Example 17 As shown in Figure 1.3.16, from the vertex $A$ of $\triangle A B C$ to the trisection points $E, F$ of the opposite side $B C$, segments $A E, A F$ are drawn. The median $B D$ from vertex $B$ is divided into the continuous ratio $x: y: z$ by these segments, with $x \geqslant y \geqslant z$. Find $x: y: z$.
Let $B D$ intersect $A E$ at $G$, and intersect $A F$ at $H$, then $B G=x, G H=y, H D=z$.
By the property of triangles with equal heights, we have
$$
x: y: z=S_{\triangle A B G}: S_{\triangle A C H}: S_{\triangle A H B}
$$

Consider $\triangle A H D$ as an inscribed triangle of $\triangle A B C$, and note the conclusion from Example 2, taking $\lambda_{1}=\frac{A A}{A B}=0, \lambda_{2}=\frac{B F}{F C}=$ $2, \lambda_{3}=\frac{C D}{D A}=1$, then we get

Similarly,
$$
\begin{array}{l}
S_{\triangle A F I D}=\frac{1}{10} S_{\triangle A B C} \\
S_{\triangle A C H}=\frac{1}{4} S_{\triangle A B C}
\end{array}
$$

Therefore,
$$
S_{\triangle A G H}=S_{\triangle A G D}-S_{\triangle A H D}=\frac{3}{20} S_{\triangle A B C}
$$

And
$$
S_{\triangle A B D}=\frac{1}{2} S_{\triangle A B C}
$$

Thus,
$$
S_{\triangle A B C}=S_{\triangle A B D}-S_{\triangle A C I}=\frac{1}{4} S_{\triangle A B C}
$$

Therefore, $x: y: z=S_{\triangle A B C}: S_{\triangle A G H}: S_{\triangle A H D}=\frac{1}{4}: \frac{3}{20}: \frac{1}{10}=5: 3: 2$