圓的綜合演練題答案之青柳念文創(chuàng)作ABOCOC=BADBDABB.1ADO2O3AB=4AD.1:,AOOE,BE,ABE=90°.EAB+E=90°.E1CE=C,C=BADOBEAB+BAD=90°.ADADO.221ABE=90°.AE=2AO=6,AB=4,BEAE2AB225.3E=C=BAD,BDAB,cosBADcosE.4ABBE.ADAEAD125.55OABACOEACECFCFCAAOEABD.1FDO2OCBEGOG2O1OCOAOC1A.

FCE1AOBDOEACAAOE90°.1AOE90°.FCAAOE1FCA90°.OCF90°.FDO.22BC

FCEOEACAEEC.GAOBDAOOBOEBCOE1BC.32OEGCBG.OGOE1.CGCB2OG2CG4.OC5O6.ABCABOBCDDDEACEIDEOIIO5BAC60DEIADODAB

AD

BCABC

D

BCOD∥ACDEACODDEDEOIIABCBAC60△ABCO5ABBC10CD1BC52ABCAB=AEABOBECCCDAEDDCABP.1PDO2AE=5BE=6DC.1OC1PDAEDDCEE=900AB=AE,OB=OCCBA=E=BCODCE=PCBBCOPCB=900PDO22AC3AB=AE=5ABOBE=6ACBEEC=BC=3AC=4CBA=EEDC=ACB=90°EDCBCA4DC=ECACABDC=3DC=1254555.RtABCC=90BC=9CA=12ABCBDACDDEDBABEOBDE

AEOBCF1:ACO;2EFEF.AC1OD-------1C90DBCBDC90

DBFC（第5題）BDABCABDDBCOBODABDODBAODBBDC90ODC90-----2EoDBFCODOACO32DEDBORtBDEBEOOrRtABCAB2BC2CA292122225AB15AAADOC90ADOACBAOOD15rrABBC159r45BE45484BEOBFE90BEFBACEFBE45354ACBA1547.ABOEABCDOADFAEFDEDAFAFCAOBE1CEO2AFFC=53AE=16OAB1CEOOD

CDFAOBEADFAECAD=DAEOA=ODODA=DAECAD=ODAODACECACODECCE

O22BFABOAFB=90°C=90°AFB=CBFECAFAC=ABAEAFFC=53AE=1658=AB16AB=105ABCAB=ACDBCBDOABPPCADE1ADO（2）若PC是圓O的切線，BC=8，求DE的長．P（1）證明：∵AB=AC，點(diǎn)D是邊BC的中點(diǎn)，．BO∴AD⊥BD．又∵BD是圓O直徑，

AEDC（第8題）∴AD是圓O的切線．2分2）解：保持OP，由BC=8，得CD=4，OC=6，OP=2．∵PC是圓O的切線，O為圓心，∴OPC90．由勾股定理，得PC42．在△OPC中，tanOP22OCP424PC在DE2,tanDCEDC4DEDCtanDCE425分2.4△DEC中，E9.如圖，已知⊙O是△ABC的外接圓，AB是C⊙O的直徑，D是AB延伸線的一點(diǎn)，AE⊥CDAOFBD交DC的延伸線于E，CF⊥AB于F，且CE＝CF．（1）求證：DE是⊙O的切線；（2）若AB＝6，BD＝3，求AE和BC的長．證明：（1）毗鄰OC,10如圖，⊙O的直徑AB4，點(diǎn)P是AB延伸線上的一點(diǎn)，過P點(diǎn)作⊙O的切線，切點(diǎn)為C，結(jié)合AC．（1）若CPA30，求PC的長；（2）若點(diǎn)P在AB的延伸線上運(yùn)動，CPA的均分線交AC于點(diǎn)M．你以為CMP的大小能否發(fā)生更改？若更改，請說明出處；若不變更，求出CMP的大小．第19題解：（1）結(jié)合OC，則OCPC．在Rt△OCP中，OC1AB2，CPA30．2∴PC3OC23．2分（2）CMP的大小不發(fā)生更改．3分1115分COPCPO9045．22211如圖，點(diǎn)P在半O的直徑BA的延伸線上，CAB2PA，PC切半O于點(diǎn)C，保持BC.1）求P的正弦值；2）若半O的半徑為1）證明：如圖，毗鄰∵PC切半O于點(diǎn)C，

PAOB2，求BC的長度.OC．CDPOBPCO90．1分A∵AB2PA，PAOAOBOCRt△PCOsinPOC12OP22OODBCDBC2BD3sinP12P30POC60OCOBBOCB30Rt△OBDOB2BDOBcos3034BC23512RtABCACB=90°ACOABDDODEBCEBBE=CEDECD.ACB=90°,ACOECOADC=90°.

CAO2EDOD,ECED.3ECD=EDC.B+ECD=BDE+EDC=90°B=BDE.BE=ED4.ABOCOBCE=CABCEABEADABECD1DEO2CE=3BE=2CD1DEODCOCABOAOBEACB=90°OA=OCOAC=ACOBCE=CABBCE=ACOABOACB=90°BCEBCO=BCOACO=OCE=90°1DEO22BCE=CABBEC=CEABECCEACEAE=BECECE=3BE=23AE=23AE=923ADABABODAOAD=CD4RtABCAD2AE2DE2CD2922CD3215CD=8514.ABOADCDBC=A.D1BCOEBOA2OCADOCBDEBD=6CE=4AD.1ABOADB=90°1ABD+A=90°DBC=AABD+DBC=90°ABC=90°BCO22OCADADB=90°OEBDOED=ADB=BEC=90°BE=1BD=342DBC=ACBEBADADBDAD6AD=95BECE34215.ABOADDAB22.5ABCACD2DABCDOAB22BC1DO1DAO=DODOC=450ACD=2DA