I'm not sure, but this is my intento of solution: Let A'' and B'' / A''CA'A and CB''B'B are paralelogramos (A'' and A' in distintos semiplanes respecto AC, and B and B'' in distintos semiplanes respecto B'C). Then A''B // ME, A'B'' // MF, and A''B=2.ME, A'B''=2.MF... then A''B=A'B'' CA=x=CA', CB=y=CB', then A''C=x.k and B''C= yk (because ACA' and BCB' are similar isosceles). Además <A''CB=A'CB''. See A''CB and A'CB triangles, we deduce that k=1 or x=y (if k no is 1, rotando the triangles obtenemos un trapecio with diagonals =, then is trapecio isosceles, and x=y). Then k=1, the triangles A''CB, A'CB are similars, and ACA' & BAB' are equilaterals... well, the angle is 60 because A''CA'=120 and A''CA'P (with P=A''B_A'B'', reference: _=intersection) is ciclic. Sorry for the redaction and word, I don't know english.

Well, I am not sure to understand all what you said above, Maybe because of the language Can anyone translate, or post a solution in English ??

Well, sorry for the difficult, It's very very difficult to me, too. However I try again, this time is with moooore details, I hope that you understand. references: GH // RT = GH paralel with RT, <ABC= angle ABC = <(AB,BC) Solution: Let R in the line A'E that RE=EA' (E bettwen R and A'). Let S in the line BF that BF=FS. Then, RAA'C and CBB'S are 'paralelogramos' (RC//AA', etc). And AC=CA', BC=BC' (because A'CB' is a rotation of ACB), then CA=RA and CB'=B'S. Let L and N the midpoints of CB and CA'. 2.EN=AA'=RC and EN//RC, 2MN=CB and MN//CB, then RCB and ENM are homotetic, then RB=2.EM and RB//EM. Analogal, 2.LF=BB'=CS and CS//LF, 2.LM=CA' and CA'//LM, then FLM and CSB are homotetic, then A'S=2.MF and MF//A'S Then RB=A'S and the angle EMF is <(RB,SA') Let x=AC and y=CB. Let 1/k=AC/AA'=AC/RC, then k=CB/BB' (because ACA' and BCB' are similar isosceles). Then RC=kx and CS=ky. If k not is 1, I will prove that AC=AB (contradiction). The rotation of A' and S, with center C and angle A'CR--> A'' and S', with A'' in the line RC and S' in the line CB. Then RS'//A''B, because RC=kx, CS'=ky, and CA''=x, CB=y. But RB=A''S', then RS'BA'' is a 'trapecio' isosceles, then x=y and contradiction. Then k=1. Finally, RCA, ACA', BCB' and CB'S are equilateral triangles. Well, not is difficult to see that <(RB,SA')=EMF=60º (because RCB and A'CS are similar triangles, with RCB=A'CS=2.60=120). Well, I hope that the problem of the language not imposibility the comunication, and I hope that the solution will be good, jeje. Yes, sorry, my redaction and words are terrible, but I never studied english.

Let $AC=2a$, $BC=2b$, $\angle ACB= \gamma$ and $\alpha = \angle ACA' = \angle BCB'$. Since $AC=A'C$ and $BC=B'C$: triangles $ACA'$ and $BCB'$ are similars, we put $AA'=2ak$ and $BB'=2bk$. Let $E'$ and $F'$ the midpoints of $A'C$ and $BC$, respect. Lemma: $\underline{ \angle EE'M= \angle FF'M}$ Since $EE'//AA'$, $E'M//CB$ ; $FF'// B'B$, $MF'//A'C$ and $\angle A'AC= \angle BB'C$, $\angle ACB=\angle B'CA'$ the lemma follows. $\blacksquare$ Clearly: $EE'=ak$, $E'M=b$ ; $FF'=bk$, $F'M=a$ ; using the lemma and the fact that $EM=MF$ we get (from the cosine law) : \[ \cos \angle EE'M= \frac{b^2+(ak)^2-EM^2}{abk}=\frac{a^2+(bk)^2-MF^2}{abk}=\cos \angle FF'M \] \[ \Rightarrow b^2+a^2k^2=a^2+b^2k^2 \] \[ \Rightarrow (k^2-1)(a^2-b^2)=0 \] since $a \neq b$ we get: $k=1$ $\Rightarrow$ $AA'=2a$ , and triangles $ACA'$ and $BCB'$ are equilaterals. Since $\angle EE'A=120^ \circ$ and $\angle A'E'M= \angle A'CB= \gamma-60^ \circ$ we get: $\angle EE'M=60^ \circ + \gamma$ , with this, clearly the triangles $EE'M$ and $ECF$ are congruents, so $EF=EM$ and since $ME=MF$, the triangle $EMF$ is equilateral, so $\angle EMF=60^ \circ$. $Tipe$

