《2025年名校题库九年级数学全一册北师大版》

1. (双流区二诊)如图,在$Rt△ABC$中,$∠ACB= 90^{\circ }$,D 为 AB 边上的一点,且点 D 到 BC 的距离等于点 D 到 AC 的距离.将$△ABC$绕点 D 旋转得到$△A'B'C'$,连接$BB',CC'$.若$\frac {CC'}{BB'}= \frac {3\sqrt {2}}{5}$,则$\frac {AC}{BC}$的值为____.

答案：
$\frac{3}{4}$ 【解析】如图,连接 $DC,DC'$,过点 $D$ 作 $DE\perp BC$ 于点 $E$. $\because\triangle ABC$ 绕点 $D$ 旋转得到 $\triangle A'B'C'$, $\therefore DB = DB',DC = DC',\angle BDB' = \angle CDC',\therefore \frac{BD}{CD} = \frac{B'D}{C'D},\therefore \triangle DBB'\backsim\triangle DCC',\therefore \frac{BD}{CD} = \frac{BB'}{CC'} = \frac{5}{3\sqrt{2}}$. 设 $DC = 3\sqrt{2}x,BD = 5x$. $\because$ 点 $D$ 到 $BC$ 的距离等于点 $D$ 到 $AC$ 的距离, $\therefore \angle ACD = \angle DCB = 45^{\circ},\therefore DE = 3x,\therefore BE = \sqrt{BD^{2}-DE^{2}} = \sqrt{(5x)^{2}-(3x)^{2}} = 4x$. 又 $\because DE// AC,\therefore \frac{DE}{AC} = \frac{BE}{BC},\therefore \frac{AC}{BC} = \frac{DE}{BE} = \frac{3x}{4x} = \frac{3}{4}$.

2. (高新区二诊)如图,在等腰$Rt△ABC$中,$AC= BC= 6\sqrt {2},∠EDF$的顶点 D 是 AB 的中点,且$∠EDF= 45^{\circ }$,现将$∠EDF$绕点 D 旋转一周,在旋转过程中,$∠EDF$的两边 DE,DF 分别交直线 AC 于点 G,H,把$△DGH$沿 DH 折叠,点 G 落在点 M 处,连接 AM.若$\frac {AH}{AM}= \frac {3}{4}$,则 AH 的长为$\frac{9\sqrt{2}}{2}$ 或 $\frac{3\sqrt{2}}{2}$ 或 $3\sqrt{2}$.

答案： $\frac{9\sqrt{2}}{2}$ 或 $\frac{3\sqrt{2}}{2}$ 或 $3\sqrt{2}$

3. (成华区一诊)如图,在$△ABC$中,$∠ACB= 90^{\circ },BC= 5,AC= 10$,点 D,E 分别在 AC,AB 边上,满足$AE= \sqrt {5}AD$.连接 DE,将$△ADE$沿 DE 翻折,得到$△FDE$.连接 CE,CF.若$△CEF的面积和△CEB$的面积相等,则 AD 的长为____.

答案：
$\frac{15 - 5\sqrt{3}}{3}$ 【解析】$\because AE = \sqrt{5}AD,\therefore$ 设 $AD = x$, 则 $AE = \sqrt{5}x$. $\because\triangle ADE$ 沿 $DE$ 翻折,得到 $\triangle FDE,\therefore DF = AD = x,\angle ADE = \angle FDE$. 如图,过点 $E$ 作 $EH\perp AC$ 于点 $H$,设 $EF$ 与 $AC$ 相交于点 $M$, 则 $\angle AHE = \angle ACB = 90^{\circ}$. 又 $\because \angle A = \angle A$,
$\therefore\triangle AHE\backsim\triangle ACB,\therefore \frac{EH}{BC} = \frac{AH}{AC} = \frac{AE}{AB}.\because CB = 5,CA = 10,\therefore AB = \sqrt{AC^{2}+BC^{2}} = \sqrt{10^{2}+5^{2}} = 5\sqrt{5},\therefore \frac{EH}{5} = \frac{AH}{10} = \frac{\sqrt{5}x}{5\sqrt{5}},\therefore EH = x,AH = 2x,\therefore DH = AH - AD = x = EH,\therefore Rt\triangle EHD$ 是等腰直角三角形, $\therefore \angle HDE = \angle HED = 45^{\circ},\therefore \angle ADE = \angle EDF = 135^{\circ},\therefore \angle FDM = 135^{\circ}-45^{\circ} = 90^{\circ}$. 在 $\triangle FDM$ 和 $\triangle EHM$ 中, $\begin{cases}\angle FDM = \angle EHM = 90^{\circ},\\\angle DMF = \angle HME,\\DF = HE,\end{cases}\therefore \triangle FDM\cong\triangle EHM(AAS),\therefore DM = MH = \frac{1}{2}x,CM = AC - AD - DM = 10 - \frac{3}{2}x,\therefore S_{\triangle CEF} = S_{\triangle CME} + S_{\triangle CMF} = \frac{1}{2}CM\cdot EH + \frac{1}{2}CM\cdot DF = \frac{1}{2}(10 - \frac{3}{2}x)\cdot x\times 2 = (10 - \frac{3}{2}x)\cdot x,S_{\triangle CBE} = S_{\triangle ABC} - S_{\triangle ACE} = \frac{1}{2}BC\cdot AC - \frac{1}{2}AC\cdot EH = 25 - 5x.\because\triangle CEF$ 的面积和 $\triangle CEB$ 的面积相等, $\therefore (10 - \frac{3}{2}x)\cdot x = 25 - 5x$, 解得 $x_{1} = \frac{15 - 5\sqrt{3}}{3},x_{2} = \frac{15 + 5\sqrt{3}}{3}$ (舍去), 则 $AD = \frac{15 - 5\sqrt{3}}{3}$.

4. (双流区一诊)如图,将一张正方形纸片进行如下操作:对折正方形 ABCD 得折痕 EF,连接 CE,将 CB 折叠到 CE 上,点 B 的对应点为 N,得折痕 CM,则$\frac {AM}{BM}$的值为____.

答案：
$\frac{\sqrt{5}-1}{2}$ 【解析】如图,延长 $DA,CM$ 相交于点 $P$. 根据折叠的性质,得 $\angle BCM = \angle ECM$. $\because$ 四边形 $ABCD$ 是正方形, $\therefore AD// BC,\therefore \angle P = \angle BCM = \angle ECM,\therefore EP = EC.\because ED = \frac{1}{2}AD,\therefore ED = \frac{1}{2}CD,\therefore CE = \sqrt{DE^{2}+CD^{2}} = \sqrt{5}DE,\therefore EP = \sqrt{5}DE,\therefore AP = PE - AE = \sqrt{5}DE - DE = (\sqrt{5}-1)DE.\because AD// BC,\therefore\triangle APM\backsim\triangle BCM,\therefore \frac{AM}{BM} = \frac{AP}{BC} = \frac{(\sqrt{5}-1)DE}{2DE} = \frac{\sqrt{5}-1}{2}$.