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初三圆的提高题

发布时间:2013-10-21 13:40:03  

1. (2009重庆)如图,⊙O是?ABC的外接圆,AB是直径,若

?BOC

??A等

A

于( )

A.60o B.50o

C.40o D.30o

B

o

O

2. (2009遂宁)如图,已知⊙O的两条弦AC,BD相交于点E,∠A=70,∠,那么sin

∠AEB的值为

A. 1 B. C. D. 322 2

C

3. (2009重庆)已知⊙O1的半径为3cm,⊙O2的半径为4cm,两圆的圆心距O1O2为7cm,

则⊙O1与⊙O2的位置关系为。

4. (2009成都)如图,A、B、c是⊙0上的三点,以BC为一边,作∠CBD=∠ABC,过BC上

一点P,作PE∥AB交BD于点E.若∠AOC=60°,BE=3,则点P到弦AB的距离为_______. 5. (2009成都)如图,△ABC内接于⊙O,AB=BC,∠ABC=120AD=6,

那么BD=_________. 1

在Rt△ABC中,的内切圆,点?C9?°,0AC,?6,B⊙?C为8△ABCO

D是斜边AB的中点,则tan?ODA?( ) .

5

),

A B C D.2 图(5)

2.(8分)如图10,⊙O的弦AD∥BC,过点D的切线交BC的延长线于点E,AC∥DE交BD于点H,DO及延长线分别交AC、BC于点G、F.

(1)求证:DF垂直平分AC; (2)求证:FC=CE;

(3)若弦AD=5㎝,AC=8㎝,求⊙O的半径.

3.(10分)如图,四边形ABCD内接于圆,对角线AC与BD相交于点E,F在AC上,AB?AD,?BFC??BAD?2?DFC.

求证:(1)CD⊥DF; (2)BC?2CD.

C

七、(本大题8分) 4.如图8,半圆的直径AB?10,点C在半圆上,BC?6. (1)求弦AC的长;

(2)若P为AB的中点,PE⊥AB交AC于点E,求PE的长.

A P (图8)

5.(本题满分10分)

如图11,在△ABC中,AB=BC,以AB为直径的⊙O与AC交于点D,过D作DF⊥BC,交AB的延长线于E,垂足为F.

(1)求证:直线DE是⊙O的切线;

(2)当AB=5,AC=8时,求cosE的值.

图11

24.(本小题满分9分)

如图8-1,已知O是锐角∠XAY的边AX上的动点,以点O为圆心、R为半径的圆与射线AY切于点B,交射线OX于点C.连结BC,作CD⊥BC,交AY于点D.

(1) (3分) 求证:△ABC∽△ACD; B

3(2) (6分) 若P是AY上一点,AP=4,且sinA=, 5

① 如图8-2,当点D与点P重合时,求R的值;

② 当点D与点P不重合时,试求PD的长(用R表示).

图8-1

8-2

24.如图,以BC为直径的⊙O交△CFB的边CF于点A,BM平分

∠ABC交AC于点M,AD⊥BC于点D,AD交BM于点N,ME⊥BC于点E,AB=AF·AC,

cos∠ABD=3,AD=12. 5

⑴求证:△ANM≌△ENM;

⑵求证:FB是⊙O的切线;

⑶证明四边形AMEN是菱形,并求该菱形的面积S.

24.如图,△ABC内接于⊙O,且∠B = 60?.过点C作圆的切线l与

直径AD的延长线交于点E,AF⊥l,垂足为F,CG⊥AD,垂足为G.

(1)求证:△ACF≌△ACG;

(2)若AF = 43,求图中阴影部分的面积.

27.已知:如图,?ABC内接于?O,AB为直径,弦CE?AB于F,C是?AD的中点,连结BD并延长交EC的延长线于点G,连结AD,分别交CE、BC于点P、Q.

(1)求证:P是?ACQ的外心;

(2)若tan?ABC?23,CF?8,求CQ的长; 4

2 (3)求证:(FP?PQ)?FP?FG.

22.证明:(1)∵DE是⊙O的切线,且DF过圆心O

∴DF⊥DE

又∵AC∥DE

∴DF⊥AC

∴DF垂直平分AC2分

(2)由(1)知:AG=GC

又∵AD∥BC

∴∠DAG=∠FCG

又∵∠AGD=∠CGF

∴△AGD≌△CGF(ASA)4分

∴AD=FC

∵AD∥BC且AC∥DE

∴四边形ACED是平行四边形

∴AD=CE

∴FC=CE5分

(3)连结AO; ∵AG=GC,AC=8cm,∴AG=4cm

在Rt△AGD中,由勾股定理得 GD=AD2-AG2=52-42=3cm6分

设圆的半径为r,则AO=r,OG=r-3

在Rt△AOG中,由勾股定理得 AO2=OG2+AG2

有:r2=(r-3)2+42解得 r=2568分

∴⊙O的半径为256cm.

