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Solution: extend EF to make a straight line parallel to b, AC and EF intersect at p,

AE = 5，EC=3，

∴AO=CE+OE, namely OE=EN= 1,

∫△DMN∽△DEO，MN= 13DM，

∴DE=3OE=3，

And ∵OE∨BP, O is the midpoint of DB, so E is also the midpoint.

∴EP=DE=3，

∴BP=2，

∵△EFC∽△PFB, the similarity ratio is 3: 2.

∴EF=EP×35= 1.8，

So df = de+ef = 3+ 1.8 = 4.8.

So the answer is: 4.8.