问题探究] (1)如图1.锐角△ABC中.分别以AB.AC为边向外作等腰△ABE和等腰△ACD.使AE=AB.AD=AC.∠BAE=∠CAD.连接BD.CE.试猜想BD与CE的大小关系.并说明理由． [深入探究] (2)如图2.四边形ABCD中.AB=7cm.BC=3cm.∠ABC=∠ACD=∠ADC=45º.求BD的长．的条件下.当△ACD在线段AC的左侧� 题目和参考答案——青夏教育精英家教网—

问题探究】

（1）如图1，锐角△ABC中，分别以AB、AC为边向外作等腰△ABE和等腰△ACD，使AE=AB，AD=AC，∠BAE=∠CAD，连接BD，CE，试猜想BD与CE的大小关系，并说明理由．

【深入探究】

（2）如图2，四边形ABCD中，AB=7cm，BC=3cm，∠ABC=∠ACD=∠ADC=45º，求BD的长．

（3）如图3，在(2)的条件下，当△ACD在线段AC的左侧时，求BD的长．

（1）答：BD =CE． ················································································································· 1分

理由：∵∠BAE=∠CAD，

∴∠BAE+∠BAC=∠CAD+∠BAC，即∠EAC=∠BAD，··························································· 2分

又∵AE=AB，AC=AD，

∴△EAC≌△BAD (SAS) ，

∴BD=CE． ··························································································································· 4分

（2）解：如图1，在△ABC的外部，以点A为直角顶点作等腰直角三角形BAE，使∠BAE=90º，AE=AB，连接EA、EB、EC． ····································································································································· 5分

∵，

∴，，

∴∠BAE=，

∴∠BAE+∠BAC=∠CAD+∠BAC，

即∠EAC=∠BAD，

∴△EAC≌△BAD (SAS) ， ·························· 7分

∴BD=CE．

∵AE=AB=7，

∴， ∠AEC=∠AEB=45º．

又∵∠ABC=45º，

∴∠ABC+∠ABE=45º+45º=90º， ···························································································· 8分

∴EC==，

∴．

答：BD长是cm． ········································································································ 9分

(3)如图2，在线段AC的右侧过点A作AE⊥AB于A，交BC的延长线于点E， ···················· 10分

∴∠BAE=90º，

又∵∠ABC=45º，

∴∠E=∠ABC=45º，

∴AE=AB=7，．····················································································· 11分

又∵∠ACD=∠ADC=45 º，

∴∠BAE= ∠DAC=90º，

∴∠BAE∠BAC=∠DAC∠BAC，

即∠EAC=∠BAD，

∴△EAC≌△BAD (SAS) ，

∴BD=CE． ····································· 13分

∵BC=3，

∴BD=CE=(cm)．

BD长是()cm．

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