625706624)--M, linewidth(1.)); draw(B--(0.5820156255452035,3.9119815625706624), linewidth(1.)); draw(M--C, linewidth(1.)); draw((0.5820156255452035,3.9119815625706624)--A, linewidth(1.)); draw(A--M, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (0.05543500867114454,6.590544901030548), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-2.1630331797794025,-0.4), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (7.0490207003109795,-0.2407138499909466), NE * labelscalefactor); dot(O,linewidth(4.pt) + dotstyle); label("$O$", (2.4497573827915518,1.6), NE * labelscalefactor); dot((0.5820156255452035,3.9119815625706624),linewidth(4.pt) + dotstyle); label("$N$", (0.1,3.912150380828061), NE * labelscalefactor); dot(M,linewidth(4.pt) + dotstyle); label("$M$", (3.6401549473259918,5.765382952887358), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

First, note that $OM$ is the external bisector of $\angle AOB$ thus it's parallel to $AB$. Similarly, $ON\parallel AC$. Construct parallelograms $ABOX$ and $ACOY$. Claim: $M,X,C,F$ are concyclic (and similarly $M,Y,B,E$ are concyclic). Proof: Note the spiral similarity $\triangle AME\stackrel{+}{\sim}\triangle AOB$ thus $\triangle AOM\stackrel{+}{\sim}\triangle ABE$. Now we ratio-chase: $$\frac{OC}{OX} = \frac{OA}{AB} = \frac{OM}{BE} = \frac{OM}{OF}$$so we are done. $\blacksquare$ To finish, note that $\{B,X\}$ and $\{C,Y\}$ are symmetric across midpoint of $AO$. Thus $BCXY$ is parallelogram which means $BY\parallel CX$. Finally, $$\angle ENO + \angle OMF = \angle YBO + \angle OCX = \angle BOC = 2\angle BAC$$as desired.

Invert about the circumcircle. Denote by $X, Y$ the images of $M, N$ and $K, L$ the images of $E, F$. It suffices to show that $\overline{KY} \parallel \overline{LX}$, as that will imply $\angle YKO + \angle OLX = \angle KOL = 2 \angle BAC$. To do so, notice that $\frac{KB}{OK} = \frac{KA}{KX}$ and similar variants. Thus, setting $OE = a$, \begin{align*}\frac{KA}{KX} + \frac{LA}{LY} &= 1 \\ \iff \frac{KB}{OK} + \frac{CL}{OL} &= 1 \\ \iff \frac{\frac{R^2}a - R}{\frac{R^2}a} +\frac{\frac{R^2}{R-a} - R}{\frac{R^2}{R-a}} &= 1 \end{align*}which is true algebraically, done.

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