40°
连接AE,BD
通过上面条件得到三角形ADB和BFC全等,三角形ABE和三角形ACF全等
∴BD=BF,AE=AF ∴∠BDF=∠DFB,∠AEF=∠AFE
∴∠AFE+∠BFD=∠AEF+∠BDF=2∠DFE+∠AFD+∠BFE ①
又∠AEF+∠FEB=∠BAF,∠BDF+∠ADF=∠ABF(三角形全等得到)
两式相加
即∠AEF+∠FEB+∠BDF+∠ADF=∠BAF+∠ABF=180°-∠AFB=130°
即∠AEF+∠BDF=130°-∠FEB-∠ADF ②
所以由①②式2∠DEF+∠AFD+∠BFE =130°-∠FEB-∠ADF
2∠DEF=130°-∠FEB-∠ADF-∠AFD-∠BFE=130°-(∠FEB+∠ADF+∠AFD+∠BFE)③
又∵∠ADF=∠DFC,∠CFE=∠FEB
所以∠FEB+∠ADF+∠AFD+∠BFE=∠AFD+∠DFC+∠CFE+∠BFE=∠AFB=50°
代入③式即2∠DEF=130°-50°=80°
∴∠DEF=40°