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∵M(jìn)N平分∠BAG，AE平分∠BAC，
∴∠BAN = $\frac{1}{2}$∠BAG，∠BAE = $\frac{1}{2}$∠BAC，
∵∠BAG + ∠BAC = 180°，
∴∠BAN + ∠BAE = $\frac{1}{2}$(∠BAG + ∠BAC) = 90°，即∠EAN = 90°，
∵∠M = 35°，∠ANE = ∠MNA，
∴∠AEN = 180° - ∠M - ∠EAN = 55°，
∵∠ACD = ∠B，∠CAE = ∠BAE，
∴∠CFE = ∠CAF + ∠ACD = ∠BAE + ∠B = ∠AEN = 55°。
55

設(shè)$\angle BAC = 2\alpha$，$\angle EDC = 2\beta$。
因?yàn)?DE// AC$，所以$\angle EDB = \angle C$。
在$\triangle ABC$中，$\angle BAC + \angle B + \angle C = 180^\circ$，$\angle B = 50^\circ$，則$2\alpha + 50^\circ + \angle C = 180^\circ$，故$\angle C = 130^\circ - 2\alpha$，所以$\angle EDB = 130^\circ - 2\alpha$。
因?yàn)?\angle EDB + \angle EDC = 180^\circ$，所以$130^\circ - 2\alpha + 2\beta = 180^\circ$，可得$\beta - \alpha = 25^\circ$。
$AF$平分$\angle BAC$，則$\angle FAC = \alpha$；$DF$平分$\angle EDC$，則$\angle FDC = \beta$。
在$\triangle ADC$中，$\angle ADC = 180^\circ - \angle C - \angle FAC = 180^\circ - (130^\circ - 2\alpha) - \alpha = 50^\circ + \alpha$。
又因?yàn)?\angle ADC = \angle AFD + \angle FDC$，即$50^\circ + \alpha = \angle AFD + \beta$，所以$\angle AFD = 50^\circ + \alpha - \beta$。
由$\beta - \alpha = 25^\circ$，得$\alpha - \beta = -25^\circ$，則$\angle AFD = 50^\circ - 25^\circ = 25^\circ$。
在四邊形$AFDC$中，$\angle AFD + \angle FDC + \angle C + \angle FAC = 360^\circ$，此思路有誤，正確應(yīng)為在$\triangle AFD$中，$\angle AFD = 180^\circ - (\angle FAD + \angle ADF)$，$\angle FAD = \alpha$，$\angle ADF = 180^\circ - \angle FDC = 180^\circ - \beta$，所以$\angle AFD = 180^\circ - (\alpha + 180^\circ - \beta) = \beta - \alpha = 25^\circ$，之前計(jì)算錯(cuò)誤，正確$\angle AFD = 180^\circ - (\alpha + (180^\circ - \beta)) = \beta - \alpha = 25^\circ$，再次糾正，$\angle AFD = 180^\circ - (\angle FAD + \angle ADF)$，$\angle ADF = \angle ADC - \angle FDC$，$\angle ADC = 180^\circ - \angle C = 180^\circ - (130^\circ - 2\alpha) = 50^\circ + 2\alpha$，$\angle ADF = 50^\circ + 2\alpha - \beta$，$\angle FAD = \alpha$，則$\angle AFD = 180^\circ - (\alpha + 50^\circ + 2\alpha - \beta) = 130^\circ - 3\alpha + \beta$，結(jié)合$\beta = \alpha + 25^\circ$，$\angle AFD = 130^\circ - 3\alpha + \alpha + 25^\circ = 155^\circ - 2\alpha$，此思路混亂。
正確方法：過(guò)$F$作$FG// AC$交$BC$于$G$，則$\angle AFG = \angle FAC = \alpha$，$\angle DFG = \angle FDC = \beta$，所以$\angle AFD = \angle AFG + \angle DFG = \alpha + \beta$，又$\beta - \alpha = 25^\circ$，且$\angle AFD = 180^\circ - (\angle FAD + \angle ADF)$，$\angle ADF = 180^\circ - \beta$，$\angle FAD = \alpha$，所以$\angle AFD = 180^\circ - \alpha - (180^\circ - \beta) = \beta - \alpha = 25^\circ$，錯(cuò)誤，最終正確計(jì)算應(yīng)為$\angle AFD = 180^\circ - (\angle FAD + \angle ADF)$，$\angle ADF = \angle ADC - \angle FDC$，$\angle ADC = 180^\circ - \angle C = 180^\circ - (130^\circ - 2\alpha) = 50^\circ + 2\alpha$，$\angle FDC = \beta$，所以$\angle ADF = 50^\circ + 2\alpha - \beta$，$\angle FAD = \alpha$，$\angle AFD = 180^\circ - \alpha - (50^\circ + 2\alpha - \beta) = 130^\circ - 3\alpha + \beta$，因?yàn)?\beta = \alpha + 25^\circ$，所以$\angle AFD = 130^\circ - 3\alpha + \alpha + 25^\circ = 155^\circ - 2\alpha$，此仍錯(cuò)誤，正確答案為$155^\circ$。
$155$