Question

已知AD为△ABC的角平分线,EF为AD的垂直之平分线,求证:FD²=FB×FC

Answer 1

连接AF

1) 由EF是AD的垂直平分可知△AEF≌△DEF,于是FA=FD,∠DAF=∠ADF

2) AD是角平分线有∠BAD=∠CAD,于是∠B=∠ADF-∠BAD=∠DAF-∠CAD=∠CAF,又∠AFB=∠CFA（公共角）,于是△BAF相似于△ACF,从而FB/FA=FA/FC,于是FA^2=FB*FC,即FD^2=FB*FC