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證明:連接AD,延長AA1交DC於O,交C1C於E,∵∠ADA1=90°-∠A1DC=∠CDC1,ADDC=DA1DC1=3,∴△AA1D∽△CC1D,∠A1AD=∠C1CD,又∵∠AOD=∠COE,∴∠ADO=∠CEO=90°,即AA1⊥CC1.
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