摘要: 证明:.. ·················································································································· 1分 在与中 ·········································· 2分 ································································ 1分 1分
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证明:
(1)如图1,△ABC中,AB=AC,延长BC至D,使CD=BC,点E在边AC上,以CE、CD为邻边作?CDFE,过点C作CG∥AB交EF于点G.连接BG、DE.
①∠ACB与∠GCD有怎样的数量关系?请说明理由.
②求证:△BCG≌△DCE.
(2)如图2,分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC=30°,EF⊥AB,垂足为F,连接DF.
①试说明AC=EF;
②求证:四边形ADFE是平行四边形.
![](http://thumb.zyjl.cn/pic3/upload/images/201211/32/db04d388.png)
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(1)如图1,△ABC中,AB=AC,延长BC至D,使CD=BC,点E在边AC上,以CE、CD为邻边作?CDFE,过点C作CG∥AB交EF于点G.连接BG、DE.
①∠ACB与∠GCD有怎样的数量关系?请说明理由.
②求证:△BCG≌△DCE.
(2)如图2,分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC=30°,EF⊥AB,垂足为F,连接DF.
①试说明AC=EF;
②求证:四边形ADFE是平行四边形.
![](http://thumb.zyjl.cn/pic3/upload/images/201211/32/db04d388.png)
证明:
(1)如图1,△ABC中,AB=AC,延长BC至D,使CD=BC,点E在边AC上,以CE、CD为邻边作?CDFE,过点C作CG∥AB交EF于点G.连接BG、DE.
①∠ACB与∠GCD有怎样的数量关系?请说明理由.
②求证:△BCG≌△DCE.
(2)如图2,分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC=30°,EF⊥AB,垂足为F,连接DF.
①试说明AC=EF;
②求证:四边形ADFE是平行四边形.
![](http://thumb.zyjl.cn/pic5/upload/201310/528566fc95ae2.png)
查看习题详情和答案>>
证明:
(1)如图1,△ABC中,AB=AC,延长BC至D,使CD=BC,点E在边AC上,以CE、CD为邻边作?CDFE,过点C作CG∥AB交EF于点G.连接BG、DE.
①∠ACB与∠GCD有怎样的数量关系?请说明理由.
②求证:△BCG≌△DCE.
(2)如图2,分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC=30°,EF⊥AB,垂足为F,连接DF.
①试说明AC=EF;
②求证:四边形ADFE是平行四边形.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202307217015995/SYS201311032023072170159023_ST/images0.png)
查看习题详情和答案>>
(1)如图1,△ABC中,AB=AC,延长BC至D,使CD=BC,点E在边AC上,以CE、CD为邻边作?CDFE,过点C作CG∥AB交EF于点G.连接BG、DE.
①∠ACB与∠GCD有怎样的数量关系?请说明理由.
②求证:△BCG≌△DCE.
(2)如图2,分别以Rt△ABC的直角边AC及斜边AB向外作等边△ACD、等边△ABE.已知∠BAC=30°,EF⊥AB,垂足为F,连接DF.
①试说明AC=EF;
②求证:四边形ADFE是平行四边形.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202307217015995/SYS201311032023072170159023_ST/images0.png)
查看习题详情和答案>>