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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 $\angle FDE$?

enter image description here

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  • $\begingroup$ Easy to show that $\measuredangle FDE=90^{\circ}.$ $\endgroup$ – Michael Rozenberg Feb 6 '18 at 20:47
  • $\begingroup$ @MichaelRozenberg: I was stuck with this for 4 hours. Requesting you to please share a solution. $\endgroup$ – kasa Feb 6 '18 at 20:52
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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 $\measuredangle FDE=90^{\circ}.$

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  • $\begingroup$ I am able to understand that $AC$ is a bisector of $\angle NAD$ and $BF$ is a bisector of $\angle ABC$ but I am unable to see how these two facts lead to "that $DF$ is a bisector of $\angle ADC$." Thank you for your patience. $\endgroup$ – kasa Feb 6 '18 at 21:10
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    $\begingroup$ Because distances from $F$ to lines $AN$ and $AD$ they are equal and distances from $F$ to lines $AN$ and $DC$ they are equal. Which gives that distances from $F$ to $AD$ and $DC$ they are equal. $\endgroup$ – Michael Rozenberg Feb 6 '18 at 21:14
  • $\begingroup$ Simply Brilliant!! $\endgroup$ – kasa Feb 6 '18 at 21:30

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