AD, BF and CE are the angle bisectors of the $\triangle{ABC}$. $\angle{BAC}=120°$. How to find $\angle{EDF}$?
[Math] Finding an angle in a triangle formed by constructions with angle bisectors
angleeuclidean-geometrygeometrytriangles
Related Solutions
I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
Hope the following sketch can help.
Best Answer
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.
and
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 $\angle ADB$.