$解：(2)已知： \triangle A^{\prime}\ \mathrm {B}^{\prime}\ \mathrm {C}^{\prime} ∽ \triangle A B C，\ $
$C D 和 C^{\prime}\ \mathrm {E} 分别为 A B 和A^{\prime}\ \mathrm {B}^{\prime} 边上的中线.$
$求证： \frac {C D}{C^{\prime}\ \mathrm {E}}=\frac {B C}{B^{\prime}\ \mathrm {C}^{\prime}}. $
$证明： \because C D 和 C^{\prime}\ \mathrm {E} 分别为 A B 和 A^{\prime}\ \mathrm {B}^{\prime} 边上的中线，$
$\therefore B D=\frac {1}{2}\ \mathrm {A}\ \mathrm {B}， B^{\prime}\ \mathrm {E}=\frac {1}{2}\ \mathrm {A}^{\prime}\ \mathrm {B}^{\prime} ，$
$\therefore \frac {B D}{A B}=\frac {B^{\prime}\ \mathrm {E}}{A^{\prime}\ \mathrm {B}^{\prime}}=-\frac {1}{2}，$
$\therefore \frac {B D}{B^{\prime}\ \mathrm {E}}=\frac {A B}{A^{\prime}\ \mathrm {B}^{\prime}}$
$\because \triangle A B C ∽ \triangle A^{\prime}\ \mathrm {B}^{\prime}\ \mathrm {C}^{\prime}，$
$\therefore \angle C B A=\angle C^{\prime}\ \mathrm {B}^{\prime}\ \mathrm {A}^{\prime}， \frac {B C}{B^{\prime}\ \mathrm {C}}=\frac {A B}{A^{\prime}\ \mathrm {B}^{\prime}}，$
$\therefore \frac {B D}{B^{\prime}\ \mathrm {E}}=\frac {B C}{B^{\prime}\ \mathrm {C}^{\prime}}，$
$\therefore \triangle B C D ∽ \triangle B^{\prime}\ \mathrm {C}^{\prime}\ \mathrm {E}，$
$\therefore \frac {C D}{C^{\prime}\ \mathrm {E}}=\frac {B C}{B^{\prime}\ \mathrm {C}^{\prime}}.$
