过点E作EM⊥BD于M,EN⊥BD于N,FO⊥AC于O已知BE是∠ABC的平分线∴∠EBM＝∠EBN∴∠MEB＝∠NEB (等角的余角相等)又BE＝BE (公共)∴△BME≌△BNE (ASA)∴ME＝NE同理,可得：NE＝OE∴ME＝OE在Rt△AME和Rt△AOE中,ME＝OE (...