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連結PO 在△APC中 ∵∠APC=90° ∴PO=1/2AC 同理,在△BPD中 ∵PO=1/2BD ∴AC=BD 又∵四邊形ABCD為平行四邊形 又對角線相等 ∴四邊形ABCD為矩形 不懂請繼續追問
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