24.(1) 由已知,CD⊥BC,∴ ∠ADC=90°–∠CBD, ····························································1分

又∵ ⊙O切AY于点B,∴ OB⊥AB,∴∠OBC=90°–∠CBD, ·····································2分

∴ ∠ADC=∠OBC.又在⊙O中,OB=OC=R,∴∠OBC=∠ACB,∴∠ACB=∠ADC. 又∠A=∠A,∴△ABC∽△ACD . ···················································································3分

3(2) 由已知,sinA=,又OB=OC=R,OB⊥AB, 5

∴ 在Rt△AOB中,AO=OBR54==R,AB

R, 3sinA3

5

58∴ AC=R+R=R . ········································································································433

由(1)已证,△ABC∽△ACD,∴

分 8RAD16?∴,因此 AD=R. ·······················································································6483RR33ACAD, ································································5?ABAC

① 当点D与点P重合时,AD=AP=4,∴163R=4,∴R=. ········································734

② 当点D与点P不重合时,有以下两种可能:

316i) 若点D在线段AP上(即0<R<),PD=AP–AD=4–R; ··········································843

ii) 若点D在射线PY上(即R>

3163)时,PD=4–R;当点D在射线PY上(即R>)434

163时,PD=R–4.又当点D与点P重合(即R=)时,PD=0,故在题设条件下,总有34

16PD=|R–4|(R>0). 3316),PD=AD–AP=R–4. ·············································943综上,当点D在线段AP上(即0<R<

24.⑴证明:∵BC是⊙O的直径

∴∠BAC=90o

又∵EM⊥BC,BM平分∠ABC,

∴AM=ME,∠AMN=EMN

又∵MN=MN,

∴△ANM≌△ENM

⑵∵AB2=AF·AC ∴AB?AF ACAB

又∵∠BAC=∠FAB=90o

∴△ABF∽△ACB

∴∠ABF=∠C

又∵∠FBC=∠ABC+∠FBA=90o

∴FB是⊙O的切线

⑶由⑴得AN=EN,AM=EM,∠AMN=EMN,

又∵AN∥ME,∴∠ANM=∠EMN,

∴∠AMN=∠ANM,∴AN=AM,

∴AM=ME=EN=AN

∴四边形AMEN是菱形

∵cos∠ABD=3,∠ADB=90o 5

∴BD?3 AB5

设BD=3x,则AB=5x,,由勾股定理AD?

而AD=12,∴x=3

∴BD=9,AB=15

∵MB平分∠AME,∴BE=AB=15

∴DE=BE-BD=6 5x2-3x2?4x

∵ND∥ME,∴∠BND=∠BME,又∵∠NBD=∠MBE

∴△BND∽△BME,则ND?BD MEBE

设ME=x,则ND=12-x,12?x?9,解得x=15 x152∴S=ME·DE=15×6=45 2

24.(1)如图,连结CD,OC,则∠ADC =∠B = 60?.

∵ AC⊥CD,CG⊥AD,∴ ∠ACG =∠ADC = 60?.

由于 ∠ODC = 60?,OC = OD,∴ △OCD为正三角形,得 ∠DCO = 60?. 由OC⊥l,得 ∠ECD = 30?,∴ ∠ECG = 30? + 30? = 60?.

进而 ∠ACF = 180?-2×60? = 60?,∴ △ACF≌△ACG.

(2)在Rt△ACF中,∠ACF = 60?,AF = 43,得 CF = 4. 在Rt△OCG中,∠COG = 60?,CG = CF = 4,得 OC =

在Rt△CEO中,OE =8. 16

160??OC232(3??)于是 S阴影 = S△CEO-S扇形COD =OE?CG?=. 23609

?, AC?CD27. (1)证明:∵C是?AD的中点,∴?

∴∠CAD=∠ABC

∵AB是⊙O的直径,∴∠ACB=90°。

∴∠CAD+∠AQC=90°

又CE⊥AB,∴∠ABC+∠PCQ=90°

∴∠AQC=∠PCQ

∴在△PCQ中,PC=PQ,

AC??AE ∵CE⊥直径AB,∴?

? AE?CD∴?

∴∠CAD=∠ACE。

∴在△APC中,有PA=PC,

∴PA=PC=PQ

∴P是△ACQ的外心。

(2)解:∵CE⊥直径AB于F,

∴在Rt△BCF中,由tan∠ABC=

得BF?CF3?,CF=8, BF4432CF?。

33

40 3∴由勾股定理,得BC??

∵AB是⊙O的直径,

∴在Rt△ACB中,由tan∠ABC=

得AC?AC340 ?,BC?BC433BC?10。 4

2易知Rt△ACB∽Rt△QCA,∴AC?CQ?BC AC215∴CQ??。 BC2

(3)证明:∵AB是⊙O的直径,∴∠ACB=90°

∴∠DAB+∠ABD=90°

又CF⊥AB,∴∠ABG+∠G=90°

∴∠DAB=∠G;

∴Rt△AFP∽Rt△GFB, ∴AFFP,即AF?BF?FP?FG ?FGBF

2易知Rt△ACF∽Rt△CBF, ∴FG?AF?BF(或由摄影定理得)

∴FC?PF?FG

由(1),知PC=PQ,∴FP+PQ=FP+PC=FC ∴(FP?PQ)?FP?FG。

22

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