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.
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.