triangle construction given side, angle and median I can't figure out the solution to this, it looks to me like it doesn't have any solution but I need some proof.
problem: Construct a triangle ABC with given $a=6 cm$ $\alpha=75^\circ $ and $m_{a}=4cm$.
 A: Find any point $P$ such that $\angle BPC = 75^\circ$ (see below), and consider the circumcircle $\bigcirc BPC$. By the Inscribed Angle Theorem, every point $Q$ on $\stackrel{\frown}{BPC}$ of that circle is such that $\angle BQC = 75^\circ$. Vertex $A$ is one of those points. Vertex $A$ is also $m_a$ units from the midpoint of $\overline{BC}$. Thus, $A$ is one of the points of intersection ---if any--- of $\stackrel{\frown}{BPC}$ with the circle of radius $m_a$ about that midpoint. 

How do you find that first point, $P$? One way is to erect a perpendicular to $\overline{BC}$ at $B$, as well as a ray making an angle of $15^\circ$ (the complement of $75^\circ$) with $\overline{BC}$ at $C$; where these lines meet is $P$. (Why?) Constructing the perpendicular is easy; for the other ray, it helps to realize that $60^\circ = 4\cdot 15^\circ$.
A: Apollonius's theorem says that, $4m_a^2=2 b^2 + 2 c^2 - a^2$. Hence, $b^2+c^2=(64+36)/2=50$. 
Now by the cosine rule, $a^2=b^2+c^2-2bc\cos\alpha$. So, $7=bc\cos 75^{\circ}$. Thus we have $bc$ and $b^2+c^2$. But since, $(b-c)^2=b^2+c^2-2bc=50-\frac{14}{\cos 75^{\circ}}<0$, $b-c$ isn't real and hence the construction isn't possible.
