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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: enter image description here

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.

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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

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  • $\begingroup$ Thanks! how to show that $\triangle AFC\sim \triangle CFB$? I can see just the same right angle and a common side. $\endgroup$ – bluemaster Aug 22 '18 at 12:16
  • $\begingroup$ I don't think that's even true? $\endgroup$ – asdf Aug 22 '18 at 12:23
  • $\begingroup$ I think my problem is seeing how did you find $|FD|=b\cos \beta$. Please help. $\endgroup$ – bluemaster Aug 22 '18 at 12:48
  • $\begingroup$ $\triangle BFD$ is not the right triangle, $\triangle ABD$ is. $\endgroup$ – g.kov Aug 22 '18 at 14:10
  • $\begingroup$ @g.kov "right" was an unnecessary word. Just apply the cosine law to the triangle after you've found the lengths of the sides, which you can do with sine rules for appropriate right triangles $\endgroup$ – asdf Aug 22 '18 at 14:40

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