1,由图可得正五边形每个角=108度,而AB=BC,所以∠CAB=∠ACB=36度.所以∠DCF=108-36=72度.因为AE=BC,∠EAB=∠CBA,AB=BA所以△EAB≌△CBA可得∠EBA=∠CAB=36度所以∠BFC=∠EBA+∠CAB=72度所以∠BFC=∠DCF推出BE‖DC,又因为∠EDC+∠DCA=180度.所以AC‖DE所以四边形CDEF是平行四边形,又CD=DE,得出四边形CDEF是菱形.
2,由上可得∠FBC=180-∠BFC-∠ACB=72=∠BFC所以BC=FC由四边形为菱形得FC=EF,所以BC=EF因为∠BAF=BEA=36度,∠EBA=∠ABF所以△EBA∽△ABF所以BA/BF=EB/AB即AB^2=BF×BE.而AB=BC=EF,所以EF^2=BF×BE