用等腰三角形三线合一.
∵AB=AC,BD=CD,
∴∠DAB=∠DAC=1/2∠BAC,
∵AE平分∠BAF,
∴∠BAE=1/2∠BAF,
∴∠DAB+∠BAF=1/2(∠BAC+∠BAF)=1/2×180°=90°,
∴AE⊥AD.
如图,在△ABC中,AB=AC,BD=DC,AE平分△ABC的外角∠BAF,求证:AE⊥AD
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