【例题1】如图①,$\triangle ABC的三个顶点在\odot O$上,$AD \perp BC$,$D$为垂足,$E是\overset{\frown}{BC}$的中点.
求证：$\angle 1 = \angle 2$.

思路导引 因为$E是\overset{\frown}{BC}$的中点,联系到圆的对称性,连接$OE$,可以得$\angle 1 = \angle 2$. 也可通过弧、圆周角间的联系及直径所对的圆周角的性质可得到更多的证明方法.
证明：方法一,如图②,连接$OE$. $\because E是\overset{\frown}{BC}$的中点,可推出$OE \perp BC$. $\because AD \perp BC于点D$,$\therefore OE // AD$. $\therefore \angle 2 = \angle E$. 又$OA = OE$,$\therefore \angle 1 = \angle E$. $\therefore \angle 1 = \angle 2$.
方法二,如图③,延长$AO交\odot O于点G$,延长$AD交\odot O于点F$,连接$FG$. $\because AG是\odot O$的直径,$\therefore \angle F = 90^{\circ}$. $\because AD \perp BC于点D$,$\therefore CB // FG$. $\therefore \overset{\frown}{BG} = \overset{\frown}{CF}$. 又$E是\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{BE} = \overset{\frown}{CE}$. $\therefore \overset{\frown}{GE} = \overset{\frown}{FE}$. $\therefore \angle 1 = \angle 2$.



方法三,如图④,延长$AO交\odot O于点F$,连接$BF$,延长$AD交\odot O于点G$. $\because AF$是直径,$\therefore \angle ABF = 90^{\circ}$. $\because \angle F = \angle C$,$AD \perp BC于点D$,$\therefore \angle 3 = \angle 4$. $\therefore \overset{\frown}{BF} = \overset{\frown}{CG}$. 又$E是\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{EB} = \overset{\frown}{CE}$. $\therefore \overset{\frown}{FE} = \overset{\frown}{EG}$. $\therefore \angle 1 = \angle 2$.
方法四,如图⑤,延长$AO交\odot O于点G$,连接$CG$. $\because AG为\odot O$的直径,$\therefore \angle ACG = 90^{\circ}$. $\because AD \perp BC$,$\angle B = \angle G$,$\therefore \angle BAD = \angle GAC$. $\therefore \angle 3 = \angle 4$. 又$E是\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{BE} = \overset{\frown}{CE}$. $\therefore \angle BAE = \angle CAE$. $\therefore \angle 1 = \angle 2$.
方法五,如图⑥,过点$O作OM \perp AB于点M$,交$\odot O于点G$,$\therefore G为\overset{\frown}{AB}$的中点. $\because AD \perp BC于点D$,$\therefore \angle C = \angle O$. $\therefore \angle 3 = \angle 4$. 又$E是\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{BE} = \overset{\frown}{CE}$. $\therefore \angle 1 + \angle 3 = \angle 2 + \angle 4$. $\therefore \angle 1 = \angle 2$.
方法六,如图⑦,延长$AO交\odot O于点F$,连接$BF$. $\because AF为\odot O$的直径,$\therefore \angle ABF = 90^{\circ}$,$\angle 3 + \angle F = 90^{\circ}$. $\because \angle C = \angle F$,$\therefore \angle 3 + \angle C = 90^{\circ}$. $\because AD \perp BC$,$\therefore \angle 4 + \angle C = 90^{\circ}$. $\therefore \angle 3 = \angle 4$. $\because E是\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{BE} = \overset{\frown}{CE}$. $\therefore \angle 1 + \angle 3 = \angle 2 + \angle 4$. $\therefore \angle 1 = \angle 2$.