How does one draw a triangle $ABC$ by using straightedge and compass knowing two sides and its median? I want to construct one triangle $ABC$ by using edge and compass. I know that $\overline{AB}=c, \overline{AC}=b$ and the measure of the median relative to side $BC$ is $m_{a}$.
Here is what I have thought. First I constructed the triangle $XYZ$ such that $\overline{XY}=c, \overline{XZ}=c$ and $\overline{YZ}=2m_{a}.$ Let $M$ be the midpoint of $YZ$. On the half-line XM I marked the point $X'$ such that $\overline{XM}=\overline{MX'}$. Therefore the triangles $MYX$ and $MX'Z$ are similar. The triangle $XZX'$ is the triangle that I am trying to construct. But there is (at least) one error here, because I do not know if I can construt the triangle $XYZ$.
I would appreciate your help.
 A: What you want is a parallelogram with sides $b,c$ and one diagonal $2m.$ So, draw the triangle with sides $b,c,2m.$ Rotate it around the midpoint of the $2m$ side to make a parallelogram. Half of the figure is your triangle $ABC.$  
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A: Although the answer below is technically correct, it probably is not the type of answer desired.  But it shows the power of the algebraic point of view: the construction can be found mechanically. 
It is easy to show using, for example, the Cosine Law, that if $m=m_a$ is the length of the median to side $BC$, then $b^2+c^2=2m^2+\frac{a^2}{2}$, and therefore
$$a=\sqrt{2b^2+2c^2-4m^2}.$$
We are given $b$, $c$, and $m$. There are standard techniques for multiplying two lengths by straightedge and compass, for adding and subtracting, and for constructing square roots. So we can construct a line segment of length $a$.
Now that we have line segments of length $a$, $b$, $c$, there is an  easy straightedge and compass construction of a triangle with sides $a$, $b$, and $c$.
Remark: For the sake of familiarity, we have used early modern algebraic notation. But precisely the same thing can be done using strictly Euclidean language. 
A: Fix $A, B$ so that $AB$ has the correct distance $c$. Let $O$ be the midpoint of AB. Construct the circle of radius $b/2$, centered at $O$. The median $M$ of side $BC$ lies on this circle since $OM$ is parallel to $AC$ and half the length. Also $M$ lies on the circle centered at $A$ of radius $m_a$. The intersection of these two circles thus gives the point $M$. So now complete the triangle, since $AC$ is parallel to $OM$.
