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證明:∵AC‖BD,∴∠CAO=∠E,∠ACO=∠F,∵∠1=∠E,∠2=∠F,∴∠1=∠CAO=12∠BAC,∠2=∠ACO=12∠ACD,∵AB‖CD,∴∠BAC+∠ACD=180°,∴∠CAO+∠ACO=90°,∴∠AOC=90°,∴AE⊥CF.
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