已知:如图,在正方形ABCD外取一点E,连接AE、BE、DE.过点A作AE的垂线交DE于点P.若AE=AP=1,PB=5.下列结论:①△APD≌△AEB;②点B到直线AE的距离为2;③EB⊥ED;④S△APD+S△APB=1+6;⑤S正
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![已知:如图,在正方形ABCD外取一点E,连接AE、BE、DE.过点A作AE的垂线交DE于点P.若AE=AP=1,PB=5.下列结论:①△APD≌△AEB;②点B到直线AE的距离为2;③EB⊥ED;④S△APD+S△APB=1+6;⑤S正](/uploads/image/z/3579094-46-4.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E5%A4%96%E5%8F%96%E4%B8%80%E7%82%B9E%2C%E8%BF%9E%E6%8E%A5AE%E3%80%81BE%E3%80%81DE%EF%BC%8E%E8%BF%87%E7%82%B9A%E4%BD%9CAE%E7%9A%84%E5%9E%82%E7%BA%BF%E4%BA%A4DE%E4%BA%8E%E7%82%B9P%EF%BC%8E%E8%8B%A5AE%EF%BC%9DAP%EF%BC%9D1%2CPB%EF%BC%9D5%EF%BC%8E%E4%B8%8B%E5%88%97%E7%BB%93%E8%AE%BA%EF%BC%9A%E2%91%A0%E2%96%B3APD%E2%89%8C%E2%96%B3AEB%EF%BC%9B%E2%91%A1%E7%82%B9B%E5%88%B0%E7%9B%B4%E7%BA%BFAE%E7%9A%84%E8%B7%9D%E7%A6%BB%E4%B8%BA2%EF%BC%9B%E2%91%A2EB%E2%8A%A5ED%EF%BC%9B%E2%91%A3S%E2%96%B3APD%EF%BC%8BS%E2%96%B3APB%EF%BC%9D1%EF%BC%8B6%EF%BC%9B%E2%91%A4S%E6%AD%A3)
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