证明:延长CE和BA交于点F,则有∠BAC=90°=∠FAC∵CE⊥BD,BE和AC交于点D∴∠DBA=∠ACF∵AB=AC∴△DBA=△FCA∴BD=CF,AF=AD∴∠ADF=∠ABC=45°∵BD是∠ABC的平分线,∠DBA=∠ACF,∠ADF=∠AFC+∠ACF∴∠AFC=∠ACF∵CE⊥BD,DE是公共边∴△FDE=△CDE∴FE=CE∵BD=CF=FE+CE∴BD=2CE
