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=B

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