Given: $\triangle ABC$ with $\angle BAC$ being obtuse. Points D, E, F are the feet of the altitudes for $\triangle ABC$ computed from $A$, $B$ and $C$, respectively. $DE\parallel CF$ and the bisector of $\angle BAC$ is parallel to $DF$.
Find: all angles of $\triangle ABC$.
Source: South African Olympiad 2014. Answer given: 108, 18, 54 degrees.
From the statement the following figure can be drawn:
In the picture $AG$ is the bisector of $\angle BAC$. As I used the answer to accurately draw the figure (a little cheating...) it is easy to see that $BE=BD$ and $AG=GC$, so that $\triangle EBD$ and $AGC$ are isosceles. And $BA$ bisects $\angle EBC$. But I'm not finding a way to prove that and finding the required angles.
Hints and solutions are welcomed.
We are going to use standard notation:
$$\angle A=\alpha, \angle B=\beta, \angle C=\gamma, |AB|=c, |AC|=b, |BC|=a$$
Looking at the triangle $BFD$, using cosine law you can calculate $|FD|=b\cos\beta$
Then use the sine rule for $\triangle FCD$ (here you are using that $FD \parallel AG$) to get that $$\cos \frac{\alpha}{2}=\cos \gamma$$ which implies that (since you know $\alpha>90^{o}$)
$$\gamma=\frac{\alpha}{2}$$
Now do the same trick for the other side:
From $\triangle CED$ using the cosine law you can find $|ED|=c\cos\gamma$ and using that $ED \parallel FC$ in the sine rule for $\triangle DEB$ you get that $$\sin \alpha= \cos \beta$$ which implies $$\beta=\alpha-90^{o}$$
Now solve this system to get your answer. QED