Finding an angle in a triangle formed by constructions with angle bisectors
Matthew Harrington 
AD, BF and CE are the angle bisectors of the $\triangle{ABC}$. $\angle{BAC}=120°$. How to find $\angle{EDF}$?
1 Answer
$\begingroup$Let $K$ be placed on the line $AC$ such that $A$ placed between $F$ and $K$.
Thus, $$\measuredangle CAD=\measuredangle DAB=\measuredangle BAK=60^{\circ},$$ which says that $AB$ is a bisector of $\angle DAK$.
But also $CE$ is a bisector of $\angle ACB$ and $$CE\cap AB=\{E\},$$ which gives that $DE$ is a bisector of $\angle ADB$.
By the same way we can show that $DF$ is a bisector of $\angle ADC,$
which says $$\measuredangle EDF=\frac{180^{\circ}}{2}=90^{\circ}.$$
I used the following facts.
Each point of an angle bisector is equidistant from the sides of the angle.
and
If a point is placed inside the angle and equidistant from the sides of the angle then this point is placed on the bisector of the angle.
Let $d(A,l)$ be the distance between a point $A$ and a line $l$.
Since $CE$ is a bisector of $\angle ACB$, we obtain: $$d(E,AC)=d(E,BC).$$ Since $AB$ is a bisector of $\angle DAK$, we obtain: $$d(E,AK)=d(E,AD).$$ But lines $AC$ and $AK$ they are the same.
Also, lines $DB$ and $BC$ they are the same.
Thus, $$d(E,AD)=d(E,AK)=d(E,AC)=d(E,BC)=d(E,DB),$$ which says $$d(E,AD)=d(E,DB)$$ and from here the ray $DE$ is a bisector of $\angle ADB$.
$\endgroup$ 5
