Question

（本题6分）已知：如图，△ABC是等边三角形，D是AB边上的点，将DB绕点D顺时针旋转60°得到线段DE，延长ED交AC于点F，连结DC、AE．

1.（1）求证：△ADE≌△DFC；

2.（2）过点E作EH∥DC交DB于点G，交BC于点H，连结AH．求∠AHE的度数；

3.（3）若BG=，CH=2，求BC的长．

 

Answer 1

 

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.

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°.

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.

解析:略