An Interesting puzzle in elementary geometry

Question

In the picture given below, $ABC$ is a triangle, with $\angle A = 120^{\circ}$. $AD$, $BF$ and $CE$ are the angule bisectors of $\angle A$, $\angle B$ and $\angle C$ respectively. What is the $\angle FDE$?

Answer 1

Let $N\in AB$ such that $A$ is placed between $B$ and $N$.

Also, let $M\in AC$ such that $A$ is placed between $C$ and $M.$

Thus, since $AC$ is a bisector of $\angle NAD$ and $BF$ is a bisector of $\angle ABC$,

we obtain that $DF$ is a bisector of $\angle ADC$.

By the same way we can show that $DE$ is a bisector of $\angle BDA$, which says that $\measuredangle FDE=90^{\circ}.$