How do I construct this triangle

Question

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.

Answer 1

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.

Answer 2

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. $\square$