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

【答案】

 

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(shè)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