证明：∵AB=AC,E是BC的中点,∴BC⊥AE,

  在△ABD和△ACD中,∠ABD=∠ACD=90°,AB=AC,AD为公共边,

∴△ABD≌△ACD,

∴BD=DC．

又∵E是BC边的中点,

∴BC⊥ED,

因AE^2=AB^2-(BC/2)^2,DE^2=DC^2-(BC/2)^2=BD^2-(BC/2)^2,AD^2=AB^2+BD^2.

所以AE^2+DE^2- -AD^2= -BC^2/2.

∴cos∠AED<0,即△AED是钝角三角形．