Let $M$ and $N$ are midpoint of $AB$ and $AC$ respectively and let $\angle{BAC}=x$. $(\Rightarrow )$ WLOG $E$ lie on segment $AM$, $N$ must lie on segment $AF$. We've $\angle{EMD}=180^{\circ}-x,\angle{DNF}=x$. So, $\frac{ME}{\sin (\angle{MDE})}=\frac{DE}{\sin (\angle{DME})}=\frac{DF}{\sin (\angle{DNF})}=\frac{NF}{\sin (\angle{NDF})}$. This gives $\angle{MDE}=\angle{NDF}$ (since the sum of those two angles is $180^{\circ}$). So $\angle{EDF}=\angle{MDN}=\angle{BAC}$. $(\Leftarrow )$ WLOG $E$ lie on segment $AM$, $N$ must lie on segment $AF$. Since $\angle{EDF}=\angle{MDN}=\angle{BAC}$, we have $\angle{MDE}=\angle{NDF}$. This gives $$\frac{ME}{\sin (\angle{MDE})}=\frac{DE}{\sin (\angle{DME})}=\frac{DF}{\sin (\angle{DNF})}=\frac{NF}{\sin (\angle{NDF})}\implies ME=NF\implies AE+AF=BE+CF.$$

ManuelKahayon wrote: Let \(D\) be the midpoint of side \(BC\) of triangle \(ABC\). Let \(E, F\) be points on sides \(AB, AC\) respectively such that \(DE = DF\). Prove that \(AE + AF = BE + CF \iff \angle EDF = \angle BAC\). Let \(G\) and \(H\) be the points on \(AC\) and \(AB\) respectively such that \(G\) lies inside segment \(AC\) and \(H\) lies on \(AC\) beyond \(A\) and \(GF = CF\) and \(EH = BE.\) Now let \(BG \cap CH \equiv I\) \(; DF \cap CH \equiv J\) and \(DE \cap AB \equiv K.\) Since, \(AE + AF = BE + CF\) therefore, \(BE - AE = AF - CF \Rightarrow HE - AE = AF - GF \Rightarrow AH = AG -[1]\). Also from the midpoint theorem, \(\Rightarrow BG = 2 \times DF = 2 \times DE = CH -[2] \) Extend \(CA\) to \(A'\) such that \(HA' = HA.\) From \(eq^{n}1\) and \(2\) \(\Rightarrow\) \(\Delta BGA \cong \Delta CHA'\) \(\Rightarrow \angle ABG = \angle A'CH = \angle ACI\) Therefore, \( \angle BAC = \angle BIC\) because the \(ABCI\) is cyclic. Now since KDGI is a parallelogram angle KDJ = EDF = KIJ = BIC = angle A.

F' and E' on AB and AC such that F'D || AB and E'D|| AC . Prove that E'E=F'F <=> agle(FDF')=agle(EDE') I think it's easy.