The following uses plain coordinate geometry.
Let the known side be $a$, and let the height be $h$. As you point out, one can find (and construct) the sum $k$ of the lengths of the other two sides.
Let one end of the known side be $(-a/2,0)$, and another end be $(a/2,0)$. Let the coordinates of the third vertex of the triangle be $(x,h)$.
We can use the usual distance formula to find the sum of the distances from $(x,h)$ to the points $(-a/2,0)$ and $(a/2,0)$. This yields the equation $\sqrt{(x+a/2)^2+h^2}+\sqrt{(x-a/2)^2+h^2}=k.\tag{$1$}$ Now comes a cute little trick. Multiply top and (virtual) bottom of the left-hand side of $(1)$ by $\sqrt{(x+a/2)^2+h^2}-\sqrt{(x-a/2)^2+h^2}$. We get after some simplification $\frac{2ax}{\sqrt{(x+a/2)^2+h^2}-\sqrt{(x-a/2)^2+h^2}}=k.$ Flip both sides over, and simplify a little. We get $\sqrt{(x+a/2)^2+h^2}-\sqrt{(x-a/2)^2+h^2}=\frac{2ax}{k}.\tag{$2$}$ "Add" Equations $(1)$ and $(2)$. We get $2\sqrt{(x+a/2)^2+h^2}=k+\frac{2ax}{k}.$ Now it is safe to square both sides and not get a mess. We get a quadratic equation in $x$. This can be solved algebraically as usual, or by compass and straightedge.