三角形证明.只需第二问的解答,已知等腰三角形ABC和等腰三角形AEF,BA=BC,EA=EF,点E在线段BC上,且∠AEF=∠ABC,AC与EF交于点P.如图2,若∠ABC=30°,AB//EF,AM⊥BC,垂足为M,过点F作FN⊥AF,FN与AM交于点N,连接PN,探
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/31 04:57:20
![三角形证明.只需第二问的解答,已知等腰三角形ABC和等腰三角形AEF,BA=BC,EA=EF,点E在线段BC上,且∠AEF=∠ABC,AC与EF交于点P.如图2,若∠ABC=30°,AB//EF,AM⊥BC,垂足为M,过点F作FN⊥AF,FN与AM交于点N,连接PN,探](/uploads/image/z/6456927-39-7.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2%E8%AF%81%E6%98%8E.%E5%8F%AA%E9%9C%80%E7%AC%AC%E4%BA%8C%E9%97%AE%E7%9A%84%E8%A7%A3%E7%AD%94%2C%E5%B7%B2%E7%9F%A5%E7%AD%89%E8%85%B0%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E5%92%8C%E7%AD%89%E8%85%B0%E4%B8%89%E8%A7%92%E5%BD%A2AEF%2CBA%3DBC%2CEA%3DEF%2C%E7%82%B9E%E5%9C%A8%E7%BA%BF%E6%AE%B5BC%E4%B8%8A%2C%E4%B8%94%E2%88%A0AEF%3D%E2%88%A0ABC%2CAC%E4%B8%8EEF%E4%BA%A4%E4%BA%8E%E7%82%B9P.%E5%A6%82%E5%9B%BE2%2C%E8%8B%A5%E2%88%A0ABC%3D30%C2%B0%2CAB%2F%2FEF%2CAM%E2%8A%A5BC%2C%E5%9E%82%E8%B6%B3%E4%B8%BAM%2C%E8%BF%87%E7%82%B9F%E4%BD%9CFN%E2%8A%A5AF%2CFN%E4%B8%8EAM%E4%BA%A4%E4%BA%8E%E7%82%B9N%2C%E8%BF%9E%E6%8E%A5PN%2C%E6%8E%A2)
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