《2025年基础训练大象出版社九年级数学全一册北师大版》

15. (★★★)如图4.4-39,在$\triangle ABC$和$\triangle A'B'C'$中,D,D'分别是AB,A'B'上一点,$\frac{AD}{AB}=\frac{A'D'}{A'B'}$.

(1)当$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AB}{A'B'}$时,求证$\triangle ABC\backsim\triangle A'B'C'$.
证明的途径可以用下面的框图表示,请填写其中的空格.

(2)当$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}$时,判断$\triangle ABC$与$\triangle A'B'C'$是否相似,并说明理由.

答案：
15.
(1) $\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AD}{A'D'}$，$\angle A = \angle A'$
(2) $\triangle ABC \sim \triangle A'B'C'$.
理由如下：如图，过点$D$，$D'$分别作$DE // BC$，$D'E' // B'C'$，$DE$交$AC$于$E$，$D'E'$交$A'C'$于$E'$.


$\because DE // BC$，
$\therefore \angle ADE = \angle B$，$\angle AED = \angle ACB$，
$\therefore \triangle ADE \sim \triangle ABC$，
$\therefore \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$
同理可得$\frac{A'D'}{A'B'}=\frac{D'E'}{B'C'}=\frac{A'E'}{A'C'}$
$\because \frac{AD}{AB}=\frac{A'D'}{A'B'}$，
$\therefore \frac{DE}{BC}=\frac{D'E'}{B'C'}$
$\therefore \frac{DE}{D'E'}=\frac{BC}{B'C'}$
同理可得$\frac{AE}{AC}=\frac{A'E'}{A'C'}$，
$\therefore \frac{AC - AE}{AC}=\frac{A'C' - A'E'}{A'C'}$，即$\frac{EC}{AC}=\frac{E'C'}{A'C'}$，
$\therefore \frac{EC}{E'C'}=\frac{AC}{A'C'}$
$\because \frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}$，
$\therefore \frac{CD}{C'D'}=\frac{DE}{D'E'}=\frac{EC}{E'C'}$
$\therefore \triangle DCE \sim \triangle D'C'E'$，
$\therefore \angle CED = \angle C'E'D'$.
$\because DE // BC$，
$\therefore \angle CED + \angle ACB = 180^{\circ}$.
同理可得$\angle C'E'D' + \angle A'C'B' = 180^{\circ}$.
$\therefore \angle ACB = \angle A'C'B'$.
$\because \frac{AC}{A'C'}=\frac{BC}{B'C'}$，
$\therefore \triangle ABC \sim \triangle A'B'C'$.