初三數(shù)學(xué)圓的有關(guān)性質(zhì)知識精講

wqixlwqixl#/16BM2.已知：如圖AABC內(nèi)接于。O,過圓心O作BC的垂線交。O于點P,Q,交AB于點D,QP、CA的延長線交于點E,求證：OA2=OD-OE。3.如圖，P為半。O直徑AB延長線上一點，PCD/P=20。，/CAD=40。，求ZACD的度數(shù)。交半。O于C、D,已知B C.已知。O的半徑二二4,AB、CD為。O的兩條弦，AB、CD的長分別是方程X2-(4y3+4)%+16M3=0的兩根，其中AB>CD且AB//CD，求AB與CD間的距離。.已知：如圖AB是。O直徑，CD是弦，AE±CD，垂足為E,BF±CD，垂足為F。(1)求證：EC=DF2)若AE=a，EF=b,BF=c,求證：氏/EAC和tg/EAD是方程ax2—bx+c=0的兩個根。參考答案.解：連結(jié)BM??.AD是BC邊上的高，，/ADB=ZADC=90°在RtAADB和RtAADC中AB=BD22+AD2=<82+62:10AC=eCD2+AD2=<32+62:3<5??.AM是直徑，，/ABM=90°，/ABM=AADC又/M=/C,:.AABM~AADCABAMADAC10AM即—=——6 3、；5??.AM=5v5.證明：延長AO交。O于F,連結(jié)BFAF是直徑，.?./ABF=90°，/BAF=90°—/FEQ1BC，??./E=90°—/C又/F=/C，.??/E=/BAF即/DAO=/E，又/DOA=/AOE??.ADOA?AAOEOA_OE…O—OA??.OA2=OD?OEmicc.解：/P--(AD—BC)

ccmcAD+BC^=180°-CD=180°-2/CAD=100°cmcm設(shè)AD^=x,BC^=y，則有11-(x-y)=20°v2x+y=100°x=70°解得v[y=30°m1cm:./ACD^=-AD^=35°24,證明：連結(jié)GF、EF,作ZADM=ZAFG,DM交AC于4,證明：連結(jié)GF、EF,作ZADM=ZAFG,DM交AC于M則AAFG?AADM,AF_AG

AD-AMAG-AD=AF-AMZAGF=ZAMD且ZAGF+ZAEF=180°.ZAEF=ZDMC易證AAEF?ACMD,AEAF_AF

Cm―Cd―ABAE-AB=AF-CM.AF?(AM+CM)=AG?AD+AE?AB即AF-AC=AG-AD+AE-AB5,解：由x2-(4v-3+4)x+16<3=0解得兩根為x=4v3，x—41 2AB、CD為方程兩根且AB>CD.AB―4<3，CD―4作OF±CD于FAB//CD，.OFLAB，設(shè)垂足為EABX貝UOE—：OB2-(—)22—v42-(2<3)2—2OF―CD —、：OD2-(C—)2―v42-22—2,：3(1)若AB、CD位于圓心O的同側(cè)如圖<1>,則EF—2a/3-2AE//OG//BFGE-GC=GF-GDGE-GC=GF-GD又04=OB,又GC=GD,即EC=DF，GE=GF(2)連結(jié)AC、BCtgZEAC=ECAE，tgZEAD=EDtgZEAC=ECAE，tgZEAD=EDAEtgZEAC+tgZEADECED + AEAEEC+EDAEDF+EDAEEFbAEatgZEAC-tg/EAD=ECEDECEDEC?EDAEAEAE2tgZEAC-tg/EAD=ECEDECEDEC?EDAEAEAE2AAEC與\CFB中/E=/F=90°?/AB為直徑，，ZACB=90°/.ZACE+/BCF=90°又ZCBF+ZBCF=90°/./ACE=ZCBF，AAEC?ACFBECAEBFCFEC-CF=AE-BF=