∵将△ABC沿线段BC向右平移得到△DEF, ∴AC∥DF,△ABC≌△DEF, ∴AC=DF, ∴四边形ACFD是平行四边形, ∴AD∥CF,AD=CF, 即AD∥AE, ∵△ABC≌△DEF, ∴BC=EF, ∴BC-EC=EF-EC, ∴BE=CF, ∴AD=BE,∴①正确; ∵△ABC≌△DEF, ∴∠ABC=∠DEF,∴②正确; ∵将△ABC沿线段BC向右平移得到△DEF, ∴AB∥ED, ∵∠BAC=90°, 即AB⊥AC, ∴ED⊥AC,∴③正确; ∵AD∥BC, ∴∠DAC=∠ACE, ∵AE=CE, ∴∠EAC=∠ACE, ∴∠DAC=∠EAC, ∵AC⊥DE, ∴∠AOE=∠AOD=90°, 在△ADO和△AEO中
∴△ADO≌△AEO, ∴AD=AE, ∵AE=CE, ∴AD=CE, ∵AD∥CE, ∴四边形AECD是平行四边形, ∵AE=EC, ∴四边形AECD是菱形,∴④正确; 即正确的个数是4个. 故选D. |