Well, sorry for the difficult, It's very very difficult to me, too. However I try again, this time is with moooore details, I hope that you understand. references: $GH // RT = GH$ paralel with $RT$, $<ABC= angle ABC = <(AB,BC)$ Solution: Let $R$ in the line $A'E$ that $RE=EA'$ ($E$ bettwen $R$ and $A'$). Let $S$ in the line $BF$ that $BF=FS$. Then, $RAA'C$ and $CBB'S$ are 'paralelogramos' ($RC//AA'$, etc). And $AC=CA'$, $BC=BC'$ (because $A'CB'$ is a rotation of $ACB$), then $CA=RA$ and $CB'=B'S$. Let $L$ and $N$ the midpoints of $CB$ and $CA'$. 2.$EN=AA'=RC$ and $EN//RC$, $2MN=CB$ and $MN//CB$, then $RCB$ and $ENM$ are homotetic, then $RB=2.EM$ and $RB//EM$. Analogal, $2.LF=BB'=CS$ and $CS//LF$, $2.LM=CA'$ and $CA'//LM$, then $FLM$ and $CSB$ are homotetic, then $A'S=2.MF$ and $MF//A'S$ Then $RB=A'S$ and the angle $EMF$ is $<(RB,SA')$ Let $x=AC$ and $y=CB$. Let $1/k=AC/AA'=AC/RC$, then $k=CB/BB'$ (because $ACA'$ and $BCB'$ are similar isosceles). Then $RC=kx$ and $CS=ky$. If k not is 1, I will prove that $AC=AB$ (contradiction). The rotation of $A'$ and $S$, with center $C$ and angle $A'CR-\to A''$ and $S'$, with $A''$ in the line $RC$ and $S'$ in the line $CB$. Then $RS'//A''B$, because $RC=kx$, $CS'=ky$, and $CA''=x$, $CB=y$. But $RB=A''S'$, then $RS'BA''$ is a 'trapecio' isosceles, then $x=y$ and contradiction. Then $k=1$. Finally, $RCA$, $ACA'$, $BCB'$ and $CB'S$ are equilateral triangles. Well, not is difficult to see that $<(RB,SA')=EMF=60º$ (because $RCB$ and $A'CS$ are similar triangles, with $RCB=A'CS=2.60=120$). Well, I hope that the problem of the language not imposibility the comunication, and I hope that the solution will be good, jeje. Yes, sorry, my redaction and words are terrible, but I never studied english. edited by Davron Latipov

Nice problem WLOG we can assume that the the angle of rotation $\theta$ is between $0$ and $\pi$.Other cases can be proven analogously. Let $c$ be the origin. Then,$a'=ae^{i\theta}$,$b'=be^{i\theta}$,$f=\frac{be^{i\theta}}{2}$,$e=\frac{a}{2}$ and $m=\frac{b+ae^{i\theta}}{2}$,$|b|\neq |a|$. The given condition is $|f-m|=|e-m|\Rightarrow (f-m)(\overline{f-m})=(e-m)(\overline{e-m})$.This simplifies to(after not-so-long calculations ) to $(|b|^2-|a|^2)(e^{i\theta}+e^{-i\theta}-1)=0$.Since $|b|\neq |a|$,this implies $e^{i\theta}+e^{-i\theta}=1\Rightarrow\cos\theta=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{3}$ since $\theta\in (0,\pi]$. Now,$\frac{f-m}{e-m}=\frac{be^{i\frac{\pi}{3}}-b-ae^{i\frac{\pi}{3}}}{a-b-ae^{i\frac{\pi}{3}}}=\frac{-b-b\omega+a}{a\omega-b\omega+a}$(multiplying the nominator and the denominator by $\omega=e^{i\frac{2\pi}{3}}$). Further,$\frac{-b-b\omega+a}{a\omega-b\omega+a}=\frac{b\omega^2+a}{-a\omega^2-b\omega}=\frac{b\omega^2+a}{-a\omega^2-b\omega^4}=\frac{-1}{\omega^2}\left(\frac{b\omega^2+a}{b\omega^2+a}\right)=\frac{e^{i\pi}}{e^{i\frac{4\pi}{3}}}=e^{-i\frac{\pi}{3}}$.So,$\angle EMF=\frac{\pi}{3}$.