例1 (教材典题)如图2-2-2,$AB$,$AC$,$BC是\odot O$的弦,$∠AOC= ∠BOC$.$∠ABC与∠BAC$相等吗?为什么?

答案:
解:$\angle ABC$与$\angle BAC$相等。
理由:在$\odot O$中,
$\because \angle AOC = \angle BOC$,
$\therefore AC = BC$(在同圆中,相等的圆心角所对的弦相等)。$\therefore \angle ABC = \angle BAC$。
理由:在$\odot O$中,
$\because \angle AOC = \angle BOC$,
$\therefore AC = BC$(在同圆中,相等的圆心角所对的弦相等)。$\therefore \angle ABC = \angle BAC$。
例2 如图2-2-3,在$\odot O$中,$\overset{\frown}{AB}= \overset{\frown}{AC}$,$∠ACB= 60^{\circ}$.
求证:$∠AOB= ∠BOC= ∠AOC$.

求证:$∠AOB= ∠BOC= ∠AOC$.
答案:
证明:$\because \overset{\frown}{AB} = \overset{\frown}{AC}$,$\therefore AB = AC$,
$\therefore \triangle ABC$是等腰三角形。
$\because \angle ACB = 60^{\circ}$,
$\therefore \triangle ABC$是等边三角形,
$\therefore AB = BC = CA$,
$\therefore \angle AOB = \angle BOC = \angle AOC$。
$\therefore \triangle ABC$是等腰三角形。
$\because \angle ACB = 60^{\circ}$,
$\therefore \triangle ABC$是等边三角形,
$\therefore AB = BC = CA$,
$\therefore \angle AOB = \angle BOC = \angle AOC$。
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