How do I construct this triangle 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.



*

*Is it possible to construct it without trigonometry or analytical geometry?

*Is it possible to show that it cannot be done without trigonometry or analytical geometry?

*(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.
 A: 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: 

Now, recall that, in a triangle, smaller angles are opposite smaller sides.
$$\begin{align}
\triangle ABE: \quad 60^\circ - \phi < 60^\circ &\quad\implies\quad p < q \\ 
\triangle A^\prime CE: \quad \phantom{60^\circ-\;}\phi < 60^\circ &\quad\implies\quad p < r
\end{align}$$
Consequently, $2p < q + r$, which contradicts the desired condition. $\square$
A: 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.
