∴ab=bc∠BAC =∠BCA = 60。 ( 1)

∫ Quadrilateral ACDE is isosceles trapezoid, ∠ EAC = 60,

∴AE=CD,∠ACD=∠CAE=60，

∴∠BAC+∠CAE= 120 =∠BCA+∠ACD，

That is ∠ BAE = ∠ BCD. (2 points)

In △ABE and △BCD, AB=BC, ∠BAE=∠BCD, AE=CD,

∴△ Abe△ ?△CBD. (3 points)

(2) existence. The answer is not unique. For example, △ ABN ∽△ CDN.

Proof: ∫∠Ban = 60 =∠DCN, ∠ANB=∠DNC,

∴△ ANB ∽△ CND。 (5 points)

The similarity ratio is abdc = 21= 2; (6 points)

(3) ANCN= ABCD=2 From (2),

∴CN= 12AN= 13AC, (8 points)

Similarly, AM= 13AC,

∴ am = Mn = NC。 (9 points)

(4) Make the extension line from DF⊥BC to BC in F,

∫∠BCD = 120，

∴∠ DCF = 60。 ( 1o)

In Rt△CDF, ∴∠ CDF = 30,

∴CF= 12CD= 12，

∴df= Cd2-cf2 = 12-( 12)2 = 32； ( 1 1)

In Rt△BDF, ∫BF = BC+CF = 2+ 12 = 52, DF= 32.

∴ BD = BF2+DF2 = (52) 2+(32) 2 = 7。 (12)