如图,AB是⊙O的直径,弦BD、CA的延长线相交于点E,EF垂直BA的延长线于点F.求证. (Ⅰ)∠DEA=∠DFA; (Ⅱ)AB2=BE•BD-AE•AC.
问题描述:
如图,AB是⊙O的直径,弦BD、CA的延长线相交于点E,EF垂直BA的延长线于点F.求证.
(Ⅰ)∠DEA=∠DFA;
(Ⅱ)AB2=BE•BD-AE•AC.
答
证明:(Ⅰ)连结AD,∵AB为圆的直径,∴∠ADB=90°,又∵EF⊥AB,∴∠EFA=90°,∴A、D、E、F四点共圆,∴∠DEA=∠DFA.(Ⅱ)∵A、D、E、F四点共圆,∴由切割线定理知BD•BE=BA•BF,连结BC,则△ABC∽△AEF,∴AB...