BC = DE，∠BED=60

∴DE=2BE

∴BC=2BE, that is, point E is the midpoint of BC.

∠ ACB = 45。

∴AE⊥BC

∴AE∥BD

And BD=√3BE, AE=BE.

∴BD=√3AE

∴ad=ab+bd=ab+√3ae=ab+√3*( 1/2)*(ab+ac)

=[(2+√3)/2]AB+(√3/2)AC (vector coincidence is omitted in this step).

∴x=(2+√3)/2,y=√3/2

Used in △ABD, but vector BD is transformed into expressions of vector AB and vector AC.