《2025年零障碍导教导学案九年级数学全一册人教版》

平行于三角形一边的直线和其他两边相交,所构成的三角形与原三角形相似.
几何语言:如图,
∵ $DE// BC$,∴ $\triangle ADE\backsim \triangle ABC$.

答案： $DE// BC$ $\triangle ADE\backsim \triangle ABC$

若两个三角形的三组对应边的比相等,则这两个三角形相似.
几何语言:如图,

∵$\frac {AB}{A'B'}=\frac {AC}{A'C'}=\frac {BC}{B'C'}$,
∴$\triangle ABC\backsim \triangle A'B'C'$.

答案： 相等 $\frac {AB}{A'B'}=\frac {AC}{A'C'}=\frac {BC}{B'C'}$ $\triangle ABC\backsim \triangle A'B'C'$

1. 如图,根据条件证明:$\triangle ABC \backsim \triangle A^{\prime} B^{\prime} C^{\prime}$.

答案： 证明：$\because \frac {AB}{A'B'}=\frac {2}{4}=\frac {1}{2},$
$\frac {AC}{A'C'}=\frac {3}{6}=\frac {1}{2},$
$\frac {BC}{B'C'}=\frac {4}{8}=\frac {1}{2},$
$\therefore \frac {AB}{A'B'}=\frac {AC}{A'C'}=\frac {BC}{B'C'}=\frac {1}{2}.$
$\therefore \triangle ABC\backsim \triangle A'B'C'.$

2. 根据下面的条件证明$\triangle ABC \backsim \triangle A^{\prime} B^{\prime} C^{\prime}$.
已知:$A B= 10, B C= 8, C A= 6, A^{\prime} B^{\prime}=5$,
$B^{\prime} C^{\prime}=4, C^{\prime} A^{\prime}=3$.
证明：$\because \frac {AB}{A'B'}=\frac {10}{5}=2,$
$\frac {BC}{B'C'}=\frac {8}{4}=2,$
$\frac {CA}{C'A'}=\frac {6}{3}=2,$
$\therefore \frac {AB}{A'B'}=\frac {BC}{B'C'}=\frac {CA}{C'A'}=2.$
$\therefore \triangle ABC\backsim \triangle A'B'C'.$

答案： 证明：$\because \frac {AB}{A'B'}=\frac {10}{5}=2,$
$\frac {BC}{B'C'}=\frac {8}{4}=2,$
$\frac {CA}{C'A'}=\frac {6}{3}=2,$
$\therefore \frac {AB}{A'B'}=\frac {BC}{B'C'}=\frac {CA}{C'A'}=2.$
$\therefore \triangle ABC\backsim \triangle A'B'C'.$

3. 如图所示的网格图中,每个方格都是边长为1的正方形. 求证:$\triangle A B C \backsim \triangle D E F$.
证明：$\because AC=\sqrt {1^{2}+1^{2}}=\sqrt {2},$$BC=\sqrt {1^{2}+3^{2}}=\sqrt {10},AB=4,$$DF=\sqrt {2^{2}+2^{2}}=2\sqrt {2},ED=8,$$EF=\sqrt {2^{2}+6^{2}}=2\sqrt {10},$$\therefore \frac {AC}{DF}=\frac {\sqrt {2}}{2\sqrt {2}}=\frac {1}{2},$$\frac {BC}{EF}=\frac {\sqrt {10}}{2\sqrt {10}}=\frac {1}{2},$$\frac {AB}{ED}=\frac {4}{8}=\frac {1}{2}.$$\therefore \frac {AC}{DF}=\frac {BC}{EF}=\frac {AB}{ED}=\frac {1}{2}.$$\therefore \triangle ABC\backsim \triangle DEF.$

答案： 证明：$\because AC=\sqrt {1^{2}+1^{2}}=\sqrt {2},$
$BC=\sqrt {1^{2}+3^{2}}=\sqrt {10},AB=4,$
$DF=\sqrt {2^{2}+2^{2}}=2\sqrt {2},ED=8,$
$EF=\sqrt {2^{2}+6^{2}}=2\sqrt {10},$
$\therefore \frac {AC}{DF}=\frac {\sqrt {2}}{2\sqrt {2}}=\frac {1}{2},$
$\frac {BC}{EF}=\frac {\sqrt {10}}{2\sqrt {10}}=\frac {1}{2},$
$\frac {AB}{ED}=\frac {4}{8}=\frac {1}{2}.$
$\therefore \frac {AC}{DF}=\frac {BC}{EF}=\frac {AB}{ED}=\frac {1}{2}.$
$\therefore \triangle ABC\backsim \triangle DEF.$

4. 如图,在正方形网格中,$\triangle A B C和\triangle D E F$的顶点都在边长为1的小正方形的顶点上.
(1)填空:$A C= $$2\sqrt{5}$,$B C= $$2\sqrt{2}$.
(2)$\triangle A B C与\triangle D E F$是否相似? 证明你的结论.

答案： 解：
(1)$2\sqrt {5}$ $2\sqrt {2}$
(2)相似. 证明如下：
$\because AC=2\sqrt {5},BC=2\sqrt {2},AB=2,$
$DF=\sqrt {10},EF=2,DE=\sqrt {2},$
$\therefore \frac {AC}{DF}=\frac {2\sqrt {5}}{\sqrt {10}}=\sqrt {2},$
$\frac {BC}{EF}=\frac {2\sqrt {2}}{2}=\sqrt {2},$
$\frac {AB}{DE}=\frac {2}{\sqrt {2}}=\sqrt {2},$
$\therefore \frac {AC}{DF}=\frac {BC}{EF}=\frac {AB}{DE}=\sqrt {2}.$
$\therefore \triangle ABC\backsim \triangle DEF.$