《2025年课堂作业武汉出版社九年级数学上册人教版》

9.如图,在$\triangle ABC$中,$\angle ACB= 90^{\circ}$,$AC= BC= 2\sqrt{2}$,点$D是AC$的中点,连接$BD$,将$\triangle BCD绕点B$旋转,得到$\triangle BEF$,连接$CF$.当$CF // AB$时,$CF= $______.

答案： $2+\sqrt{6}$或$\sqrt{6}-2$

10.(1)如图①,点$E$,$F分别在正方形ABCD的边BC$,$CD$上,$\angle EAF= 45^{\circ}$.把$\triangle ABE绕点A逆时针旋转90^{\circ}至\triangle ADG$,从而发现$EF= BE+DF$,请利用图①证明上述结论.
(2)如图②,四边形$ABCD$中,$\angle BAD \neq 90^{\circ}$,$AB= AD$,$\angle B+\angle D= 180^{\circ}$,点$E$,$F分别在边BC$,$CD$上,则当$\angle EAF与\angle BAD$满足 ______ 关系时,有$EF= BE+DF$.
(3)如图③,在某公园的同一水平面上,四条通道(忽略通道宽)围成四边形$ABCD$.已知$AB= AD= 80\ m$,$\angle B= 60^{\circ}$,$\angle ADC= 120^{\circ}$,$\angle BAD= 150^{\circ}$.道路$BC$,$CD上分别有景点E$,$F$,且$AE \perp AD$,$DF= 40(\sqrt{3}-1)\ m$.现要在景点$E$,$F$之间修一条笔直的道路,求这条道路$EF$的长.

答案：

(1)证明:如图①,将$\triangle ABE$绕点 A 逆时针旋转$90^{\circ}$至$\triangle ADG$,使 AB 与 AD 重合.

$\because \triangle ADG\cong \triangle ABE$,
$\therefore AG=AE$,$\angle DAG=\angle BAE$,$DG=BE$.
$\because \angle EAF=45^{\circ}$,即$\angle DAF+\angle BAE=\angle EAF=45^{\circ}$,
$\therefore \angle GAF=\angle EAF$.
在$\triangle GAF$和$\triangle EAF$中,
$\left\{\begin{array}{l} AG=AE,\\ \angle GAF=\angle EAF,\\ AF=AF,\end{array}\right.$
$\therefore \triangle AFG\cong \triangle AFE(SAS)$.
$\therefore GF=EF$.
又$\because DG=BE$,
$\therefore GF=BE+DF$,
$\therefore EF=BE+DF$.
(2)$\angle BAD=2\angle EAF$.
理由如下:如图②,延长 CB 至 M,使$BM=DF$,连接 AM,

$\because \angle ABC+\angle D=180^{\circ}$,$\angle ABC+\angle ABM=180^{\circ}$,
$\therefore \angle D=\angle ABM$.
在$\triangle ABM$和$\triangle ADF$中,
$\left\{\begin{array}{l} AB=AD,\\ \angle ABM=\angle D,\\ BM=DF,\end{array}\right.$
$\therefore \triangle ABM\cong \triangle ADF(SAS)$,
$\therefore AF=AM$,$\angle DAF=\angle BAM$.
$\because \angle BAD=2\angle EAF$,
$\therefore \angle DAF+\angle BAE=\angle EAF$,
$\therefore \angle BAE+\angle BAM=\angle EAM=\angle EAF$.
在$\triangle FAE$和$\triangle MAE$中,
$\left\{\begin{array}{l} AE=AE,\\ \angle EAF=\angle EAM,\\ AF=AM,\end{array}\right.$
$\therefore \triangle FAE\cong \triangle MAE(SAS)$,
$\therefore EF=EM=BE+BM=BE+DF$,
即$EF=BE+DF$.
(3)如图③,把$\triangle ABE$绕点 A 逆时针旋转$150^{\circ}$至$\triangle ADG$,连接 AF,过点 A 作$AH\perp GD$,垂足为 H.

$\because \angle BAD=150^{\circ}$,$\angle DAE=90^{\circ}$,
$\therefore \angle BAE=60^{\circ}$.
又$\because \angle B=60^{\circ}$,
$\therefore \triangle ABE$是等边三角形,
$\therefore BE=AB=80m$.
根据旋转的性质得到:$\angle ADG=\angle B=60^{\circ}$,
又$\because \angle ADC=120^{\circ}$,
$\therefore \angle GDF=180^{\circ}$,即点 G 在 CD 的延长线上.
$\because \triangle ADG\cong \triangle ABE$,
$\therefore AG=AE$,$\angle DAG=\angle BAE$,$DG=BE$.
$\because AH=80× \frac{\sqrt{3}}{2}=40\sqrt{3}(m)$,$HF=HD+DF=40+40(\sqrt{3}-1)=40\sqrt{3}(m)$,
$\therefore AH=HF$,
$\therefore \angle HAF=45^{\circ}$,
$\therefore \angle DAF=\angle HAF-\angle HAD=45^{\circ}-30^{\circ}=15^{\circ}$,
$\therefore \angle EAF=\angle DAE-\angle DAF=90^{\circ}-15^{\circ}=75^{\circ}$.
$\because \angle BAD=150^{\circ}=2× 75^{\circ}=2\angle EAF$,
$\therefore$由
(2)得$EF=BE+DF$,
$\therefore EF=80+40(\sqrt{3}-1)=(40+40\sqrt{3})(m)$,
即这条道路 EF 的长为$(40+40\sqrt{3})m$.