小题1:（1）证明：如图，
∵线段 DB 顺时针旋转60°得线段 DE ，
∴∠ EDB =60°， DE = DB .
∵△ ABC 是等边三角形，
∴∠ B =∠ ACB =60°.
∴∠ EDB =∠ B .
∴ EF ∥ BC .····································· 1分
∴ DB = FC ，∠ ADF =∠ AFD =60°.
∴ DE=DB=FC ，∠ ADE =∠ DFC =120°，△ ADF 是等边三角形.
∴ AD=DF .
∴△ ADE ≌△ DFC .
小题2:（2）由△ ADE ≌△ DFC ，
得 AE = DC ，∠1 = ∠2.
∵ ED ∥ BC ， EH ∥ DC ，
∴四边形 EHCD 是平行四边形.
∴ EH=DC ，∠3 = ∠4.
∴ AE=EH . ················································································· 3分
∴∠ AEH =∠1+∠3 = ∠2+∠4 = ∠ ACB =60°.
∴△ AEH 是等边三角形.
∴∠ AHE= 60°.
小题3:（3）设 BH = x ，则 AC = BC = BH ＋ HC = x ＋2，
由（2）四边形 EHCD 是平行四边形，
∴ ED=HC .
∴ DE=DB=HC=FC =2.
∵ EH ∥ DC ，
∴△ BGH ∽△ BDC .······································································· 5分
∴
.即
.
解得
.
∴ BC =3.
略