如图所示,AD是圆O的直径,BC切圆O于点D,AB,AC与圆O相交于E.F,求证AE×AB=AF×AC

夜猫猫_涵er,（图见参考资料.）1）如图1．连接DE、DF,AD为直径,则∠AED＝90°＝∠ADB；又∠BAD＝∠BAD.则△AED∽△ADB,AD/AE＝AB/AD,AD^2＝AE×AB⑴；同理△AFD∽△ADC,AD/AF＝AC/AD,AD^2＝AF×AC⑵.∴AE×AB＝AF×A...