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∵ABCD是平行四邊形 ∴BC=AD=8cm AC、BD相交於點O 則AO=OC BO=OD=6cm 在直角三角形AOD中 OA²;=AD²;+OD²; OA²;=8²;+6²; OA=10cm AC=2OA=20cm ∴BC為8cm AC為20cm
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