10. 将$△ ABC$的纸片按如图所示的方式折叠，使点$B$落在边$AC$上，记为点$B'$，折痕为$EF$，已知$AB=AC=8$，$BC=10$，如果以点$B'$，$F$，$C$为顶点的三角形与$△ ABC$相似，那么$BF$的长度是。

10.$\frac{40}{9}$ 或 5 【点拨】设 $BF=x$，由题意知 $B'F=BF=x$，$\therefore FC=BC-BF=10-x$.①若$△ CFB'∽△ CBA$,则$\frac{CF}{CB}=\frac{B'F}{AB}$，$\therefore \frac{10-x}{10}=\frac{x}{8}$,解得 $x=\frac{40}{9}$；②若$△ CFB'∽△ CAB$,则$\frac{CF}{CA}=\frac{FB'}{AB}$，$\therefore \frac{10-x}{8}=\frac{x}{8}$,解得 $x=5$.综上所述，当$BF=\frac{40}{9}$或5时，以点$B'$，$F$，$C$为顶点的三角形与$△ ABC$相似.

11.如图,网格中的小方格是边长为1的正方形,A,B,C,D四点都在格点(网格线的交点)上.
(1)找出图中一组相似三角形,并给予证明;
(2)作∠ABC和∠ACD的平分线BM,CM,求∠BMC的度数.

11.【解】(1)$△ ADC∽△ ACB$.证明如下：由题意得$AD=2\sqrt{5},AC=5\sqrt{2},DC=\sqrt{10},BC=5,AB=5\sqrt{5}$，$\therefore \frac{DC}{BC}=\frac{AC}{AB}=\frac{AD}{AC}=\frac{\sqrt{10}}{5},\therefore △ ADC∽△ ACB$.(2)如图，取格点$H$，连接$CH$.由$△ ADC∽△ ACB$得$∠ ACD=∠ ABC$.又$\because BM$，$CM$分别平分$∠ ABC$和$∠ ACD$，$\therefore ∠ ABM=∠ CBM=∠ DCM=∠ ACM$，$\therefore ∠ BMC=180°-∠ MBC-∠ BCD-∠ DCM=180°-∠ ABM-∠ CBM-∠ BCD-∠ BDC$.由勾股定理可得$CH=\sqrt{10},CD=\sqrt{10},DH=2\sqrt{5}$.$\therefore CH^2+CD^2=20=DH^2,\therefore ∠ DCH=90°$.又$\because CH=CD,\therefore ∠ BDC=45°,\therefore ∠ BMC=∠ BDC=45°$.

12. 如图,在$△ ABC$和$△ A'B'C'$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:$△ ABC ∽ △ A'B'C'$.
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$△ ABC$与$△ A'B'C'$是否相似,并说明理由.

12.【解】(1)$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AD}{A'D'}$；$∠ A=∠ A'$(2)$△ ABC$与$△ A'B'C'$相似.理由如下：如图，过点$D,D'$分别作$DE// BC,D'E'// B'C'$,$DE$交$AC$于点$E$,$D'E'$交$A'C'$于点$E'$. $\because DE// BC,\therefore △ ADE∽△ ABC$.$\therefore \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$.同理可证$\frac{A'D'}{A'B'}=\frac{D'E'}{B'C'}=\frac{A'E'}{A'C'}$.又$\because \frac{AD}{AB}=\frac{A'D'}{A'B'},\therefore \frac{DE}{BC}=\frac{D'E'}{B'C'},\frac{AE}{AC}=\frac{A'E'}{A'C'}$.$\therefore \frac{DE}{D'E'}=\frac{BC}{B'C'}$,$\frac{AC-AE}{AC}=\frac{A'C'-A'E'}{A'C'}$，即$\frac{EC}{AC}=\frac{E'C'}{A'C'}$.$\therefore \frac{EC}{E'C'}=\frac{AC}{A'C'}$.又$\because \frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{BC}{B'C'},\therefore \frac{CD}{C'D'}=\frac{DE}{D'E'}=\frac{EC}{E'C'}$.$\therefore △ DCE∽△ D'C'E'$.$\therefore ∠ CED=∠ C'E'D'$.$\because DE// BC,\therefore ∠ CED+∠ ACB=180°$.同理$∠ C'E'D'+∠ A'C'B'=180°$.$\therefore ∠ ACB=∠ A'C'B'$.又$\because \frac{AC}{A'C'}=\frac{BC}{B'C'},\therefore △ ABC∽△ A'B'C'$.