《2025年通城学典非常课课通九年级数学上册人教版江苏专版》

示例6 如图①，$D$是以$AB$为直径的$\odot O$上一点，过点$B$作$\odot O$的切线，交$AD$的延长线于点$C$，$E$是$BC$的中点，连接$DE$并延长，与$AB$的延长线交于点$F$.
（1）求证：$DF$是$\odot O$的切线；
（2）若$OB = BF$，$EF = 4$，求$AD$的长.

答案： 解：（1）如图②，连接$OD$，$BD$.$\because AB$为$\odot O$的直径，$\therefore\angle ADB = 90^{\circ}=\angle BDC$.在$\mathrm{Rt}\triangle BDC$中，$\because E$是$BC$的中点，$\therefore DE = CE = BE$.$\therefore\angle 1=\angle 3$.$\because BC$是$\odot O$的切线，$\therefore AB\perp BC$.$\therefore\angle 3+\angle 4 = 90^{\circ}$.$\because OD = OB$，$\therefore\angle 2=\angle 4$.$\therefore\angle 1+\angle 2 = 90^{\circ}$.$\therefore OD\perp DF$.$\because OD$是$\odot O$的半径，$\therefore DF$是$\odot O$的切线.
（2）$\because OB = BF$，$\angle ODF = 90^{\circ}$，$\therefore BD = OB$.又$\because OD = OB$，$\therefore BD = OD = OB$.$\therefore\triangle ODB$是等边三角形.$\therefore\angle BOD = 60^{\circ}$.