Example 7. As shown in the figure, on the sides $BC$, $CA$ of the equilateral $\triangle ABC$, there are points $E$, $F$ respectively, with $BE = CF = a$, $EC = FA = b$.
When $BF$ bisects $AE$, try to compare the sizes of $\frac{2a+b}{b}$, $\frac{a+b}{a-b}$, $\frac{(a+b)^2}{ab}$, $\frac{a^3}{b^3}$.
To compare the sizes of the above four expressions, it is necessary to determine what kind of relationship exists between $a$ and $b$. Draw $E K!$ perpendicular to $C A$ at $K$, easily obtaining $E K = A F = b$. From $\triangle B E K \sim \triangle B C F \Rightarrow \frac{a}{a+b} = \frac{b}{a} \Rightarrow a^2 = b(a+b)$. This implicit equation indicates that $E$ is the golden section point of side $B C$, and $F$ is the golden section point of side $C A$.

With this equation's help, comparing the above four expressions becomes easier. Please see:
$$
\begin{array}{l}
\frac{2a + b}{b} = \frac{a^2 + a(a + b)}{ab} \\
= \frac{b(a + b) + a(a + b)}{ab} \\
= \frac{(a + b)^2}{ab}, \\
\frac{a + b}{a - b} = \frac{(a + b)^2}{a^2 - b^2} = \frac{(a + b)^2}{b(a + b) - b^2} \\
= \frac{(a + b)^2}{ab}, \\
\frac{a^3}{b^3} = \frac{ab(a + b)}{b^3} = \frac{a(a + b)}{b^2} = \frac{a^2(a + b)}{ab^2} \\
= \frac{b(a + b)(a + b)}{ab^2} = \frac{(a + b)^2}{ab}. \\
\text{Therefore, } \frac{2a + b}{b} = \frac{a + b}{a - b} = \frac{(a + b)^2}{ab} = \frac{a^3}{b^3}. \\
\end{array}
$$