过点$M$作$MK$∥$BC$，交$AF$，$AE$分别于$K$，$N$，

$\because M$是$AC$的中点，

$\therefore \dfrac{MN}{EC}=\dfrac{NK}{EF}=\dfrac{AN}{AE}=\dfrac{AM}{AC}=\dfrac{1}{2}$，

$\because E$、$F$是$BC$的三等分点，

$\therefore BE=EF=FC$，

$\therefore MN=2NK$，

$\because \dfrac{MH}{BH}=\dfrac{MK}{BF}=\dfrac{1}{4}$，$\dfrac{MG}{BG}=\dfrac{MN}{BE}=1$，

$\therefore MH=\dfrac{1}{4}BH$，$MG=BG$，

设$MH=a$，$BH=4a$，$BG=GM=\dfrac{5a}{2}$，

$\therefore GH=GM-MN=\dfrac{3a}{2}$，

$\therefore BG:GH:HM=\dfrac{5a}{2}:\dfrac{3a}{2}:a=5:3:2$.

故答案为：$5:3:2$.

过点$M$作$MK$∥$BC$，交$AF$，$AE$分别于$K$，$N$，

$\because M$是$AC$的中点，

$\therefore \dfrac{MN}{EC}=\dfrac{NK}{EF}=\dfrac{AN}{AE}=\dfrac{AM}{AC}=\dfrac{1}{2}$，

$\because E$、$F$是$BC$的三等分点，

$\therefore BE=EF=FC$，

$\therefore MN=2NK$，

$\because \dfrac{MH}{BH}=\dfrac{MK}{BF}=\dfrac{1}{4}$，$\dfrac{MG}{BG}=\dfrac{MN}{BE}=1$，

$\therefore MH=\dfrac{1}{4}BH$，$MG=BG$，

设$MH=a$，$BH=4a$，$BG=GM=\dfrac{5a}{2}$，

$\therefore GH=GM-MN=\dfrac{3a}{2}$，

$\therefore BG:GH:HM=\dfrac{5a}{2}:\dfrac{3a}{2}:a=5:3:2$.

故答案为：$5:3:2$.