如图,已知AB/AD=BC/DE=AE/AC,点B、D、E在同一直线上,试说明∠BAD=∠CBE=∠EAC
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如图,已知AB/AD=BC/DE=AE/AC,点B、D、E在同一直线上,试说明∠BAD=∠CBE=∠EAC
![](http://img.wesiedu.com/upload/f/eb/feb3e5bbe6110da1fab530c634d3e977.jpg)
![](http://img.wesiedu.com/upload/f/eb/feb3e5bbe6110da1fab530c634d3e977.jpg)
![如图,已知AB/AD=BC/DE=AE/AC,点B、D、E在同一直线上,试说明∠BAD=∠CBE=∠EAC](/uploads/image/z/8444116-28-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5AB%2FAD%3DBC%2FDE%3DAE%2FAC%2C%E7%82%B9B%E3%80%81D%E3%80%81E%E5%9C%A8%E5%90%8C%E4%B8%80%E7%9B%B4%E7%BA%BF%E4%B8%8A%2C%E8%AF%95%E8%AF%B4%E6%98%8E%E2%88%A0BAD%3D%E2%88%A0CBE%3D%E2%88%A0EAC)
∵AB/AD=BC/DE=AE/AC
∴△ABC∽△ADE
∴∠BAC=∠DAE
即∠BAD+∠DAC=∠DAC+∠EAC
∴∠BAD=∠EAC
2、
∵△ABC∽△ADE
∴∠ADE=∠ABC
∵∠ADE=∠BAD+∠ABE
∠ABC=∠ABE+∠CBE
∴∠BAE=∠CBE=∠EAC
![](http://img.wesiedu.com/upload/5/75/575dac20034c45ce28a8fb0280e510c9.jpg)
∴△ABC∽△ADE
∴∠BAC=∠DAE
即∠BAD+∠DAC=∠DAC+∠EAC
∴∠BAD=∠EAC
2、
∵△ABC∽△ADE
∴∠ADE=∠ABC
∵∠ADE=∠BAD+∠ABE
∠ABC=∠ABE+∠CBE
∴∠BAE=∠CBE=∠EAC
![](http://img.wesiedu.com/upload/5/75/575dac20034c45ce28a8fb0280e510c9.jpg)
如图,已知AB/AD=BC/DE=AE/AC,点B、D、E在同一直线上,试说明∠BAD=∠CBE=∠EAC
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