# (South Africa 2014) Finding angles in an obtuse triangle

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 $\triangle ABC$.

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 $$\beta=\alpha-90^{o}$$

• Thanks! how to show that $\triangle AFC\sim \triangle CFB$? I can see just the same right angle and a common side. – bluemaster Aug 22 '18 at 12:16
• I think my problem is seeing how did you find $|FD|=b\cos \beta$. Please help. – bluemaster Aug 22 '18 at 12:48
• $\triangle BFD$ is not the right triangle, $\triangle ABD$ is. – g.kov Aug 22 '18 at 14:10