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證明:連接AC,BD相交於點O 連接OP ∵四邊形ABCD是平行四邊形 ∴AO=OC,BO=OD ∵∠APC=90° ∴OP=1/2AC(直角三角形斜邊中線等於斜邊一半) 同理可得 OP=1/2BD ∴AC=BD ∴四邊形ABCD是平行四邊形(對角線相等的平行四邊形是矩形)
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