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(zhuǎn)60°得到線段DE，延長ED交AC于點F，連結(jié)DC、AE．

1.（1）求證：△ADE≌△DFC；

2.（2）過點E作EH∥DC交DB于點G，交BC于點H，連結(jié)AH．求∠AHE的度數(shù)；

3.（3）若BG=，CH=2，求BC的長．

 

Answer 1

 

1.（1）證明：如圖，

∵ 線段DB順時針旋轉(zhuǎn)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.

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°.

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:略