Question

（本題6分）已知：如圖，△ABC是等邊三角形，D是AB邊上的點，將DB繞點D順時針旋轉60°得到線段DE，延長ED交AC于點F，連結DC、AE．

【小題1】（1）求證：△ADE≌△DFC；
【小題2】（2）過點E作EH∥DC交DB于點G，交BC于點H，連結AH．求∠AHE的度數(shù)；
【小題3】（3）若BG=，CH=2，求BC的長．

Answer 1

【小題1】（1）證明：如圖，
∵線段DB順時針旋轉60°得線段DE，
∴∠EDB =60°，DE=DB.
∵△ABC是等邊三角形，
∴∠B=∠ACB =60°.
∴∠EDB =∠B.
∴EF∥BC.····································· 1分
∴DB=FC，∠ADF=∠AFD =60°.
∴DE=DB=FC，∠ADE=∠DFC =120°，△ADF是等邊三角形.
∴AD=DF.
∴△ADE≌△DFC.
【小題2】（2）由△ADE≌△DFC，
得AE=DC，∠1=∠2.
∵ED∥BC， EH∥DC，
∴四邊形EHCD是平行四邊形.
∴EH=DC，∠3=∠4.
∴AE=EH. ················································································· 3分
∴∠AEH=∠1+∠3=∠2+∠4 =∠ACB=60°.
∴△AEH是等邊三角形.
∴∠AHE=60°.
【小題3】（3）設BH=x，則AC= BC =BH＋HC= x＋2，
由（2）四邊形EHCD是平行四邊形，
∴ED=HC.
∴DE=DB=HC=FC=2.
∵EH∥DC，
∴△BGH∽△BDC.······································································· 5分
∴.即.
解得.
∴BC=3.

解析