题目

（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．  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 

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