《2025年通城学典课时作业本九年级数学下册苏科版江苏专版》

9. （2024·上海改编）如图，在矩形ABCD中，E为边CD上一点，且AE⊥BD. 求证：AD² = DE·DC.
　　

答案： $\because$ 四边形$ABCD$是矩形，$\therefore \angle ADE = \angle BAD = 90^{\circ}$，$BA = DC$.$\therefore$ 在$Rt\triangle DAB$中，$\angle ABD + \angle ADB = 90^{\circ}$. 设$AE$、$BD$交于点$H$.$\because AE \perp BD$，$\therefore$ 在$Rt\triangle DHA$中，$\angle DAE + \angle ADB = 90^{\circ}$.$\therefore \angle DAE = \angle ABD$.$\therefore \triangle ADE \backsim \triangle BAD$.$\therefore \frac{AD}{BA} = \frac{DE}{AD}$.$\therefore AD^{2} = DE \cdot BA$.$\therefore AD^{2} = DE \cdot DC$

10. 如图，O为线段PB上一点，以点O为圆心，OB长为半径的⊙O交PB于点A，点C在⊙O上，连接PC，满足PC² = PA·PB，连接AC、BC.
（1）求证：PC是⊙O的切线；
（2）若AB = 3PA，求$\frac{AC}{CB}$的值.
　　　　　

答案：
(1) 连接$OC$.$\because PC^{2} = PA \cdot PB$，$\therefore \frac{PA}{PC} = \frac{PC}{PB}$.$\because \angle P = \angle P$，$\therefore \triangle PAC \backsim \triangle PCB$.$\therefore \angle PCA = \angle B$.$\because AB$是$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}$.$\therefore \angle CAB + \angle B = 90^{\circ}$.$\because OA = OC$，$\therefore \angle CAB = \angle OCA$.$\therefore \angle PCA + \angle OCA = 90^{\circ}$，即$\angle PCO = 90^{\circ}$.$\therefore OC \perp PC$.$\because OC$为$\odot O$的半径，$\therefore PC$是$\odot O$的切线
(2) 设$PA = x$，则$AB = 3PA = 3x$.$\therefore PB = PA + AB = 4x$，$OA = OC = \frac{3}{2}x$.$\therefore PO = PA + OA = \frac{5}{2}x$.$\because \angle PCO = 90^{\circ}$，$\therefore$ 由勾股定理，得$PC = \sqrt{PO^{2} - OC^{2}} = 2x$.$\because \triangle PAC \backsim \triangle PCB$，$\therefore \frac{AC}{CB} = \frac{PC}{PB} = \frac{2x}{4x} = \frac{1}{2}$

11. 如图，在Rt△ABC中，∠C = 90°，AC = 3，BC = 4，D、E分别是斜边AB、直角边BC上的点，把△ABC沿着DE折叠. 当折叠后点B落在边AC上的点P处，且四边形PEBD是菱形时，求折痕DE的长.
　　　　　

答案： 在$Rt\triangle ABC$中，$\because \angle C = 90^{\circ}$，$AC = 3$，$BC = 4$，$\therefore$ 由勾股定理，得$AB = \sqrt{AC^{2} + BC^{2}} = 5$. 连接$BP$.$\because$ 四边形$PEBD$是菱形，$\therefore PE = BE$，$PE // AB$. 设$CE = x$，则$BE = PE = 4 - x$.$\because PE // AB$，$\therefore \triangle PEC \backsim \triangle ABC$.$\therefore \frac{CE}{CB} = \frac{PE}{AB}$，即$\frac{x}{4} = \frac{4 - x}{5}$，解得$x = \frac{16}{9}$.$\therefore CE = \frac{16}{9}$，$BE = PE = 4 - \frac{16}{9} = \frac{20}{9}$.$\because$ 在$Rt\triangle PCE$中，$PE = \frac{20}{9}$，$CE = \frac{16}{9}$，$\therefore$ 由勾股定理，得$PC = \sqrt{PE^{2} - CE^{2}} = \frac{4}{3}$.$\therefore$ 在$Rt\triangle PCB$中，由勾股定理，得$BP = \sqrt{PC^{2} + BC^{2}} = \frac{4\sqrt{10}}{3}$. 又$\because$ 易得$S_{菱形PEBD} = BE \cdot PC = \frac{1}{2}DE \cdot BP$，$\therefore DE = \frac{2BE \cdot PC}{BP} = \frac{4\sqrt{10}}{9}$