How do I construct this triangle [closed]

I was trying to draw the following triangle in latex tikz and I just could not find a way to do it with respect to the given conditions.

1. Is it possible to construct it without trigonometry or analytical geometry?
2. Is it possible to show that it cannot be done without trigonometry or analytical geometry?
3. (Regrettably) I tried to do it via solving some trigonometrical equations, then I realized I am getting nowhere. So, how would I approach with trigonometry?

Note 1: In the actual drawing, $$c=6$$, but that does not matter, obviously. Note 2: I am not looking for LaTeX help.

closed as off-topic by Cesareo, TheSimpliFire, A. Pongrácz, José Carlos Santos, max_zornJan 20 at 22:45

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• What concrete is given in the exercise? – Dr. Sonnhard Graubner Jan 20 at 10:13
• Are you looking for $\LaTeX$ help or geometrical help such as $\psi = 60^\circ-\phi$ – Henry Jan 20 at 10:15
• And $$\angle {CEB}=120^{\circ}$$ – Dr. Sonnhard Graubner Jan 20 at 10:18
• @Henry Yes, Ican see all thse, buıt those do not help to draw the triangle with all the conditions met. – blackened Jan 20 at 10:20
• "figure not to scale"... – David Mitra Jan 20 at 10:21

As noted in the comments, a little angle-chasing shows that $$\phi + \psi = 60^\circ$$. As a result, we can find $$A^\prime$$ on $$\overline{BC}$$ such that $$\triangle ABE\cong \triangle A^\prime BE$$, as shown:
\begin{align} \triangle ABE: \quad 60^\circ - \phi < 60^\circ &\quad\implies\quad p < q \\ \triangle A^\prime CE: \quad \phantom{60^\circ-\;}\phi < 60^\circ &\quad\implies\quad p < r \end{align}
Consequently, $$2p < q + r$$, which contradicts the desired condition. $$\square$$
This triangle cannot be constructed for any $$\phi$$. If we assume $$\phi$$ can be any value and still have the triangle be constructed, we can see that length $$BE = 2 \cdot a$$ as $$\phi \to 60^\circ$$. Then applying the sine rule on lengths BE and AE gives: $$\frac{2\cdot a}{\sin{(\phi + 60^\circ)}}=\frac{a}{\sin{(60^\circ - \phi)}}$$ For which the only solutions are $$a=0$$ (invalid) or $$\phi = 30^\circ$$. If $$\phi = 30^\circ$$, then length $$AB =\sqrt{3} \cdot a$$ by the sine rule and we are left with a right angled triangle ABC. By Pythagoras' we then have $$AB^2+BC^2=AC^2$$ which is not true with the given lengths - a contradiction. So there is no way to construct the given triangle.