证明：（1） 连结AC，如图10 G ∵C是弧BD的中点
C ∴∠BDC=∠DBC ················································ 1分
D 又∠BDC=∠BAC
F 在三角形ABC中，∠ACB=90°，CE⊥AB
∴ ∠BCE=∠BAC B A O E ∠BCE=∠DBC ·············································· 3分 ∴ CF=BF ······················································· 4分 因此，CF=BF．
图10 （2）证法一：作CG⊥AD于点G，
∵C是弧BD的中点
∴ ∠CAG=∠BAC ， 即AC是∠BAD的角平分线． ··········································· 5分 ∴ CE=CG，AE=AG ···························································································· 6分 在Rt△BCE与Rt△DCG中，CE=CG ， CB=CD ∴Rt△BCE≌Rt△DCG ∴BE=DG ·············································································································· 7分 ∴AE=AB-BE=AG=AD+DG 即 6-BE=2+DG
∴2BE=4，即 BE=2 ··························································································· 8分
又 △BCE∽△BAC
·AB?12 ·∴ BC?BE···················································································· 9分
BC??23（舍去负值）
∴BC?23 ··································································································· 10分 （2）证法二：∵AB是⊙O的直径，CE⊥AB
∴∠BEF=?ADB?90?， ······························· 5分 在Rt△ADB与Rt△FEB中， ∵?ABD??FBE
2C D F A O E B
ADAB?∴△ADB∽△FEB，则 EFBF26?即， ∴BF?3EF ···················· 6分 EFBF又∵BF?CF， ∴CF?3EF
利用勾股定理得：
图10
······································································ 7分 BE?BF2?EF2?22EF ·又∵△EBC∽△ECA 则
CEBE2?，即则CE?AE?BE ································································ 8分 AECE2∴(CF?EF)?(6?BE)?BE
即(3EF?EF)2?(6?22EF)?22EF
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∴EF?2 ···································································································· 9分 2········································································ 10分 BE2?CE2?23 ·
∴BC?
17、如图，A、P、B、C是⊙O上的四点，∠APC =∠BPC = 60?，AB与PC交于Q点． （1）判断△ABC的形状，并证明你的结论；（2）求证：
APAQ；（3）若∠ABP = 15?，△ABC?PBQBP A Q O B C 的面积为43，求PC的长．
解：（1） ∵ ∠ABC =∠APC = 60?，∠BAC =∠BPC = 60?，
∴ ∠ACB = 180?－∠ABC－∠BAC = 60?， ∴ △ABC是等边三角形．
（2）如图，过B作BD∥PA交PC于D，则 ∠BDP =∠APC = 60?． 又 ∵ ∠AQP =∠BQD，∴ △AQP∽△BQD，
AQAP． ?QBBDP A Q O F C H AQAP?∵ ∠BPD =∠BDP = 60?， ∴ PB = BD． ∴ ． QBPB（3）设正△ABC的高为h，则 h = BC· sin 60?．
B R E 11∵ BC · h = 43， 即BC · BC· sin 60? = 43，解得BC = 4．
22连接OB，OC，OP，作OE⊥BC于E．
由△ABC是正三角形知∠BOC = 120?，从而得∠OCE = 30?， ∴ OC?G M N CE4?．
cos30?3由∠ABP = 15? 得 ∠PBC =∠ABC +∠ABP = 75?，于是 ∠POC = 2∠PBC = 150?． ∴ ∠PCO =（180?－150?）÷2 = 15?．
如图，作等腰直角△RMN，在直角边RM上取点G，使∠GNM = 15?，则∠RNG = 30?，作GH⊥RN，垂足为H．设GH = 1，则 cos∠GNM = cos15? = MN．
∵ 在Rt△GHN中，NH = GN · cos30?，GH = GN · sin30?．
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于是 RH = GH，MN = RN · sin45?，∴ cos15? =
2?6． 426． 3在图中，作OF⊥PC于E，∴ PC = 2FD = 2 OC ·cos15? =22?
