在正方形ABCD中,E是对角线AC上的一点,AB=AE,P是EB上任意一点,PF⊥AB,PG⊥AC,垂足分别为F、G.求证:PF+PG+二分之一AC
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![在正方形ABCD中,E是对角线AC上的一点,AB=AE,P是EB上任意一点,PF⊥AB,PG⊥AC,垂足分别为F、G.求证:PF+PG+二分之一AC](/uploads/image/z/10008653-5-3.jpg?t=%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CE%E6%98%AF%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E4%B8%8A%E7%9A%84%E4%B8%80%E7%82%B9%2CAB%3DAE%2CP%E6%98%AFEB%E4%B8%8A%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2CPF%E2%8A%A5AB%2CPG%E2%8A%A5AC%2C%E5%9E%82%E8%B6%B3%E5%88%86%E5%88%AB%E4%B8%BAF%E3%80%81G.%E6%B1%82%E8%AF%81%EF%BC%9APF%2BPG%2B%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80AC)
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