2019-2020年高中数学 第二讲 直线与圆的位置关系单元测评2 新人教A版选修4-1.doc

2019-2020年高中数学 第二讲 直线与圆的位置关系单元测评2 新人教A版选修4-11.若ABCD为O的内接四边形,BOD =100,则BAD为()A.130B.100C.50D.50或130思路解析:运用圆周角定理,注意不同的情况,BAD =50或130.答案:D2.圆内接四边形的4个角中,如果没有直角,那么一定有()A.2个锐角2个钝角B.1个锐角3个钝角C.1个钝角3个锐角D.都是锐角或钝角思路解析:圆内接四边形的对角互补,由于圆内接四边形的4个角中没有直角,所以必有2个锐角,2个钝角.答案:A3.如图2-16,ABC内接于直径为d的圆,AC=b,BC=a,则ABC的高CD等于()图2-16A.B. C. D. 思路解析:过C点作出直径CE,连结AE,容易知道CDBCAE,从而有=,所以CD = =.答案:D4.如图2-17,点D、E在以AB为直径的半圆O上,点F、C在AB上,CDEF为正方形,若正方形的边长为1,AC=a,BC=b,则下列式子中不正确的是()图2-17A.a2+b2=5B.a+b=C.ab=1D.a-b=1思路解析:容易知道OF =OC =,圆的半径为,所以AC =OA +OC,a = +,整理得a -b =1.由于DC2=ACCB,所以1 =ab.a +b =.答案: A5.PT切O于点T,PB是O的割线,与O交于A、B两点,且过圆心O,若OPT=30,PT=20 cm,则PB等于 .思路解析:连结OT,则OTPT,在RtPOT中,OPT =30,PT =20,所以OT =OB =,PO =,从而PB =PO +OB = + =.答案: 6.如图2-18,AB为O的直径,直线MN切O于C点,AMMN,BNMN,若AM =a,BN =b,则O的半径是 .图2-18思路解析:由题意,连结OC,则OCMN,因为AMMN,BNMN,所以AMOCBN.又AO =OB,所以=.答案: 7.如图2-19,CD是O的直径,AE切O于B点,DC的延长线交AB于A,A=20,则DBE = .图2-19思路解析:连结OB,则OBAB.在RtOBA中,A =20,AOB =70,D=AOB =35.又DBE =A +D,DBE =20+35=55.答案:558.如图2-20,AB为O的直径,延长AB到D点,使BD=OB,DC切O于C点,则ACAD = .图2-20思路解析:连结BC,则ACB是直角三角形.设AO =BO =BD =R,则由切割线定理得CD2=DBDA =R3R =3R2,CD =R.又CBD ACD,= = =,TanA = =,AB =2R.A =30.AC =.= =.答案:39.如图2-21,过O外一点P作O的两条切线PA、PB,切点分别为A、B,连结AB,在AB、PB、PA上分别取一点D、E、F,使AD =BE,BD =AF,连结DE、DF、EF,则EDF与P的关系是 .图2-21思路解析:考虑切线长定理得PA =PB,由条件容易证明ADFBED,从而ADF =BED,EDF =180-(ADF +BDE)=180-(BED+BDE )=180-(180-EBD)=EBD =90-P.答案:EDF =90-P10.一条弦分圆成13的两部分,过这条弦的一个端点引圆的切线,则所成的弦切角为 .思路分析:注意两种情况:一种是45,另一种是135.答案:45或13511.如图2-22,正方形ABCD中,AE切以BC为直径的半圆O于点E,交DC于F.求CFFD.图2-22思路分析:用代数法,设正方形的边长为a,FC =x,运用勾股定理、切线定理构造一元二次方程,确定x与a的关系即可.解:设正方形边长为a,FC =x,ABC =90,AB与半圆O相切,又AF切半圆O于点E,AB =AE.同理,FC =FE.AF =x +a.在RtADF中,AF2=AD2+DF2,(x +a)2=a2+(a -x)2.