14 [2025 济南]如图, A B 是 $\odot O$ 的直径, C 为 $\odot O$ 上一点, P 为 $\odot O$ 外一点, $O P / / A C$, 且 $∠ O B P=$ $90°$, 连接 $P C$.
(1) 求证: $P C$ 与 $\odot O$ 相切;
(2) 若 $A O=3, O P=5$, 求 $A C$ 的长.

14. (1) 连接$OC$。$\because OC=OA$，$\therefore ∠ OAC=∠ OCA$。$\because OP// AC$，$\therefore ∠ OAC=∠ BOP$，$∠ OCA=∠ COP$，$\therefore ∠ COP=∠ BOP$。$\because OP=OP$，$OC=OB$，$\therefore △ COP≌△ BOP$，$\therefore ∠ OCP=∠ OBP=90°$，$\therefore OC⊥ PC$。$\because OC$是$\odot O$的半径，$\therefore PC$与$\odot O$相切 (2) 连接$BC$交$OP$于点$D$。$\because △ COP≌△ BOP\ (\mathrm{SAS})$，$\therefore PC=PB$。$\because OB=OC$，$\therefore OP$垂直平分$BC$，$\therefore BC=2BD$。$\because ∠ OBP=90°$，$AO=BO=3$，$OP=5$，$\therefore$在$\mathrm{Rt}△ OBP$中，$BP=\sqrt{OP^2-OB^2}=\sqrt{5^2-3^2}=4$。$\because S_{△ OBP}=\dfrac{1}{2}OB· BP=\dfrac{1}{2}OP· BD$，$\therefore BD=\dfrac{OB· BP}{OP}=\dfrac{3×4}{5}=\dfrac{12}{5}$，$\therefore BC=\dfrac{24}{5}$。$\because AB$是$\odot O$的直径，$\therefore AB=2AO=6$，$∠ ACB=90°$，$\therefore$在$\mathrm{Rt}△ ACB$中，$AC=\sqrt{AB^2-BC^2}=\sqrt{6^2-(\dfrac{24}{5})^2}=\dfrac{18}{5}$

15 [2025 南京]如图①,$O$是$□ ABCD$的对称中心,$BC$与$\odot O$相切于点$E$.

(第 15 题)
(1) 求证:直线$AD$是$\odot O$的切线. 如图②,选择其中一名同学的想法,完成证明.
(2) 当$AB$与$\odot O$相切时,$□ ABCD$是菱形吗? 请说明理由.

15. (1) 答案不唯一，如选择右侧同学的想法 如图①，连接$BD$，$EO$，延长$EO$交$AD$于点$F$。$\because O$是$□ ABCD$的对称中心，$\therefore BD$过点$O$，$OB=OD$。$\because$四边形$ABCD$是平行四边形，$\therefore AD// BC$，$\therefore ∠ ODF=∠ OBE$，$∠ DFO=∠ BEO$，$\therefore △ DOF≌△ BOE\ (\mathrm{AAS})$，$\therefore OF=OE$。$\because BC$与$\odot O$相切于点$E$，$\therefore OE⊥ BC$，$\therefore ∠ BEO=90°$，$\therefore ∠ DFO=90°$，$\therefore OF⊥ AD$。$\because OF$是$\odot O$的半径，$\therefore$直线$AD$是$\odot O$的切线 (2) 当$AB$与$\odot O$相切时，$□ ABCD$是菱形 理由：如图②，设$AB$与$\odot O$相切于点$H$，连接$OH$，$OE$，$BD$，$\because$点$O$是$□ ABCD$的对称中心，$\therefore BD$过点$O$。$\because AB$，$BC$是$\odot O$的切线，$\therefore OH⊥ AB$，$OE⊥ BC$。$\because OH=OE$，$\therefore ∠ ABD=∠ CBD$。$\because$四边形$ABCD$是平行四边形，$\therefore AD// BC$，$\therefore ∠ CBD=∠ ADB$，$\therefore ∠ ABD=∠ ADB$，$\therefore AB=AD$，$\therefore □ ABCD$是菱形。