證明:(1)對(duì)于y=3x+3,令y=0,得3x+3=0,x=-1,
∴B(-1,0).
∵C(1,0),
∴OB=OC,
∴AO垂直平分BC,
∴AB=AC,
∴∠ABC=∠ACB;
解:(2)∵AO⊥BC,DE⊥AC,
∴∠1+∠C=∠2+∠C=90°,
∴∠1=∠2.
∵AB=AC,
∴AO平分∠BAC,
∴∠2=∠3,
∴∠1=∠3.
對(duì)于y=3x+3,當(dāng)x=0時(shí),y=3,
∴A(0,3),
又∵D(-3,0),
∴DO=AO.
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∵∠AOB=∠DOF=90°,
∴△DOF≌△AOB(ASA),
∴OF=OB,
∴F(0,1).
設(shè)直線DE的解析式為y=kx+b,
∴
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,
解得
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,
∴y=
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x+1,
聯(lián)立
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,
解得
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,
所以,點(diǎn)G(-
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,
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);
解:(3)OM的長(zhǎng)度不會(huì)發(fā)生變化,過P點(diǎn)作PN∥AB交BC于N點(diǎn),
則∠1=∠Q,∠ABC=∠PNC,
∵∠ABC=∠ACB,
∴∠PNC=∠PCB,
∴PN=PC,
∵CP=BQ,
∴PN=BQ,
∵∠2=∠3,
∴△QBM≌△PNM(AAS),
∴MN=BM.
∵PC=PN,PO⊥CN,
∴ON=OC,
∵BM+MN+ON+OC=BC,
∴OM=MN+ON=
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BC=1.
分析:(1)先求出點(diǎn)B的坐標(biāo),然后根據(jù)點(diǎn)B、點(diǎn)C的坐標(biāo)求出OB=OC,再根據(jù)線段垂直平分線上的點(diǎn)到線段兩端點(diǎn)的距離相等得到AB=AC,然后根據(jù)等邊對(duì)等角的性質(zhì)即可證明;
(2)根據(jù)等角的余角相等求出∠FDO=∠BAO,然后利用“角邊角”證明△DOF和△AOB全等,根據(jù)全等三角形對(duì)應(yīng)邊相等可得OF=OB,從而求出點(diǎn)F的坐標(biāo),再根據(jù)待定系數(shù)法求直線解析式求出直線DF的解析式,與直線l
1的解析式聯(lián)立求解即可得到點(diǎn)G的坐標(biāo);
(3)過點(diǎn)P作PN∥AB交BC于點(diǎn)N,根據(jù)平行線的性質(zhì)可得∠MPN=∠Q,然后證明PN=BQ,再利用“角角邊”證明△QBM和△PNM全等,根據(jù)全等三角形對(duì)應(yīng)邊相等可得MN=BM,再根據(jù)等腰三角形三線合一的性質(zhì)可得ON=OC,從而證明OM=
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BC,是定值.
點(diǎn)評(píng):本題綜合考查了一次函數(shù),待定系數(shù)法求直線解析式,兩直線的交點(diǎn)的求解,全等三角形的判定與性質(zhì),以及等角對(duì)等邊,等邊對(duì)等角的性質(zhì),綜合性較強(qiáng),關(guān)系比較復(fù)雜,但難度不大,只要仔細(xì)分析,認(rèn)真求解,便不難解答.