解之,得.DF =a-x = a.CFFD =a =13.12.如图2-23,等腰梯形ABCD的内切圆圆心为点O,点EH、G为切点,AH、BO的长度分别为关于x的方程x2- mx + 9m =0的两根,又cosBAO =.图2-23(1)求证:AOBO;(2)求梯形ABCD的面积.思路分析:(1)运用切线长定理;(2)依据一元二次方程“根与系数的关系”求出m的值,进一步求得AD、BC、EF的长,求出梯形面积.(1)证明:由切线长定理知OA平分DAB,OB平分ABC,又ADBC,AOB=90,即AOBO.(2)解:连结OH,则OHAB.cosBAO =,设OA =3k,AB =5k,OB =4k,在RtAOH中, , , ,由韦达定理得OB +AH = m,OBAH =9m,即=,4k5k =9m.解之,得k1=5,k2 =0(舍去).BO =20,AH =9.又AD =2AH =18,BC =2BF =2BH =32,EF =2OH =24,即梯形的高EF =24.S梯形ABCD =(18+32)24=600.13.如图2-24,ABC内接于O,AE切O于点A,BD平分ABC,交O于点D,交AF的延长线于点E,DFAE于点F.图2-24求证:(1) =;(2)AC =2AF.思路分析:(1)运用弦切角定理证ABEDAE即可;(2)依据圆周角定理得点D为AC的中点.结合条件容易得到AD平分EAC,从而联系“角平分线上的点到角两边距离相等”,过点D作DHAC,垂足为H,结合垂径定理得结论.证明:(1)AE切O于点A,EAD =EBA.又E =E,DAEABE.=,即=.(2)过点D作DHAC,垂足为H.EAD ABD,DAC DBC,BD平分ABC,EAD DAC.又DFAE,DHAC,DF DH.在RtDFA和RtDHA中,DF =DH,DA DA,RtDFARtDHA.AF =AH.DC =DA,DHAC,AH =CH,AC =2AH =2AF.14.如图2-25,经过点A、C的圆O切AB于点A,交CB于点D,ABC的平分线BG交O于点G,作GAE =GAB,AE交CG于F点,交CD于E点,求证:图2-25(1)EDG =EFG;(2)AG2 =CGDG.思路分析:(1)要证EDG =EFG,只需证GDB =AFG,而GDB =CAG,AFG =GCA +CAF,由GAB为弦切角,用弦切角定理证明即可;(2)由(1)易证AFGCAG,从而得AG2=CFCG,故再证FG =DG,连结EG,证EFGEDG就能达到目的.证明:(1)四边形AGDC是圆内接四边形,EDG +CAG =180.EFG =AFC,EFG +ACF +CAF =180.ACF =BAG,BAG =FAG,EFG +BAG+CAF =EFG+FAG +CAF =180.EFG +CAG =180.EDG =EFG.(2)连结EG,AFG =ACF +CAF,ACF =BAG =FAG,AFG =CAG.又AGC公用,AGFCGA.=.GA平分EAB,BG平分ABD,G到AE、B、BE的距离相等(即G为ABE的内心).G在AEB的角平分线上.FEG =DEG.EG =EG,EDG =EFG,FEGDEG.FG =DG.=,即AG2=CGDG.15.如图2-26,O的半径为r,MN切O于点A,弦BC交OA于点Q,BPBC,交MN于点P,求证:(1)PQAC;图2-26(2)若AQ=a,AC=b,则.思路分析:(1)连结AB,应用弦切角定理得CAN =ABC,运用圆内接四边形的判定和性质说明QPA =ABC,从而得出结论;(2)过点A作直径AE,连结CE,证PAQECA就可达到目的.证明:(1)连结AB,MN切O于点A,OAMN.又BPBC,BA、Q四点共圆,QPA =ABC.又CAN =ABC,CAN =QPA.PQAC.(2)过点A作直径AE,连结CE,则ECA为直角三角形.CAN =E,CAN =QPA,E =QPA.RtPAQRtECA.=, =.