2. In isosceles $\triangle A B C$, the vertex angle $\angle A=108^{\circ}, B C$ $=m, A B=A C=n$, let $x=\frac{m+n}{m-n}, y=$ $\frac{(m+n)^{2}}{m n}, z=\frac{m^{3}}{n^{3}}$. Then the size relationship of $x 、 y 、 z$ is ( ).
(A) $x>y>z$
(B) $x<y<z$
(C) $x=y \neq z$
(D) $x=y=z$
2. (D).

Take $D$ on $BC$ such that $DC = AC = n$. It is easy to get
$$
\begin{array}{l} 
\triangle D B A \sim \triangle A B \\
\Rightarrow (m-n): n = n: m \\
\Rightarrow n^{2} = m(m-n) \\
\Rightarrow m^{2} = n(m+n) \\
\Rightarrow \frac{m+n}{m-n} = \frac{m^{3}}{n^{3}} . \\
\text { Also } \frac{(m+n)^{2}}{m n} = \frac{[(m+n) n]^{2}}{m n^{3}} = \frac{m^{4}}{m n^{3}} = \frac{m^{3}}{n^{3}}, \\
\therefore x = y = z .
\end{array}
$$