解:(1)△ADC≌△ABE,△CDF≌△EBF;(2)连接CE,∵Rt△ABC≌Rt△ADE,∴AC=AE,∴∠ACE=∠AEC(等边对等角),又∵Rt△ABC≌Rt△ADE,∴∠ACB=∠AED,∴∠ACE-∠ACB=∠AEC-∠AED,即∠BCE=∠DEC,∴CF=EF。
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