This problem is there on AOPS, I have seen it before,I have forgotten its source. Edit: Found it, : Macedonia MO 2019 P1

Nice problem! Firstly we obtain that $\angle ADB = 90^\circ - \frac{\angle AMB}{2}$ and $\angle AEC = 90^\circ - \frac{\angle AMC}{2}$. Let $P$ and $Q$ be intersections of $(ADB)$ and $(AEC)$ with line $AM$ where $P$ and $Q$ are not $A$. It's easy to prove that $M$ is not on $(ADB)$ nor $(AEC)$, so $P$ and $Q$ lie on ray $AM$. Now we can see that $\angle BDA = 90^\circ - \frac{\angle PMB}{2}$ $\implies$ $MP = MB$. Similarly $MQ = MP = MB$ $(1)$, so $Q = P$. Now by power of point $P$ we obtain $MB * MF = MP * MA = MC * MG$, from $(1)$ it implies $MF= MG = MA$, so $BF=CG$ and we are done!

Let $I_1$ and $I_2$ denotes incenters of $\triangle ABM$ and $\triangle AMC$ respectively. Clearly $I_1 \in \odot(ABD)$ and $I_2 \in \odot(ACE)$. Also it is obvious that points $M,I_1,D$ and $M,I_2,E$ are collinear. Claim: Quadrilateral $I_1I_2DE$ is cyclic. Proof: Quick angle chasing reveals that $\triangle MAI_1 \sim \triangle MDB$. This implies that: $$ MA \cdot MB =MI_1 \cdot MD $$Similarly we can get that $MA \cdot MC = MI_2 \cdot ME$. This implies that: $$ ME \cdot MI_2 = MI_1 \cdot MD $$Consequently quadrilateral $I_1I_2DE$ is cyclic as desired. Now by radical axis theorem on $\odot(ADBI_1)$, $\odot(EAI_2C)$ and $\odot(I_1I_2DE)$ we get that $\odot(ADBI_1)$ and $\odot(EAI_2C)$ intersects on line $AM$. Denote this point of intersection by $X$. Then from PoP follows that: $$MX \cdot MA = MB \cdot MF = MC \cdot MG$$This implies that $BF = CG$ as desired.

From Junior Malaysia 2015 G2 https://artofproblemsolving.com/community/c6h1116026p5108235 we know $ACM$ and $ABM$ intersect at $S$ on $AM$ so we have $MB.MF = MS.MA = MC.MG$ so $MG = MF$ so $BF = CG$.

I used inversion.

Barycentric coordinates: $A(1,0,0),B(0,1,0),M(0,0,1)$ and $a=BM,b=AM,c=AB$. It is well known that $D(a:b:-c)$ \begin{align*} (ADE):&A\in(ADE) \Rightarrow u=0 \\ &B\in (ADE) \Rightarrow v=0 \\ &D\in (ADE) \Rightarrow bca^2+acb^2-abc^2-cw(a+b-c)=0 \Leftrightarrow \\ & abc-cw=0 \Rightarrow w= ab \end{align*}Since $F \in BM$, $F(0,f,1-f)$.Putting the coordinates to the circle equation: $$ -a^2f+ab=0 \Rightarrow f= \frac{b}{a}$$ Therefore $F(0,\frac{b}{a},\frac{b-a}{a})$. $\overrightarrow{FC}(0,\frac{b-a}{a},\frac{a-b}{a})$, Putting these on distance formula: $$|FC|^2: b(b-a)$$ Similiarly, Let $AMC$ be the reference triangle and $AM=b,MC=MB=a,AC=c'$ then $$|CG|^2:b(b-MC)=b(b-a)$$ Trivially they are equal, hence we are done. $\blacksquare$

