如图所示,三角形ABC中,∠ABC、∠ACB的角平分线交于点D,过D作DE平行BC交AB于E,交AC于F,求证：EF=BE+CF

难道我画的图和你的不同?
我证出这个
∵BD平分∠ABC
∴∠ABD = ∠DBC
∵DE //BC
∴∠EDB = ∠DBC
∴∠ABD = ∠EDB
∴EB = ED
∵CD是∠ACG的平分线
∴∠ACD = ∠DCG
∵ED//BC
∴∠FDC = ∠DCG
∴∠ACD = ∠FDC
∴FC = FD
∵EF= ED-FD
∴EF = EB - FC