5. In $\triangle A B C$, points $E$ and $F$ on side $B C$ trisect $B C$, $B M$ is the median of side $A C$, and $A E, A F$ intersect $B M$ at points $G, H$ respectively. Then $B G: G H: H M$ equals ( ).
(A) $4: 3 \div 2$
(B) $4: 2: 3$
(C) $5 \pm 3: 2$
(D) $3: 3: 2$
5. C

As shown in the figure, draw $C H^{\prime} / / A H, C G^{\prime} / / A G$, intersecting the extension of $B M$ at $H^{\prime}, G^{\prime}$, then $G H = G^{\prime} H^{\prime}, H M = H^{\prime} M$. For convenience, let $B G = x, G H = y$, $H M = z$, then $\frac{x}{2 y + 2 z} = \frac{B E}{E C} = \frac{1}{2}$. Therefore, $x = y - z, \frac{x + y}{2 z} = \frac{B F}{\bar{P} C} = \frac{2}{1}$. Thus, $x + y = 4 z$. Hence, $x := \frac{5}{2} z, y = \frac{3}{2} z$. Therefore, $B G : G H : H M = x : y : z = 5 : 3 : 2$.