已知三角形ABC的角ACB的外角平分线交三角形ABC的角平分线于点D.角A等于60度求角D

题目应当是这样的吧?“已知三角形ABC的角ACB的外角平分线交三角形ABC的∠B的平分线于点D.角A等于60度求角D”,如是,
设BD交AC于E,则,∠DEC=180-∠A-∠B/2=180-60-∠ABC/2=120-∠ABC/2
∠DCE=∠ACF(F是BC延长线上任意一点)
∵∠DCE=∠ACF/2=∠A+∠ABC/2=60°+∠ABC/2
∴∠D=180°-∠DCE-∠DEC
=180-120+∠ABC/2-60-∠ABC/2
=30°