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證明:∵AB=AC,CD=BD,∴∠1=∠2,∠B=∠C,AD⊥BC,又∵AE是△ABC的外角平分線,∴∠3=∠4=12(∠B+∠C)=∠C,∴AE‖BC,∠DAE+∠ADB=180°,又∵AD⊥BC,∴∠DAE=∠ADC=90°.∴AE⊥AD.
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