1. 顶点在
圆上
,并且两边都和圆
相交的角叫做圆周角.
答案:
圆上 圆
2. 圆周角的度数等于它所对弧上的圆心角度数的
一半
,同弧或等弧所对的圆周角相等
.
答案:
一半 相等
3. 圆周角的度数等于它所对弧的度数的
一半
.
答案:
一半
1. 如图,A,B,P 是⊙O 上的三点,若∠AOB = 40°,则∠APB 的度数为(
A.80°
B.140°
C.20°
D.50°
C
)A.80°
B.140°
C.20°
D.50°
答案:
C
2. 如图,CD⊥AB 于点 E,若∠B = 60°,则∠A = (

A.20°
B.30°
C.40°
D.60°
B
)A.20°
B.30°
C.40°
D.60°
答案:
B
3. 如图,点 A,B,C 在⊙O 上,∠BAC = 35°,则∠BOC = __
70°
__.
答案:
$70^{\circ}$
4. 如图,在⊙O 中,∠AOB = 100°,C,D 是$\overset{\frown}{AmB}$上任意两点,则∠C + ∠D 的度数是
$100^{\circ}$
.
答案:
$100^{\circ}$
5. 如图,在⊙O 中,AC//OB,∠BAO = 25°,求∠BOC 的度数.

答案:
解:$\because OA = OB$,$\therefore \angle B = \angle BAO = 25^{\circ}$.
$\because OB// AC$,$\therefore \angle CAB = \angle B = 25^{\circ}$,
$\therefore \angle BOC = 2\angle CAB = 50^{\circ}$.
$\because OB// AC$,$\therefore \angle CAB = \angle B = 25^{\circ}$,
$\therefore \angle BOC = 2\angle CAB = 50^{\circ}$.
6. 如图,在⊙O 中,AB = CD,求证:∠ADC = ∠BAD.

答案:
证明:$\because AB = CD$,$\therefore \overset{\frown}{AB} = \overset{\frown}{CD}$,$\therefore \overset{\frown}{AB} - \overset{\frown}{BC} = \overset{\frown}{CD} - \overset{\frown}{BC}$,即$\overset{\frown}{AC} = \overset{\frown}{BD}$,$\therefore \angle ADC = \angle BAD$.
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