let $\omega_1, \omega2$ be the two circumcircles mentioned in the problem, respectively. It is quite clear, since $MC_MB$, that $M$ lying on the radical axis of the two circles suffices to finish (this also would show that $MG=MF$, but this is actually a slightly stronger statement that follows from the other one, so the other one should be easier to show) let $P$ be the intersection of $\overline{AM}$ and $\omega_1$. It suffices to show that $P$ lies on $\omega_2$ let $H$ lie on $\overline{AM}$ beyond $A$ let $L$ be the intersection of $\overline{MD}$ and $\omega_1$ let $I$ be the intersection of $\overline{MD}$ and $\overline{PB}$ let $N$ be the intersection of $\overline{ME}$ with $\omega_2$ let $\alpha = \angle HAD$ let $\beta = \angle PCN$ $L$ is the incentre of $\triangle MBA$ by incentre-excentre lemma, by the same lemma, $LD$ is a diametre of $\omega_1$ Now, angle chasing gives: \[ \alpha = \angle DAB = \angle DPB = \angle PBD = \angle PLD = \angle DLB\]where we used angle bisectors, inscribed angle theorem and opposite angles in cyclic quadrilateral summing up to $180°$ further angle chase gives $\angle IPL = \angle BPL = \angle BDL = 90° - \alpha = \angle LDP = \angle LBP = \angle LBI$ From this conclusion, $I$ is the midpoint of $\overline{PB}$, together with $\triangle DPB$ being isosceles, this means $\overline{DI} \perp \overline{PB}$ which implies $\overline{MI} \perp \overline {PB}$, and since $\overline{PI}$ bisects $\angle PMB$, this implies $\triangle PMB$ also being isosceles. We now know $MB=MP=MC$ Additionally, since $\angle EMD = 90°$, $\angle MIP = 90°$ and $\triangle PMC$ is isosceles, $\overline{PC} \perp \overline{ME}$ because $\triangle PMC$ is isosceles, $\triangle PNC$ is also isosceles, therefore $\beta = \angle PCN = \angle NPC$ But we can do more, namely $\angle CNE = 90°- \beta = \angle ENP = \angle ECP$ This implies that $P$ lies on $\omega_2$ since $\angle ECN = 90°$

[asy][asy] pointpen=black+linewidth(2); size(12cm); pair A=dir(120),B=dir(210),C=dir(330); filldraw(A--B--C--cycle,invisible,heavygreen); pair M=(B+C)/2; pair I1=incenter(A,M,B),I2=incenter(A,M,C); pair D=2*circumcenter(A,B,I1)-I1,E=2*circumcenter(A,C,I2)-I2; pair F=IP(circumcircle(A,B,I1),B--(2*B-C),0); pair G=IP(circumcircle(A,C,I2),C--(10*C-9*B),1); filldraw(circumcircle(A,B,I1),invisible,pathpen+dashed); filldraw(circumcircle(A,C,I2),invisible,pathpen+dashed); pair P=IP(CP(M,B),A--M,0); D(F--B,heavygreen); D(G--C,heavygreen); D(A--M); D("A",D(A),dir(35)); D("B",D(B),S); D("C",D(C),S); D("I_1",D(I1),unit(I1-D)); D("I_2",D(I2),unit(I2-E)); D("D",D(D),unit(D-I1)); D("E",D(E),unit(E-I2)); D("M",D(M),S); D("F",D(F),S); D("G",D(G),S); D("P",D(P),dir(12)); [/asy][/asy] Let $I_1$ and $I_2$ be the incenters of triangles $AMB$ and $AMC$, respectively. Further let $\gamma_1 = (ABD)$, and $\gamma_2 = (ACE)$. Note that $\gamma_1$ passes through $I_1$, and that $\gamma_2$ passes through $I_2$, by incenter-excenter lemma. Now let $P$ be the point on ray $\overrightarrow{MA}$ such that $MP = MB = MC$. Clearly $P$ actually lies on segment $\overline{MA}$, as $\angle BPC = 90^\circ$ but $\angle BAC < 90^\circ$. Claim: $P$ lies on $\gamma_1$. Proof. Note \begin{align*} \angle APB &= 180^\circ - \angle BPM \\ &= 180^\circ - \left(90^\circ - \frac{1}{2} \angle PMB\right) \\ &= 90^\circ + \frac{1}{2} \angle AMB \\ &= 90^\circ + \frac{1}{2} (180^\circ - \angle BAM - \angle ABM) \\ &= 180^\circ - \frac{1}{2} \angle BAM - \frac{1}{2} \angle ABM \\ &= 180^\circ - \angle BAI_1 - \angle ABI_1 \\ &= \angle AI_1 B \end{align*}and we are done as $P$ and $I_1$ clearly lie on the same side of line $AC$. $\blacksquare$ Therefore by symmetry $P$ lies on $\gamma_2$. Thus $\gamma_1$ and $\gamma_2$ intersect at $A$ and $P$. Now \[MF = \frac{MB \cdot MF}{MB} = \frac{MP \cdot MA}{\frac{1}{2} BC} = \frac{MC \cdot MG}{MC} = MG.\]Subtracting $\frac{1}{2} BC$ from both sides gives us $BF = CG$, as desired.