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20 17 Nankai district mathematics module 2
∫△ABC is an equilateral triangle,

∴∠B=∠C=∠BAC=60,

∠∠ADC =∠b+∠BAD,∠ADE=60,

∴∠BAD=∠CDE,

∴△ABD∽△DCE,

∴ABDC=CEBD,

AB = BC = CA = 6,CD=2BD

∴BD=2,CD=4,

∴64=2CE,

∴CE=43,

∴AE=6-43= 143,

∫△ADC∽△AED,

∴AEAD=ADAC,

∴AD2=AE×AC= 143×6=28,

∴AD=27.

So the answer is: 27.