(6分)如图,点P为△ABC的内心,延长AP交△ABC的外接圆于D,在AC延
(6分)如图,点P为△ABC的内心,延长AP交△ABC的外接圆于D,在AC延长线上有一点E,满足AD

=AB·AE,求证:DE是⊙O的切线.

第20题图
证明略解析:
证明:连结DC,DO并延长交⊙O于F,连结AF.∵AD

=AB·AE,∠BAD=∠DAE,∴△BAD∽△DAE,∴∠ADB=∠E. 又∵∠ADB=∠ACB,∴∠ACB=∠E,BC∥DE,∴∠CDE=∠BCD=∠BAD=∠DAC,又∵∠CAF=∠CDF,∴∠FDE=∠CDE+∠CDF=∠DAC+∠CDF=∠DAF=90°,故DE是⊙O的切线