(1)∵AD∥BC,∴∠MBC=∠AMB=60°
∵BM∥CN,∴∠MBE=∠CEB,
∵CE=BC,∴∠CEB=∠CBE
∴∠MBE=∠CBE=30°,
CB=CD=CE=2√(3),∴tan30°=CF/BC=CF/2√(3)=√(3)/3
∴CF=2,∴DF=2√(3)-2
(2)∠AMB=∠CND=∠BFC=60°
∠BAM=∠BCF=∠CDN=90°
AB=BC=CD,∴△ABM?△CBF?△DCN(AAS)
∴CF=DN=AM
∵AM=BM/2
∴BM=2AM=DN+CF
(1)∵AD∥BC,∴∠MBC=∠AMB=60°
∵BM∥CN,∴∠MBE=∠CEB,
∵CE=BC,∴∠CEB=∠CBE
∴∠MBE=∠CBE=30°,
CB=CD=CE=2√(3),∴tan30°=CF/BC=CF/2√(3)=√(3)/3
∴CF=2,∴DF=2√(3)-2
(2)∠AMB=∠CND=∠BFC=60°
∠BAM=∠BCF=∠CDN=90°
AB=BC=CD,∴△ABM?△CBF?△DCN(AAS)
∴CF=DN=AM
∵AM=BM/2
∴BM=2AM=DN+CF