《2026年课堂作业武汉出版社九年级数学下册人教版》

在圆中，三角形相似的常见基本图形如下：

答案： △PAB∽△PCA，△ACD∽△ABC∽△CBD

例 如图，$\triangle ABC$内接于$\odot O$，且$AB$为$\odot O$的直径.$ \angle ACB$的平分线交$\odot O$于点$D$，过点$D$作$\odot O$的切线$PD$交$CA$的延长线于点$P$，过点$A$作$AE \perp CD$于点$E$，过点$B$作$BF \perp CD$于点$F$.
(1)试猜想线段$AE$，$EF$，$BF$之间有何数量关系，并加以证明.
(2)若$AC = 6$，$BC = 8$，求线段$PD$的长.

答案：
(1) $BF - AE = EF$.证明如下：
∵$AB$为$\odot O$直径，$\therefore \angle ACB = 90°$.
∵$CD$平分$\angle ACB$，$\therefore \angle ACD = \angle BCD = 45°$.
∵$AE \perp CD$，$BF \perp CD$，$\therefore \triangle AEC$，$\triangle BFC$为直角三角形.
$\therefore \angle CAE = 90° - \angle ACD = 45°$，$\angle CBF = 90° - \angle BCD = 45°$.
$\therefore \triangle AEC$，$\triangle BFC$为等腰直角三角形，$\therefore AE = CE$，$BF = CF$.
∵$CF = CE + EF$，$\therefore BF = AE + EF$，即$BF - AE = EF$.
(2) 在$Rt\triangle ACB$中，$AB = \sqrt{AC^2 + BC^2} = \sqrt{6^2 + 8^2} = 10$.
∵$CD$平分$\angle ACB$，$\therefore \overset{\frown}{AD} = \overset{\frown}{BD}$，$\therefore AD = BD$.
∵$AB$为直径，$\therefore \angle ADB = 90°$，$\triangle ADB$为等腰直角三角形，$\therefore AD = \frac{AB}{\sqrt{2}} = 5\sqrt{2}$.
∵$AE \perp CD$，$\triangle AEC$为等腰直角三角形，$AC = 6$，$\therefore AE = CE = \frac{AC}{\sqrt{2}} = 3\sqrt{2}$.
在$Rt\triangle AED$中，$DE = \sqrt{AD^2 - AE^2} = \sqrt{(5\sqrt{2})^2 - (3\sqrt{2})^2} = 4\sqrt{2}$，$\therefore CD = CE + DE = 3\sqrt{2} + 4\sqrt{2} = 7\sqrt{2}$.
∵$PD$为$\odot O$切线，$\therefore \angle PDA = \angle PCD$（弦切角等于所夹弧对圆周角）.
又$\angle P = \angle P$，$\therefore \triangle PDA \backsim \triangle PCD$，$\therefore \frac{PD}{PC} = \frac{PA}{PD} = \frac{AD}{DC} = \frac{5}{7}$.
设$PD = x$，则$PA = \frac{5}{7}x$，$PC = \frac{7}{5}x$.
∵$PC = PA + AC$，$\therefore \frac{7}{5}x = \frac{5}{7}x + 6$，解得$x = \frac{35}{4}$.
$\therefore PD = \frac{35}{4}$.