Vtyjkyuk

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. $DEparallel CF$ 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 $|FD|=bcosbeta$

Then use the sine rule for $triangle FCD$ (here you are using that $FD parallel AG$) to get that $$cos fracalpha2=cos gamma$$ which implies that (since you know $alpha>90^o$)

$$gamma=fracalpha2$$

Now do the same trick for the other side:

From $triangle CED$ using the cosine law you can find $|ED|=ccosgamma$ 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