连接BD。∵AD⊥AB∴∠BAF=∠BAD=90°∴BD是⊙O的直径又∵AF=AE，AB=AB∴△BAF≌△BAE（SAS）∴∠ABF=∠ABE∵AB=AC∴∠ABE=∠C∵∠C=∠D（同弧所对的圆周角相等）∴∠D=∠ABF∵AD/BD=cos∠D=cos∠ABF=0.8∴BD=AD/0.8=...
如图△ABC内接于圆O且AB=AC，点D在圆上AD⊥AB于点A，AD与BC交于点E，
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