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

【答案】

 

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