18、已知：如图，在Rt△ABC中，∠ABC＝90°，以AB上的点O为圆心，OB的长为半径的圆与AB交于点E，与AC切于点D． （1）求证：BC＝CD； （2）求证：∠ADE＝∠ABD；
（3）设AD＝2，AE＝1，求⊙O直径的长．
CDAEO?B 解：（1）∵∠ABC＝90°，∴OB⊥BC． ···················································································· 1分
∵OB是⊙O的半径，∴CB为⊙O的切线． ················ 2分 又∵CD切⊙O于点D，∴BC＝CD； ·························· 3分 （2）∵BE是⊙O的直径，∴∠BDE＝90°．
∴∠ADE＋∠CDB ＝90°． ······································· 4分 又∵∠ABC＝90°，∴∠ABD＋∠CBD＝90°． 由（1）得BC＝CD，∴∠CDB ＝∠CBD． ∴∠ADE＝∠ABD；
（3）由（2）得，∠ADE＝∠ABD，∠A＝∠A．
∴△ADE∽△ABD． ························································································· 7分 ∴
CDAEO?BADAE＝． ································································································· 8分 ABAD21∴＝，∴BE＝3，················································································ 9分
21?BE∴所求⊙O的直径长为3． ·········································································· 10分
19、如图，⊙O是△ABC的外接圆，AB?AC，过点A作AP∥BC，交BO的延长线于点P．
（1）求证：AP是⊙O的切线；
（2）若⊙O的半径R?5，BC?8，求线段AP的长． A P
O
C B
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解：（1）证明：过点A作AE⊥BC，交BC于点E． ?AB?AC，?AE平分BC．
······································ （2分） ?点O在AE上． ·
又?AP∥BC， ?AE⊥AP．
?AP为⊙O的切线． ································ （4分） （2）?BE?A P O 1C B BC?4，?OE?OB2?BE2?3． E 2 又??AOP??BOE，
?△OBE∽△OPA． ··································································································· （6分） BEOE43???． ． 即APOAAP520?AP?． ················································································································· （8分）
3
20、如图，在△ABC中，∠C=90°，AC=3，BC=4．0为BC边上一点，以0为圆心，OB为半径作半圆与BC边和AB边分别交于点D、点E，连结DE．
(1)当BD=3时，求线段DE的长； (2)过点E作半圆O的切线，当切线与AC边相交时，设交点为F．求证：△FAE是等腰三角形．
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21、如图，Rt△ABC中，?ABC?90°，以AB为直径作⊙O交AC边于点D，E是边BC的中点，连接DE．
（1）求证：直线DE是⊙O的切线；
（2）连接OC交DE于点F，若OF?CF，求tan?ACO的值．
C
D F E 证明：（1）连接OD、OE、BD．
?AB是⊙O的直径，??CDB??ADB?90°， ?E点是BC的中点，?DE?CE?BE． ?OD?OB，OE?OE，△?ODE≌△OBE． ??ODE??OBE?90°，?直线DE是⊙O的切线． （2）作OH⊥AC于点H，
由（1）知，BD⊥AC，EC?EB．
A
O
B
C
D F H A
O
B
1AC． 2??CDF??OEF，?DCF??EOF．
?CF?OF，?△DCF≌△EOF，?DC?OE?AD． ?BA?BC，??A?45°． ?OH⊥AD，?OH?AH?DH．
OH1?CH?3OH，?tan?ACO??．
CH3?OA?OB，?OE∥AC，且OE?E
22、如图9所示，在△ABC中，AB＝AC＝2，∠A＝90°，O为BC的中点，动点E在BA边上自由移动，动点F在AC边上自由移动.
（1）点E，F的移动过程中，△OEF是否能成为∠EOF＝45°的等腰三角形？若能，请指出△OEF为等腰三角形时动点E，F的位置.若不能，请说明理由. （2）当∠EOF＝45°时，设BE＝x，CF＝y，求y与x之间的函数解析式，写出x的取值范围. （3）在满足（2）中的条件时，若以O为圆心的圆与AB相切（如图10），试探究直线EF与⊙O的位置关系，并证明你的结论.
A A
E E F F
B
C B
C O 图9
O 图10
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