(1)證明:∵M(jìn)N∥BC,
∴∠OEC=∠ECB,∠OFC=∠FCD.
又∵CE平分∠ACB,F(xiàn)C平分∠ACD.
∴∠ECB=∠OCE,∠OCF=∠FCD,
∴∠OEC=∠OCE,∠OFC=∠OCF,
∴EO=OC,F(xiàn)O=OC,
∴EO=FO;
(2)解:當(dāng)點(diǎn)O運(yùn)動到AC中點(diǎn)時,四邊形AECF為正方形.理由如下:
由(1)知,OE=OC=OF,
當(dāng)OC=OA,即點(diǎn)O為AC的中點(diǎn)時,
∴OE=OC=OF=OA,
∴四邊形AECF是平行四邊形,AC=EF,
∴這時四邊形AECF是矩形;
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又∵∠ACB=90°,MN∥BC,
∴∠AOE=∠ACB=90°,
∴AC⊥EF,
∴矩形AECF是正方形.
∴AE=CE=
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,∠AEC=90°,
∴AC=2,OA=OE=1.
在Rt△ABC中,∵∠ACB=90°,AB=4,AC=2,
∴sin∠B=
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=
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=
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,
∴∠B=30°,
∴∠AGO=∠B=30°,OG=
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OA=
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.
過E作EH⊥AB于H,設(shè)EH=x,則GE=2x,
∵GE+OE=OG,
∴2x+1=
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,
∴x=
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.
在Rt△AHE中,sin∠HAE=
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=
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=
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,
∴sin∠BAE=
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.
分析:(1)由直線MN∥BC,MN交∠BCA的平分線于點(diǎn)E,交∠BCA的外角平分線于點(diǎn)F,易證得△EOC與△FOC是等腰三角形,即可得OE=OF;
(2)由(1)知,OE=OC=OF,當(dāng)OC=OA,即點(diǎn)O為AC的中點(diǎn)時,可得OE=OC=OF=OA,證得四邊形AECF是矩形;再由∠ACB=90°,MN∥BC,得出AC⊥EF,從而證明矩形AECF是正方形;根據(jù)正方形的性質(zhì)及勾股定理求出AC=2,OA=OE=1,在Rt△ABC中,由正弦函數(shù)的定義得到∠B=30°,則∠AGO=30°,OG=
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.過E作EH⊥AB于H,設(shè)EH=x,由GE+OE=OG,列出方程2x+1=
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,解方程求出x=
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,然后在Rt△AHE中,利用正弦函數(shù)的定義求出sin∠HAE的值,即可得到sin∠BAE的值.
點(diǎn)評:此題考查了平行線的性質(zhì),角平分線的定義,等腰三角形的判定與性質(zhì),正方形、矩形的判定與性質(zhì),解直角三角形.此題綜合性較強(qiáng),難度適中,解題的關(guān)鍵是注意數(shù)形結(jié)合思想的應(yīng)用.