证明:(Ⅰ)连结AD因为AB为圆的直径,所以∠ADB=90°,又EF⊥AB,∠EFA=90°则A、D、E、F四点共圆(4分)∴∠DEA=∠DFA (5分)
(Ⅱ)由(Ⅰ)知,BD•BE=BA•BF(6分),又△ABC∽△AEF∴
即:AB•A
F=AE•AC(8分)∴ BE•BD-AE•AC =BA•BF-AB•AF=AB(BF-AF)=AB 2(10分)
略
证明:(Ⅰ)连结AD因为AB为圆的直径,所以∠ADB=90°,又EF⊥AB,∠EFA=90°则A、D、E、F四点共圆(4分)∴∠DEA=∠DFA (5分)
(Ⅱ)由(Ⅰ)知,BD•BE=BA•BF(6分),又△ABC∽△AEF∴
即:AB•A
F=AE•AC(8分)∴ BE•BD-AE•AC =BA•BF-AB•AF=AB(BF-AF)=AB 2(10分)
略