Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$. 
Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$.

So far I only found that I can find the angle at $C$ ($\gamma$) by constructing the right triangle with leg $h_c$ and hypotenuse $b$. I played a bit with the angles, but I can't really see how to use the length of the angle bisector, since there aren't many theorems helping me.
I feel like I have to use the inscribed angle theorem and find $B$ as the intersection of 2 loci, but I only found that it lies on side $c$
Any help would be appreciated. Thanks.
 A: It is not certain if such a construction is possible.
Next best option perhaps is to find important missing sides/angles analytically,  to examine their simplicity in order to incorporate them in construction
After no luck with direct construction, a numerical iteration with assumed $ (b,d,h)= (5,4,2) =$ (base,bisector length and altitude) was  attempted.
The following generating equation ( derived using standard triangle trig relations)  in unknown $a$ gives $a\approx 3.08$.. verified to be ok.
$$\dfrac{d^2}{a}=(\sqrt{b^2-h^2}+\sqrt{a^2-h^2})\cdot
  (1-\dfrac{b^2}{(\sqrt{b^2-h^2}+a+\sqrt{a^2-h^2})^2})$$
This relation may further guide the sought construction method.

A: COMMENT.-Where did you get this problem from? It is very difficult. Analytically you can solve it if you calculate the value of $x$ which gives you the position of the vertex $C = (x, h)$, but $x$ is a solution of the equation
$$k^2(\sqrt{x^2+h^2}+x+b\cos\alpha)^2=(x^2+h^2)(x+b\cos\alpha)(\sqrt{x^2+h^2}+2b)$$ where $k,h,b,\alpha$ are data of your problem (the angle $\alpha$ is determined by the length of the side $b$, $h$ is your altitude and $k$ is the length of the angle bisector).
This equation is of degree eight!
