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. |