三角形ABC,AB=AC角BAC等于90度,BE平分角ABC,CE垂直BE,垂足为E,BE交AC于D,求2CE=BD

延长CE、BA,相交于点F.因为,∠BEC = 90°= ∠BEF ,BE为公共边,∠CBE = ∠FBE ,所以,△BCE ≌ △BFE ,可得：CE = EF ,即有：CF = 2CE ；因为,∠ACF = 90°-∠AFC = ∠ABD ,AC = AB ,∠CAF = 90°= ∠BAD ,所以,△CAF ...