BD,CE分别是△ABC的外角平分线,过点A作AF⊥BD,AG⊥CE,延长AF,AG与直线BC相交,证明证明FG=1/2(AB+BC+CA)若BD,CE为内角平分线,FG与△ABC三边有什么关系,理由若BD是内角平分线,CE是外角平分线,又有什么关
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![BD,CE分别是△ABC的外角平分线,过点A作AF⊥BD,AG⊥CE,延长AF,AG与直线BC相交,证明证明FG=1/2(AB+BC+CA)若BD,CE为内角平分线,FG与△ABC三边有什么关系,理由若BD是内角平分线,CE是外角平分线,又有什么关](/uploads/image/z/10204959-39-9.jpg?t=BD%2CCE%E5%88%86%E5%88%AB%E6%98%AF%E2%96%B3ABC%E7%9A%84%E5%A4%96%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E8%BF%87%E7%82%B9A%E4%BD%9CAF%E2%8A%A5BD%2CAG%E2%8A%A5CE%2C%E5%BB%B6%E9%95%BFAF%2CAG%E4%B8%8E%E7%9B%B4%E7%BA%BFBC%E7%9B%B8%E4%BA%A4%2C%E8%AF%81%E6%98%8E%E8%AF%81%E6%98%8EFG%3D1%2F2%28AB%2BBC%2BCA%29%E8%8B%A5BD%2CCE%E4%B8%BA%E5%86%85%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%2CFG%E4%B8%8E%E2%96%B3ABC%E4%B8%89%E8%BE%B9%E6%9C%89%E4%BB%80%E4%B9%88%E5%85%B3%E7%B3%BB%2C%E7%90%86%E7%94%B1%E8%8B%A5BD%E6%98%AF%E5%86%85%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%2CCE%E6%98%AF%E5%A4%96%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E5%8F%88%E6%9C%89%E4%BB%80%E4%B9%88%E5%85%B3)
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