《2026年通城学典全国中考试题分类精粹中考数学》

17. (2025·南充)在矩形ABCD中，$AB = 10$，$AD = 17$，E是线段BC上异于点B的一个动点，连接AE，把$\triangle ABE$沿直线AE折叠，使点B落在点P处。
【初步感知】(1) 如图①，当E为BC的中点时，延长AP交CD于点F，求证：$FP = FC$。
【深入探究】(2) 如图②，点M在线段CD上，$CM = 4$。求在点E的移动过程中，PM长的最小值。
【拓展运用】(3) 如图②，点N在线段AD上，$AN = 4$。点E在移动过程中，点P在矩形内部，当$\triangle PDN$是以DN为斜边的直角三角形时，求BE的长。

答案：

(1) 连接$EF$。$\because$四边形$ABCD$为矩形，$\therefore \angle D = \angle B = \angle C = 90°$，$CD = AB = 10$。由折叠，可得$\angle APE = \angle B = 90°$，$PE = BE$，$\therefore \angle FPE = 90°$。$\because E$为$BC$的中点，$\therefore BE = EC$。$\therefore PE = EC$。在$Rt \triangle EPF$和$Rt \triangle ECF$中，$EP = EC$，$EF = EF$，$\therefore Rt \triangle EPF \cong Rt \triangle ECF$。$\therefore FP = FC$。
(2) 由折叠，可得$AP = AB = 10$，$\therefore$点$E$在移动过程中，$AP$的长不变。$\therefore$点$P$在以点$A$为圆心，$10$为半径的$\odot A$的弧上。连接$AM$，当点$P$在线段$AM$上时，$PM$长有最小值。$\because CM = 4$，$\therefore DM = 6$。$\therefore AM = \sqrt{AD^2 + DM^2} = \sqrt{17^2 + 6^2} = \sqrt{325} = 5\sqrt{13}$。$\therefore PM$长的最小值为$AM - AP = 5\sqrt{13} - 10$
(3) 如图，过点$P$作$PH \perp AD$于点$H$，延长$HP$交$BC$于点$G$，$\therefore \angle NHP = 90°$。$\because \triangle PDN$是以$DN$为斜边的直角三角形，$\therefore \angle NPD = 90°$，即$\angle 1 + \angle 2 = 90°$。$\because \angle 1 + \angle 3 = 90°$，$\therefore \angle 3 = \angle 2$。$\because \angle PHN = \angle DHP = 90°$，$\therefore \triangle PHN \sim \triangle DHP$。$\therefore \frac{HP}{HD} = \frac{HN}{HP}$。$\therefore HP^2 = HN · HD$。$\because AN = 4$，$AD = 17$，$\therefore DN = 13$。设$HN = x$，则$HD = 13 - x$，$AH = x + 4$。$\therefore HP^2 = x(13 - x)$。$\because AP = 10$，$\therefore HP^2 = AP^2 - AH^2$，$\therefore HP^2 = 10^2 - (x + 4)^2$。$\therefore x(13 - x) = 10^2 - (x + 4)^2$，解得$x = 4$。$\therefore HP = 6$，$AH = 8$。$\because$易得四边形$ABGH$为矩形，$\therefore HG = AB = 10$，$BG = AH = 8$。$\therefore PG = 4$。设$BE = m$，则$PE = m$，$GE = 8 - m$。在$Rt \triangle PGE$中，$PE^2 = FG^2 + PG^2$，即$m^2 = (8 - m)^2 + 4^2$，解得$m = 5$。$\therefore BE = 5$。

18. (2025·长春)如图，在$\triangle ABC$中，$\angle C = 90^{\circ}$，$AC = BC = 4$，D为边AC的中点，E为边AB上一动点，连接DE，将线段ED绕点E顺时针旋转$45^{\circ}$得到线段EF。
(1) 线段AB的长为$4\sqrt{2}$；
(2) 当$EF // AC$时，求AE的长；
(3) 当点F在BC上时，求证：$\triangle ADE \cong \triangle BEF$；
(4) 当点E到BC的距离是点F到BC距离的2倍时，直接写出AE的长。

答案：

(1) $4\sqrt{2}$
(2) 如图①，在$\triangle ABC$中，$\angle C = 90^{\circ}$，$AC = BC = 4$，$D$为边$AC$的中点，$\therefore \angle A = \angle B = 45^{\circ}$，$AD = CD = 2$。$\because EF // AC$，$\therefore \angle FEB = \angle A = 45^{\circ}$。$\because$由题意，知$\angle DEF = 45^{\circ}$，$\therefore \angle DEB = 90^{\circ} = \angle AED$。$\therefore AE = AD · \cos 45^{\circ} = 2 × \frac{\sqrt{2}}{2} = \sqrt{2}$。
(3) $\because$将线段$ED$绕点$E$顺时针旋转$45^{\circ}$得到线段$EF$，$\therefore DE = EF$，$\angle DEF = 45^{\circ}$。如图②，$\because \angle A + \angle ADE = \angle DEB = \angle DEF + \angle BEF$，$\angle DEF = \angle A = 45^{\circ}$，$\therefore \angle ADE = \angle BEF$。$\because \angle A = \angle B = 45^{\circ}$，$DE = EF$，$\therefore \triangle ADE \cong \triangle BEF$
(4) 过点$E$作$EG \perp BC$于点$G$，过点$F$作$FQ \perp BC$于点$Q$，过点$D$作$DH \perp AB$于点$H$。① 当点$F$在$BC$的左边时，如图③，过点$F$作$FK \perp EG$于点$K$，$\therefore$易得四边形$FKGQ$为矩形。$\because EG = 2FQ$，$\therefore FQ = GK = EK$。易得$DH = AH = \sqrt{2}$。$\because EG \perp BC$，$\angle B = 45^{\circ}$，$\therefore \angle GEB = \angle B = 45^{\circ}$。$\therefore GB = GE = 2EK$。$\because \angle DEF = 45^{\circ}$，$\therefore \angle DEF + \angle GEB = 90^{\circ}$。$\because \angle DHE + \angle KEF = 90^{\circ}$。$\because \angle DHE = 90^{\circ}$，$\therefore \angle HDE + \angle HED = 90^{\circ}$。$\therefore \angle HDE = \angle KEF$。$\because DE = EF$，$\therefore \triangle DHE \cong \triangle EKF$。$\therefore DH = EK = \sqrt{2}$。$\therefore EG = BG = 2\sqrt{2}$。$\therefore BE = \sqrt{EG^2 + BG^2} = 4$。$\because AB = 4\sqrt{2}$，$\therefore AE = 4\sqrt{2} - 4$。② 当点$F$在$BC$的右边时，如图④，过点$F$作$FK \perp EG$，交$EG$的延长线于点$K$。同理①，可得$EK = DH = \sqrt{2}$。$\therefore$易得四边形$FKGQ$为矩形，$\therefore FQ = GK$。$\because GE = 2FQ$，$\therefore GE = 2GK$。$\therefore EG = \frac{2\sqrt{2}}{3}$。易得$EG = \frac{2\sqrt{2}}{3}$，$BG = \frac{2}{3}$。$\therefore BE = \sqrt{EG^2 + BG^2} = \frac{4}{3}$。$\because AB = 4\sqrt{2}$，$\therefore AE = 4\sqrt{2} - \frac{4}{3}$。综上所述，$AE$的长为$4\sqrt{2} - 4$或$4\sqrt{2} - \frac{4}{3}$。