D
分析:①由弦AC=BD,可得
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=
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,繼而可得
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=
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,然后由圓周角定理,證得∠ABD=∠BAC,即可判定AE=BE;
②連接OA,OD,由AE=BE,AC⊥BD,可求得∠ABD=45°,繼而可得△AOD是等腰直角三角形,則可求得AD=
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R;
③設(shè)AF與BD相交于點G,連接CG,易證得△BGF是等腰三角形,CE=DE=EG,繼而求得答案.
解答:①∵弦AC=BD,
∴
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=
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,
∴
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=
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,
∴∠ABD=∠BAC,
∴AE=BE;
②連接OA,OD,
∵AC⊥BD,AE=BE,
∴∠ABE=∠BAE=45°,
∴∠AOD=2∠ABE=90°,
∵OA=OD,
∴AD=
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R;
③設(shè)AF與BD相交于點G,連接CG,
∵
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=
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,
∴∠FAC=∠DAC,
∵AC⊥BD,
∵在△AGE和△ADE中,
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,
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∴△AGE≌△ADE(ASA),
∴AG=AD,EG=DE,
∴∠AGD=∠ADG,
∵∠BGF=∠AGD,∠F=∠ADG,
∴∠BGF=∠F,
∴BG=BF,
∵AC=BD,AE=BE,
∴DE=CE,
∴EG=CE,
∴BE=BG+EG=BF+CE,
∵AB=
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,
∴BE=AB•cos45°=1,
∴BF+CE=1.
故其中正確的是:①②③.
故選D.
點評:此題考查了圓周角定理、弧與弦的關(guān)系、等腰直角三角形的性質(zhì)與判定以及全等三角形的判定與性質(zhì)等知識.此題難度較大,注意掌握輔助線的作法,注意數(shù)形結(jié)合思想的應(yīng)用.