$解：(2)△ABC∽△A'B'C'，理由如下：$
$过点D，D'分别作 D E / / B C， D^{\prime}\ \mathrm {E}^{\prime}\ \mathrm {/}\ \mathrm {/} B^{\prime}\ \mathrm {C}^{\prime}， D E 交 A C 于点 E，$
$D^{\prime}\ \mathrm {E}^{\prime} 交 A^{\prime}\ \mathrm {C}^{\prime} 于点 E^{\prime}.$
$因为 D E / / B C，$
$所以 \triangle A D E ∽\triangle A B C，$
$所以 \frac {A D}{A B}=\frac {D E}{B C}=\frac {A E}{A C}.$
$同理可得 \frac {A^{\prime}\ \mathrm {D}^{\prime}}{A^{\prime}\ \mathrm {B}^{\prime}}=\frac {D^{\prime}\ \mathrm {E}^{\prime}}{B^{\prime}\ \mathrm {C}^{\prime}}=\frac {A^{\prime}\ \mathrm {E}^{\prime}}{A^{\prime}\ \mathrm {C}^{\prime}}.$
$因为 \frac {A D}{A B}=\frac {A^{\prime}\ \mathrm {D}^{\prime}}{A^{\prime}\ \mathrm {B}^{\prime}}，$
$所以 \frac {D E}{B C}=\frac {D^{\prime}\ \mathrm {E}^{\prime}}{B^{\prime}\ \mathrm {C}^{\prime}}，$
$所以 \frac {D E}{D^{\prime}\ \mathrm {E}^{\prime}}=\frac {B C}{B^{\prime}\ \mathrm {C}^{\prime}}.$
$同理可得 \frac {A E}{A C}=\frac {A^{\prime}\ \mathrm {E}^{\prime}}{A^{\prime}\ \mathrm {C}^{\prime}}，$
$所以 \frac {A C-A E}{A C}=\frac {A^{\prime}\ \mathrm {C}^{\prime}-A^{\prime}\ \mathrm {E}^{\prime}}{A^{\prime}\ \mathrm {C}^{\prime}}， 即 \frac {E C}{A C}=\frac {E^{\prime}\ \mathrm {C}^{\prime}}{A^{\prime}\ \mathrm {C}^{\prime}}，$
$所以 \frac {E C}{E^{\prime}\ \mathrm {C}^{\prime}}=\frac {A C}{A^{\prime}\ \mathrm {C}^{\prime}}.$
$因为 \frac {C D}{C^{\prime}\ \mathrm {D}^{\prime}}=\frac {A C}{A^{\prime}\ \mathrm {C}^{\prime}}=\frac {B C}{B^{\prime}\ \mathrm {C}^{\prime}}，$
$所以 \frac {C D}{C^{\prime}\ \mathrm {D}^{\prime}}=\frac {E C}{E^{\prime}\ \mathrm {C}^{\prime}}=\frac {D E}{D^{\prime}\ \mathrm {E}^{\prime}}，$
$所以 \triangle D C E \backsim \triangle D^{\prime}\ \mathrm {C}^{\prime}\ \mathrm {E}^{\prime}，$
$所以 \angle C E D=\angle C^{\prime}\ \mathrm {E}^{\prime}\ \mathrm {D}^{\prime}.$
$因为 D E / / B C，$
$所以 \angle C E D+\angle A C B=180^{\circ}.$
$同理可得 \angle C^{\prime}\ \mathrm {E}^{\prime}\ \mathrm {D}^{\prime}+\angle A^{\prime}\ \mathrm {C}^{\prime}\ \mathrm {B}^{\prime}=180^{\circ}，$
$所以 \angle A C B=\angle A^{\prime}\ \mathrm {C}^{\prime}\ \mathrm {B}^{\prime}.$
$又 \frac {A C}{A^{\prime}\ \mathrm {C}^{\prime}}=\frac {B C}{B^{\prime}\ \mathrm {C}^{\prime}}，$
$所以 \triangle A B C \sim \triangle A^{\prime}\ \mathrm {B}^{\prime}\ \mathrm {C}^{\prime}.$
