延长CE,BA交于点F;
∵BF=BF,∠AEF=∠AEC,∠FBE=CBE(BE是角平分线)
∴△BEF≌△BCE,即E是CF中点;
∵∠BAD=DEC,∠ADB=EDC,
∴∠ECD=ABD;
又∵∠BAC=CAF=90,AB=AC;
∴△ABD≌△AFC,即BD=CF,
则BD=2CE
延长CE,BA交于点F;
∵BF=BF,∠AEF=∠AEC,∠FBE=CBE(BE是角平分线)
∴△BEF≌△BCE,即E是CF中点;
∵∠BAD=DEC,∠ADB=EDC,
∴∠ECD=ABD;
又∵∠BAC=CAF=90,AB=AC;
∴△ABD≌△AFC,即BD=CF,
则BD=2CE