证明：延长CE,BA交于F所以△BCE≌△BFE(ASA),∴CE=FE,∴CF=2CE ∵∠ADB=∠=EDC,∵等角的余角相等所∠ABD=∠ACF∴△ABD≌△ACF(ASA),所以BD=CF=2CE ∴CE=1/2BD俊狼精英为您解答
