《2026年授之以渔河北各地市中考试题汇编数学》

26. （本小题满分13分）
如图1和图2，平面上，四边形$ABCD$中，$AB = 8$，$BC = 2\sqrt{11}$，$CD = 12$，$DA = 6$，$\angle A = 90^{\circ}$，点$M$在$AD$边上，且$DM = 2$.将线段$MA$绕点$M$顺时针旋转$n^{\circ}(0 ＜ n\leqslant 180)$到$MA'$，$\angle A'MA$的平分线$MP$所在直线交折线$AB - BC$于点$P$，设点$P$在该折线上运动的路径长为$x(x ＞ 0)$，连接$A'P$.
（1）若点$P$在$AB$上，求证：$A'P = AP$；
（2）如图2，连接$BD$.
①求$\angle CBD$的度数，并直接写出当$n = 180$时，$x$的值；
②若点$P$到$BD$的距离为2，求$\tan\angle A'MP$的值；
（3）当$0 ＜ x\leqslant 8$时，请直接写出点$A'$到直线$AB$的距离（用含$x$的式子表示）.

答案： 26.
(1)证明：由旋转的性质，得$MA^{\prime} = MA$.$\because MP$平分$\angle A^{\prime}MA$,$\therefore \angle A^{\prime}MP = \angle AMP$.又$\because MP = MP$,$\therefore \triangle A^{\prime}MP \cong \triangle AMP(SAS)$.$\therefore A^{\prime}P = AP$.
(2)解：①$\because \angle A = 90^{\circ}$,$\therefore BD^{2} = AB^{2} + DA^{2} = 8^{2} + 6^{2} = 100$.$\because BD^{2} + BC^{2} = 100 + (2\sqrt{11})^{2} = 144$,$CD^{2} = 12^{2} = 144$,$\therefore BD^{2} + BC^{2} = CD^{2}$.$\therefore \triangle BCD$是直角三角形，且$\angle CBD = 90^{\circ}$.当$n = 180$时，$x$的值为$13$. ②如图1，当点$P$在$AB$上时，过点$P$作$PE \perp BD$于点$E$,则$PE = 2$.$\because \sin \angle DBA = \frac{PE}{BP} = \frac{DA}{BD} = \frac{3}{5}$,$\therefore BP = \frac{10}{3}$.$\therefore AP = AB - BP = \frac{14}{3}$.$\therefore \tan \angle A^{\prime}MP = \tan \angle AMP = \frac{AP}{AM} = \frac{7}{6}$. 如图2，当点$P$在$BC$上时，则$PB = 2$,过点$P$作$PF \perp AB$交$AB$延长线于点$F$,$PG \perp AD$于点$G$.$\because \angle CBD = 90^{\circ}$,$\therefore \angle PBF + \angle ABD = 90^{\circ}$.$\because \angle A = 90^{\circ}$,$\therefore \angle ABD + \angle ADB = 90^{\circ}$.$\therefore \angle PBF = \angle ADB$.又$\because \angle F = \angle A = 90^{\circ}$,$\therefore \triangle PBF \backsim \triangle BDA$.$\therefore \frac{PF}{AB} = \frac{BF}{AD} = \frac{PB}{BD} = \frac{2}{10} = \frac{1}{5}$.$\therefore PF = \frac{8}{5}$,$BF = \frac{6}{5}$.易得四边形$PFAG$是矩形，$\therefore AG = PF = \frac{8}{5}$,$PG = AF = BF + AB = \frac{46}{5}$.$\therefore MG = DA - DM - AG = \frac{12}{5}$.$\therefore \tan \angle A^{\prime}MP = \frac{PG}{MG} = \frac{\frac{46}{5}}{\frac{12}{5}} = \frac{23}{6}$. 综上，$\tan \angle A^{\prime}MP$的值为$\frac{7}{6}$或$\frac{23}{6}$.
(3)解：点$A^{\prime}$到直线$AB$的距离为$\frac{8x^{2}}{x^{2} + 16}$.