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 $\angle{FDE} = \frac {1}{2}(\angle{BDA} + \angle{CDA}) = 90^\circ$.

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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, $2\angle EDF=\angle BDC=180.$ $\square$