解:(1)∵矩形紙片沿直線AF折疊,使得點(diǎn)D與OC上的點(diǎn)E重合,
∴∠DAF=∠EAF.
故答案為=;
(2)∵AE平分∠OAF,
∴∠OAE=∠EAF,
而∠DAF=∠EAF,
∴∠DAF=∠EAF=∠OAE=30°,
在Rt△OAE中,OA=
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,
∴OE=OA•tan30°=1,
∴A(0,
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)、E(1,0),
設(shè)直線AE的解析式為y=kx+b(k≠0),
∴
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,解得:
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∴直線AE的解析式為y=-
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x+
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;
∵∠AEO=60°,∠AEF=90°,
∴∠FEC=30°
設(shè)點(diǎn)F的坐標(biāo)(x,y),則CF=y,
∴EF=DF=2y
又DF=DC-DF,
∴DF=
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-y,
∴2y=
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-y,解得
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,
又EC=
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CF=1,
∴OC=2,
∴F(2,
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);
(3)
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存在.理由如下:
如圖,作DN∥AM交y軸于點(diǎn)N,過(guò)點(diǎn)N作MN⊥y軸交直線AE于點(diǎn)M,
則MN∥AD,
∴四邊形MADN是平行四邊形.
∴MN=AD=2,
又∠OAE=∠MAN=30°.
∴AN=
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AD=2
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,
∴點(diǎn)
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;
延長(zhǎng)DC交直線AE于點(diǎn)M',則DM'∥AO,
作M'N'⊥y軸于點(diǎn)N',則M'N'∥AD,
∴四邊形AN'M'D是平行四邊形.
∴N'M'=OC=2
又點(diǎn)M'在直線
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上,當(dāng)x=2時(shí),
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,
∴點(diǎn)
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綜上,存在2個(gè)符合條件的點(diǎn)M坐標(biāo),它們是
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或
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.
分析:(1)根據(jù)折疊的性質(zhì)直接得到∴∠DAF=∠EAF;
(2)由AE平分∠OAF,得到∠OAE=∠EAF,而∠DAF=∠EAF,則∠DAF=∠EAF=∠OAE=30°,根據(jù)含30°的直角三角形三邊的關(guān)系得到OE=1,即A(0,

)、E(1,0),再利用待定系數(shù)法即可求出直線AE的解析式;設(shè)點(diǎn)F的坐標(biāo)(x,y),利用折疊的性質(zhì)和含30°的直角三角形三邊的關(guān)系可得到CF=
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,EC=1,即可得到F點(diǎn)的坐標(biāo).
(3)作DN∥AM交y軸于點(diǎn)N,過(guò)點(diǎn)N作MN⊥y軸交直線AE于點(diǎn)M,則四邊形MADN是平行四邊形,利用平行四邊形的性質(zhì)得到MN=AD=2,根據(jù)含30°的直角三角形三邊的關(guān)系
得AN=
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AD=2
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,即可得到M點(diǎn)的坐標(biāo);同理可得當(dāng)延長(zhǎng)DC交直線AE于點(diǎn)M',則DM'∥AO,作M'N'⊥y軸于點(diǎn)N',則M'N'∥AD,求出點(diǎn)M′的坐標(biāo).
點(diǎn)評(píng):本題考查了利用待定系數(shù)法求直線解析式的方法;也考查了折疊的性質(zhì)、含30°的直角三角形三邊的關(guān)系以及平行四邊形的性質(zhì).