We should only prove that $ \angle{EDF} = 90^\circ$. Lemma. Let $ ABC$ be a triangle with $ \angle{A} = 60^\circ$. $ BD$ is bisector of $ \angle{B}$. A line, which passes through $ D$ intersect side $ AB$ at $ F$. Then $ \angle{FDB} = 30^\circ \Rightarrow CF$ is bisector of $ \angle{C}$.

It is clear that $ \angle{FIB} = 30^\circ$. Apply $ Lemma$ in triangle $ ADB$ for bisector $ BI$, line $ FI$ and obtain that $ DF$ is bisector of angle $ BDA$. By analogy $ DE$ is bisector of angle $ ADC$. So $ \angle{FDE} = \frac {1}{2}(\angle{BDA} + \angle{CDA}) = 90^\circ$.

see here

IWW wrote: In triangle $ ABC$ , $ m(\widehat {BAC}) = 120^{\circ}$ . Let the bisectors of angles $ \widehat A$ , $ \widehat B$ and $ \widehat C$ meet the opposite sides at $ D$ , $ E$ and $ F$ respectively. Prove that $ DE\perp DF$ . Proof I. Denote $ S\in BE$ , $ AS\perp AE$ and $ S'\in BE\cap DF$ . Observe that the ray $ [AS$ is the bisector of $ \widehat {IAB}$ and $ AS\perp AE$ $ \implies$ the division $ (B,S,I,E)$ is harmonically. Is well-known that the division $ (A,S',I,E)$ is harmonically. Thus, $ S'\equiv S$ , i.e. the ray $ [\overline {DSF}$ is the bisector of $ \widehat {ADB}$ . Analogously prove that the ray $ [DE$ is the bisector of $ \widehat {ADC}$ . In conclusion, $ DF\perp DE$ . See and here . Proof II. $ \left\|\begin{array}{ccc} A = 120^{\circ} & \implies & AD = \frac {bc}{b + c} \\ \\ \left\|\begin{array}{c} X\in DF\ ,\ Y\in DE \\ \\ A\in XY\ ,\ \overline {XAY}\parallel BC\end{array}\right\| & \implies & AX = AY = \frac {bc}{b + c}\end{array}\right\|$ $ \implies$ $ DE\perp DF$ . Proof III. If denote the intersection $ L\in CF\cap DE$ , then $ D(F,I,L,C)$ is harmonical pencil, i.e. $ (\infty , X,A,Y)$ is harmonical division $ \implies$ $ AX = AY = \frac {bc}{b + c}$ .If denote the second intersection $ A'$ of $ AI$ with the circumcircle of $ ABC$ , then $ BCA'$ is equilateral. Therefore, $ AA' = b + c$ (Pompeiu's theorem). But $ AA'\cdot AD = bc$ ( $ ABA'\sim ADC$ ). In conclusion, $ AD = \frac {bc}{b + c}$ . Otherwise, you can construct outside of $ ABC$ equilateral triangles a.s.o.

Since $\overline{AC}$ bisects the exterior angle $\angle BAD$, $E$ must be the $B$-excenter of $\triangle ABD$. By symmetry, $F$ must be the $C$-excenter of $\triangle ACD$. Thus, $\overline{DE}$ bisects $\angle ADC$, while $\overline{DF}$ bisects $\angle ADB$, from which we can find that $\angle FDE = \tfrac{1}{2}\angle BDC = 90^\circ$. The desired conclusion follows immediately.

Notice $\measuredangle (\overline{CA},\overline{AB})=60=\measuredangle DAC$ so $E$ is the $B$-Excenter of $\triangle ABD.$ Hence, $\angle CDE=\angle EDA,$ and similarly $\angle ADF=\angle FDB.$ Thus, $2\angle EDF=\angle BDC=180.$ $\square$