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【不知圖,設AD在∠BAC間】 證明: ∵AB⊥AC ∴∠BAD+∠DAC=90º; ∵AD⊥AE ∴∠CAE+∠DAC=90º; ∴∠BAD=∠CAE 又∵AB=AC,AD=AE ∴⊿ABD≌⊿ACE(SAS)
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