∵弦CD⊥AB于E， ∴CE=
1CD． ···················································································································· 3 2在Rt△OCE中，∠CEO=90°，∠COB=30°，OC=2，
∴CE=1． ··························································································································· 4 ∴CD=2． ·························································································································· 5
7．（大兴18期末21）已知： 如图，⊙O的直径AB的长为5cm，C为⊙O上的一个点，∠ACB
的平分线交⊙O于点D，求BD的长．
21. 解：∵ AB为直径，
∴ ∠ADB=90°， ……………………………… 1分 ∵ CD平分∠ACB， ∴ ∠ACD=∠BCD，
⌒ =BD⌒ ．………………………………… 2分 ∴ AD
∴ AD=BD ……………………………………… 3分 在等腰直角三角形ADB中， 2 5BD=ABsin45°=5×2 =
2∴ BD=
522 ……………… 5分
2 .
8．（通州18期末19）如图，△ABC内接于⊙O.若⊙O的半径为6，?B?60?，求AC的长．
9．（顺义18期末24）已知：如图，AB为⊙O直径，CE⊥AB于E，BF∥OC，连接BC，CF． 求证：∠OCF=∠ECB．
24．
证明： 延长CE交⊙O于点G．
∵AB为⊙O的直径，CE⊥AB于E， ∴BC=BG， ∴∠
G=∠2，……………………………………………..2分
∵BF∥OC，
∴∠1=∠F，………………………………………………3分 又∵∠G=∠F，………………………………………..….5分 ∴∠1=∠2．…………………………………………….…6分
（其它方法对应给分）
10．（燕山18期末19）如图，AB为⊙ O的直径，弦 CD ⊥ AB于点E ，连 接BC．若AB＝6，
∠ B＝30°，求：弦CD的长.
19．如图，AB为⊙O的直径，弦CD⊥AB于点E，连接BC．若AB＝6，∠B＝30°，求：弦CD的长． 解：连结AC , ∵AB为⊙O的直径 ,
∴∠ACB=90° ……………………..……………..1′
又AB＝6∠B＝30°
∴AC=3 ∠CAE＝60° ……………………..……………..2′ ∵弦CD⊥AB,AB为⊙O的直径
∴CE=ED ……………………..……………..3′
∵Rt△CEA中CE=3 sin60°=33…………………………………………………………..5′
2
