题目内容


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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