(1), ∫In△ABC, AB=AC, P is a point above BC, ∠ BAC = ∠ BDP = α = 60,

Abc = ACB = BAC = 60, △Abc is equilateral△, m is the midpoint of BC, ∠ DPB = (1/2) ∠ ABC = 30, MF⊥BC, let CK⊥AB be in K, and the extension line of MF be in G,.

G is the center of △ABC (inner center, outer center, center of gravity, vertical center), ∠ BCG = 30,

∴PD∥CK, PD⊥AB in E,

∴∠DBP= 180 -60 -30 =90，

At Rt⊿PEB, BE=( 1/√3)PE,

In Rt⊿PDB, according to the projective theorem, yes? =DE？ PE，

[( 1/√3)PE)？ =DE？ PE，

∴DE=( 1/3)PE。

(2)∠BDP=∠BAC=90，∠ABC=∠ACB=45，∠DPB=( 1/2)∠ABC=22.5，

∠PBD=90 -22.5 =67.5，∠DBE==67.5 -45 =22.5，

∠DBE=∠DPB=22.5，

DE/BD = tan 22.5 = BD/PD = 0.4 142 1，DE？ PD=BD？ ,

0.4 142 1PD= BD，

∴de=0.4 142 1 BD =(0.4 142 1)？ PD=0. 17 157 PD，

∫DE+PE = PD，

∴DE=0. 17 1573(DE+PE)，

0.82843 DE = 0. 17 157 PE，

DE=0207 1PE .