Solution: connect AO and extend the intersection BC to point h, connect OC,

AB = AC，

∴AB=AC，

O is the center of the circle,

∴AH⊥BC,BH=HC，

∴HC=3，

Radius OC=5,

∴OH=4,AH=9，

∴, tan∠HAC=HCAH=39= 13 in Rt△AHC, that is, tan∠OAE= 13.

D and e are the midpoint of AB side and AC side, respectively.

∴DE∥BC，

∴AH⊥DE，

∴∠OAE+∠AED=90，

∫E is the midpoint of the side AC, and o is the center of the circle.

∴OE⊥AC，

∴∠AED+∠OED=90，

∴∠OAE=∠OED，

∴tan∠OED=tan∠OAE= 13.

The tangent of ∴∠OED is: 13.