圓的綜合練習(xí)題及答案2

OCO=于.OO.:,O,,.∴+.∵E,，∴+.∴O.1分2分ECOB.∵AD∴BEAEAB25.3分22∵=,,E.4分∴∴.4256即∴.AD1255.5分ADOCO與OFC1EA..AOBD，..O.2分FCE又，GAOBD1OEBC且3分2∴.ABCO于BCOBAC60，OODADBC，DBCABC是．DEACODDE．OABCBAC60△ABC1O2O交于C作于，P.O.分∴且∴分∴125即∴=5分45在ABC，BCCAABCBD交AC，DEDB交ABOBDE交BC于點FEACO.O分∵C90DBCBDC90．BFCBDABCABDDBC．A∴ODBBDC90ODC90EODOACO3o分DEO是BDEBEOO，BCCA912225在ABCAB2∴AB15，2222∵∴9．84BEOBFE90BEFBAC∴分OE是DOC交．DFAOBECD．F=．=．．AOBE，．．∵O∴．O==D．APEBCDOP22PC4242,分4EC）OODA23.13.分是.2分1AB3.2在RtOCD中，3,BD6,D3060.分在RtADEADABBD9,0019AEAD.4分220BC3.5分4POOAC．CPA30點P在CPAACCPCMPCOCOC．1在Rt△OCPOC．2OC23．第題CMPAMPA111COPCPO9045．222PO2PCOC，BC.CP2BCOPAOB）證明：如圖，連接OC．C∵PCOC，PCO901分D∵2PAOAOBOC．，PAOBOC1在Rt△PCOsinP．·······························································2分OP2O作ODBCBC2BD．·············································3分D1sinP，2P，POC60．∵OCOB，BOCB．在RtOBDOB2，BDOBcos303．·············································································4分BC23．······························································································5分O交DO交．B．D.ECOAO.2分OO.3分.，..分OCO=交．OO．∵O∴DAC，EOB=．，．∵O∴1分O2分==，∴．=．∵，．9=．3分2OO．4分在，22292∴．CDCD32228=5分O.OCB交于∵O∴1分∴+DEAO．∴+∴O2分∵，∴=1∴=．4分2，∴．9ADBDBECEAD6∴．=．5分234如圖：AB是⊙O的直徑，AD是弦，DAB22.5，延長AB到點C，使得