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∵在△ABC中,CF⊥AB於F,BE⊥AC於E,M為BC的中點,∴BC=2MF,BC=2EM,∴MF=EM,△EFM的周長=MF+EM+EF=BC+EF,∴EF=5,BC=8,∴△EFM的周長=8+5=13.故答案為:13.
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