,/.OP=OB<AZABO=ZAPO-AZAPO+ZDCO=ZABO+ZOBD=ZABD,VZBAC=120u,AB=AC・・\ZABD=-^(180°—ZBAC)=30c,/.ZAPO+ZDCO=30°. 6分⑵解:AOPC是等边三角形.理由:VZABD=30°,.\ZBPC+ZBCP=180°—ZABD=150°,VZAPO+ZDCO=30°,,\ZOPC+ZOCP=I50°—30°=120°,VOP=OC,AZOPC=ZOCT=60°,AAOPC是等边三角形.?12分23•探究:证明:如图②,作DE丄AB于E.DF丄AC的延长线于F,TAD平分ZBAC>ADE=DEVZB+ZACD=I8O°•ZACD+ZDCF=180o•AZB=ZDCF.VZDEB=ZF=90°,AADBE^ADCF(AAS)••:DB二DC.?5分应用:解:如图③,连接AD,作DE丄ABFE,DF丄AC的延长线1*F.VZB+ZACD=180°,ZACD+ZDCF=180°•AZB=ZDCF.VZDEB=ZF=90°•TDB=DC,•'•△DBE竺△DCF(AAS)••IDE二DF.BE=CF,在RtZ^ADE和RtZkADF中•VAD=AD.DE二DF.ARtAADE^RtAADF(HL)>AAE=AF,.\ABAC=(AE+BE)-(AF-CF)2BE.在RtADBE中.ZDEB=90n,ZB=60°•ZBDE=30°•ABE=丄BD・VBD=a>ABE=-a./.ABAC=a.?12?分22解:⑴如图所示 3分AC〃BD 5分由柚对称性知,ZABD=ZCDB.VZABD=2ZADB>AZADC=ZADB.由(2)知AC〃BD・ZADC=ZCAD・TAO二CO.AB二CD.AO二AB・AAC=AO=CO>•••△AOC是等边三角形,AZAOC=60a,?10分ZDAO:ZDAB=-. 14分
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