7. [2025南阳宛城区模拟]如图,在7×7的正方形网格中,每个小正方形的边长都为1,点A,B,C,D,E,F均在格点上,则到线段AB的两个端点距离相等的点所在的直线是 ()

A.直线 CD
B.直线 CE
C.直线 DE
D.直线 DF

7 C 如图,连接BC,AD,BD,BE,BF. 根据勾股定理,得
$BC = \sqrt{1^2 + 1^2} = \sqrt{2}, AD = BD = \sqrt{2^2 + 1^2} = \sqrt{5}, BE = \sqrt{3^2 + 4^2} = 5, \therefore AE = BE = 5, \therefore$ 点D和点E在线段AB的垂直平分线上,
$\therefore$ 到线段AB的两个端点距离相等的点所在的直线是直线DE.

8. [2026南通期末]）如图,在$△ ABC$中,$AB = AC$,
AD是边BC上的高,动点$M,N$分别在$AD,AC$
上,连接$BM,MN$.若$AB = 10\ \mathrm{cm},BC = 12\ \mathrm{cm},$
则$BM + MN$的最小值是$\mathrm{cm}.$

8 9.6 解题思路:根据“两点之间线段最短”,可将$BM + MN$的最小值转化为线段BN的长. 根据“垂线段最短”可得,当$BN ⊥ AC$时,线段BN取得最小值,再用等面积法求出线段BN的长即可.
$\because AB = AC = 10\ \mathrm{cm}$,$AD$是边BC上的高,$\therefore ∠ ADC = 90°$,$AD$是等腰三角形$ABC$的中线.$\because BC = 12\ \mathrm{cm}, \therefore BD = CD = 6\ \mathrm{cm}$.
在$\mathrm{Rt}△ ADC$中,由勾股定理得,$AD = \sqrt{AC^2 - CD^2} = \sqrt{10^2 - 6^2} = 8(\mathrm{cm})$.根据“两点之间线段最短”可得,当点B,M,N共线时,$BM + MN$取得最小值,即线段BN的长,根据“垂线段最短”可得,当$BN ⊥ AC$时,$BN$取得最小值. 如图,过点B作$BN ⊥ AC$于点N,交AD于点M,$\therefore \frac{1}{2}AC · BN = \frac{1}{2}BC · AD$(等面积法),$\therefore BN = \frac{AD · BC}{AC} = \frac{8 × 12}{10} = 9.6(\mathrm{cm})$,
$\therefore BM + MN$的最小值是9.6 cm.

9. [2026福州期末]如图，△ACB和△ECD都是等腰直角三角形，且$CA=CB$，$CE=CD$，点A在DE上，若$AE=2$，$AD=4$，则AC的长为。

9 $\sqrt{10}$ 解题思路:连接BD,证明$△ BCD ≌ △ ACE$,从而得$∠ BDA = 90°, DB = AE = 2$. 利用勾股定理得$AB^2 = AD^2 + DB^2 = AC^2 + BC^2$,即可求出AC的长.
如图,连接BD(根据“手拉手模型”构造全等三角形).
$\because CA = CB, CE = CD, ∠ ACB = ∠ ECD = 90°, \therefore ∠ ADC = ∠ E = 45°, ∠ ACB - ∠ ACD = ∠ ECD - ∠ ACD$,即$∠ DCB = ∠ ECA$. 在$△ BCD$和$△ ACE$中,$BC = AC, ∠ DCB = ∠ ECA, DC = EC$,
$\therefore △ BCD ≌ △ ACE(\mathrm{SAS}), \therefore BD = AE = 2, ∠ BDC = ∠ E = 45°$,
$\therefore ∠ BDA = ∠ BDC + ∠ ADC = 90°$. 在$\mathrm{Rt}△ ADB$中,$AD = 4$,$BD = 2, \therefore AB^2 = AD^2 + DB^2 = 4^2 + 2^2 = 20$. 在$\mathrm{Rt}△ ABC$中,$\because BC = AC, \therefore AB^2 = AC^2 + BC^2 = AC^2 + AC^2 = 20, \therefore AC^2 = 10$,
$\therefore AC = \sqrt{10}$.

10. [2025枣庄峄城区期中]平面内将长度为8,5,6的三根木棒按如图所示的方式连接成折线A→B→C→D,其中AB可以绕点B任意旋转,∠C=90°,将点A,D用绷直的皮筋连接.设皮筋的长为d,求d的取值范围.

10 解:如图1,连接BD.
$\because ∠ C = 90°, BC = 8, CD = 6$,
$\therefore BD = \sqrt{BC^2 + CD^2} = \sqrt{8^2 + 6^2} = 10$.
$\because AB$可以绕点B任意旋转,$\therefore$ 分情况讨论如下:

①如图2,当点A在线段BD上时,$d = 10 - 5 = 5$.
②如图3,当点A在线段DB的延长线上时,$d = 5 + 10 = 15$.
综上所述,$5 ≤ d ≤ 15$.

11. 新考法 [2026揭阳期末] 用一张正方形纸板制作无盖正方体纸盒.
步骤1:如图1,将正方形纸板的边长三等分,画出九个相同的小正方形,并剪去四个角上的小正方形;
步骤2:如图2,把剪好的纸板折成无盖正方体纸盒.
(1)直接写出纸板上∠ABC与纸盒上∠A₁B₁C₁的大小关系;
(2)证明(1)中你发现的结论.

11 (1)解:$∠ ABC = ∠ A_1B_1C_1$.
(2)证明:如图,连接AC,设小正方形的边长为1,
则$AC = BC = \sqrt{1^2 + 2^2} = \sqrt{5}, AB = \sqrt{1^2 + 3^2} = \sqrt{10}$.
$\because AC^2 + BC^2 = 5 + 5 = AB^2, \therefore △ ABC$为等腰直角三角形.
$\because A_1C_1 = B_1C_1 = 1, A_1C_1 ⊥ B_1C_1, \therefore △ A_1B_1C_1$为等腰直角三角形,
$\therefore ∠ ABC = ∠ A_1B_1C_1 = 45°$.

12. 推理能力 在$△ ABC$中，$AB = 15$，$AC = 20$，$D$是$BC$边所在直线上的点，$AD = 12$，$BD = 9$，求$BC$的长。

12 解:①如图1,当点D在线段BC上时,
$\because AD = 12, BD = 9, AB = 15, \therefore AD^2 + BD^2 = AB^2$,
$\therefore △ ABD$是直角三角形,且$∠ ADB = 90°, \therefore ∠ ADC = 90°$,
$\therefore DC = \sqrt{AC^2 - AD^2} = \sqrt{20^2 - 12^2} = 16$,
$\therefore BC = BD + CD = 9 + 16 = 25$.

②如图2,当点D在CB的延长线上时,
同理可得$DC = 16, BD = 9$,
$\therefore BC = CD - BD = 16 - 9 = 7$.

③$\because AC ＞ AB, \therefore$ 点D不在BC的延长线上.
综上所述,$BC$的长为25或7.