Triangle Construction being known its perimeter, height and angle "Construct a triangle $ABC$ knowing his perimeter, the angle $\widehat{A}$ and the height relative to $BC$, i.e., $h_a$."
It really looks to be an easy one, but I wasn't able to do it... :( Any hint?
(I only done two paralel lines with distance $h_a$, since the vertex $A$ and the basis $BC$ must lie in them...)
@Rory Daulton?
 A: This problem is discussed in George Polya's How to Solve It in the section "Auxiliary Problem" (pages 48-50 in my edition). Polya does not give a complete solution, just a hint to a certain point... so it looks ideal for this question. I use some quotes and a diagram from the book in this answer.
The first idea is that the problem is symmetrical in sides $b$ (opposite point $B$) and $c$ (opposite point $C$), so we should draw a figure with that same symmetry. With that in mind, we get this figure.

We want triangle $ABC$. We are given the angle at $A$ (with size $\alpha$), the altitude from point $A$ to side $a$ (with length $h$) , and the perimeter $a+b+c$. (These given items are in red). The hard part is where to draw the perimeter as one length. Here we added to side $a$ the segment $\overline{CE}$ of length $b$ on one side and the segment $\overline{BD}$ of length $c$ on the other side so that the perimeter appears as length $ED$ or $b+a+c$.
We are quickly led to add segments $\overline{AE}$ and $\overline{AD}$, each of which is a base of an isosceles triangle.
Examining this figure and figuring out some angles using those isosceles triangles we will see that
$$\angle DAE=\frac{\alpha}2+90^{\circ}$$
We now have a new and simpler problem: construct the triangle $ADE$ given the angle at $A$, the side opposite $A$, and the altitude from $A$.
This is as far as Polya goes, but this new problem is truly simpler. Once you get triangle $ADE$ you can find points $B$ and $C$ since they are on the perpendicular bisectors of $\overline{AE}$ and $\overline{AD}$.
Let me know if you need more.
