Solution: (1) Because ∠ Abe = 30, BE = DE. So ∠ ADB = 30 Let AE =x BE =DE=2x AB =√3x BD=2√3x Because PQ∨BD, EP/ED=PQ/BD is (2x- 1)/2x=3√3/2√3x, so x=2.

(2) PE = 4-2√3 exists