Example 6. Given $\triangle A B C$ is constant, $E, F$ are the trisection points of $B C$, $M$ is the midpoint of $A C$, $B M$ intersects $A E, A F$ at $G, H$ respectively. Find $B C: C H: H M$. (1982 Tianjin Junior High School Competition)
Let $ABM$
$$
=\sim, \angle M B C=\beta \text {. }
$$

Since $M$ is the midpoint of $AC$,
by the Angle Bisector Theorem, we have

thus $\frac{\sin \alpha}{\sin \beta}=\frac{BC}{AB}$.
By the Angle Bisector Theorem again, we get
$$
\begin{aligned}
\frac{AG}{CE} & =\frac{AB \cdot \sin \alpha}{BE \cdot \sin \beta}=\frac{AB}{BE} \cdot \frac{BC}{AB}=\frac{BC}{BE}=3, \\
\frac{AH}{HF} & =\frac{AB \cdot \sin \alpha}{BF \cdot \sin \beta}=\frac{AB}{BF} \cdot \frac{BC}{AB}=\frac{BC}{BF} \\
& =\frac{3}{2} .
\end{aligned}
$$

Considering $MF$, it is easy to see that $GH: HM=AH: HF=3: 2$.
By the Angle Bisector Theorem again, we easily know
$$
\sin \angle BAE / \sin \angle EAF=AF / AB \cdot
$$

Thus, using the Angle Bisector Theorem again, we get
$$
\begin{array}{l}
=\frac{AF}{AH} . \\
\end{array}
$$

Noting that $AH: HF=3: 2$, we know
$$
BG: GH=5: 3 .
$$

Therefore, $BG: GH: HM=5: 3: 2$.