Solution: (1) solution: As shown in the figure, the crossing point F is FG∑BC and AE is in G,

Then ∠DFG =∠DCE∠DGF =∠ Dirk,

D is the midpoint of CF,

∴CD=DF，

At △DCE and △DFG,

∠DFG=∠DCE？ DF=CD？ ∠GDF=∠EDC？ ,

∴△DCE≌△DFG(ASA)，

∴EC=GF，

∫BFAF = Mn，

∴AFAB=nm+n，

∫FG∨BC，

∴△AFG∽△ABE，

∴AFAB=FGBE=nm+n，

∴beec=m+nn；

(2) Proof: If BE=2EC, then be: ec = 2,

According to (1), m+nn=2,

Solution: m=n,

Point f is the midpoint of AB,

AC = BC，

∴CF⊥AB.