如图1,图2,图3,在△ABC中,分别以AB,AC为边,向△ABC外作正三角形,正四边形,正五边形,BE,CD相交于点O.1)如图1,求证△ABE≌△ADC.(2)如图1,∠BOC=?°,(3)如图2,∠BOC=?°,(4)如图3,∠BOC=?°.如图4,已知AB,AD是以AB
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![如图1,图2,图3,在△ABC中,分别以AB,AC为边,向△ABC外作正三角形,正四边形,正五边形,BE,CD相交于点O.1)如图1,求证△ABE≌△ADC.(2)如图1,∠BOC=?°,(3)如图2,∠BOC=?°,(4)如图3,∠BOC=?°.如图4,已知AB,AD是以AB](/uploads/image/z/13429599-15-9.jpg?t=%E5%A6%82%E5%9B%BE1%2C%E5%9B%BE2%2C%E5%9B%BE3%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E5%88%86%E5%88%AB%E4%BB%A5AB%2CAC%E4%B8%BA%E8%BE%B9%2C%E5%90%91%E2%96%B3ABC%E5%A4%96%E4%BD%9C%E6%AD%A3%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E6%AD%A3%E5%9B%9B%E8%BE%B9%E5%BD%A2%2C%E6%AD%A3%E4%BA%94%E8%BE%B9%E5%BD%A2%2CBE%2CCD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9O.1%29%E5%A6%82%E5%9B%BE1%2C%E6%B1%82%E8%AF%81%E2%96%B3ABE%E2%89%8C%E2%96%B3ADC.%282%29%E5%A6%82%E5%9B%BE1%2C%E2%88%A0BOC%3D%3F%C2%B0%2C%283%29%E5%A6%82%E5%9B%BE2%2C%E2%88%A0BOC%3D%3F%C2%B0%2C%284%29%E5%A6%82%E5%9B%BE3%2C%E2%88%A0BOC%3D%3F%C2%B0.%E5%A6%82%E5%9B%BE4%2C%E5%B7%B2%E7%9F%A5AB%2CAD%E6%98%AF%E4%BB%A5AB)
如图1,图2,图3,在△ABC中,分别以AB,AC为边,向△ABC外作正三角形,正四边形,正五边形,BE,CD相交于点O.1)如图1,求证△ABE≌△ADC.(2)如图1,∠BOC=?°,(3)如图2,∠BOC=?°,(4)如图3,∠BOC=?°.如图4,已知AB,AD是以AB
如图1,图2,图3,在△ABC中,分别以AB,AC为边,向△ABC外作正三角形,正四边形,正五边形,BE,CD相交于点O.
1)如图1,求证△ABE≌△ADC.
(2)如图1,∠BOC=?°,(3)如图2,∠BOC=?°,(4)如图3,∠BOC=?°.
如图4,已知AB,AD是以AB为边向△ABC外所作正N边形的一组邻边,AC,AE是以AC为边向△ABC外所作正N边形的一组邻边,BE,CD的延长线相交于点O
(5)如图4∠BOC= 用含有n的式子表示
(6)根据图4证明你的猜想
如图1,图2,图3,在△ABC中,分别以AB,AC为边,向△ABC外作正三角形,正四边形,正五边形,BE,CD相交于点O.1)如图1,求证△ABE≌△ADC.(2)如图1,∠BOC=?°,(3)如图2,∠BOC=?°,(4)如图3,∠BOC=?°.如图4,已知AB,AD是以AB
(1)①证法一
∵△ABD与△ACE均为等边三角形,
∴AD=AB,AC=AE,
且∠BAD=∠CAE=60°,
∴∠BAD+∠BAC=∠CAE+∠BAC,
即∠DAC=∠BAE,
∴△ABE≌△ADC.
证法二:
∵△ABD与△ACE均为等边三角形,
∴AD=AB,AC=AE,
且∠BAD=∠CAE=60°,
∴△ADC可由△ABE绕着点A按顺时针方向旋转60°得到,
∴△ABE≌△ADC,
②120°,90°,72°.
(2)①360° n .
②证法一:依题意,知∠BAD和∠CAE都是正n边形的内角,
AB=AD,AE=AC,
∴∠BAD=∠CAE=(n-2)180° n ,
∴∠BAD-∠DAE=∠CAE-∠DAE,
即∠BAE=∠DAC,
∴△ABE≌△ADC,
∴∠ABE=∠ADC,
∵∠ADC+∠ODA=180°,
∴∠ABO+∠ODA=180°,
∵∠ABO+∠ODA+∠DAB+∠BOC=360°,
∴∠BOC+∠DAB=180°,
∴∠BOC=180°-∠DAB=180°-(n-2)180° n =360° n ;
证法二:同上可证△ABE≌△ADC.
∴∠ABE=∠ADC,如图,延长BA交CO于F,
∵∠AFD+∠ABE+∠BOC=180°,∠AFD+∠ADC+∠DAF=180°,
∴∠BOC=∠DAF=180°-∠BAD=360° n ;
证法三:同上可证△ABE≌△ADC.
∴∠ABE=∠ADC.
∵∠BOC=180°-(∠ABE+∠ABC+∠ACB+∠ACD),
∴∠BOC=180°-(∠ADC+∠ABC+∠ACB+∠ACD),
∵∠ABC+∠ACB=180°-∠BAC,∠ADC+∠ACD=180°-∠DAC,
∴∠BOC=180°-(360°-∠BAC-∠DAC),
即∴∠BOC=180°-∠BAD=360° n ;
证法四:同上可证△ABE≌△ADC.
∴∠AEB=∠ACD.如图,连接CE,
∵∠BEC=∠BOC+∠OCE,
∴∠AEB+∠AEC=∠BOC+∠ACD-∠ACE,
∴∠BOC=∠AEC+∠ACE.
即∴∠BOC=180°-∠CAE=360° n .
兄弟,你问得太好了,你图都没给,我怎么答?去问上帝吧!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!...
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兄弟,你问得太好了,你图都没给,我怎么答?去问上帝吧!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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)①用SAS可得△ABE≌△ADC;②∠BOC=360÷n
(2)①猜想∠BOC=360÷n
②先用外角得出∠BOC=∠ADC+∠BOE+∠DAE=∠ADC+∠DCA+∠DAE
=180°-∠DAE-∠EAC+∠DAE=180°-∠EAC
=360÷n