Constructing a triangle given an angle, the side opposed to this angle, and the median to the given side. I came across the following problem in my Euclidean Geometry text:

Construct a triangle having given an angle, the side opposed to this angle, and the median to the given side.

I have tried solving this problem using method of loci, with Locus 1 being the circumcircle of the sought for triangle and Locus 2 being the circle of radius of the given  median. However I don't know how to start this construction.
 A: If the given angle is acute: To construct the circumscribed circle, observe that the inscribed angle is half of the central angle (see here for the terms and a picture). So you can start with your given side, construct an isosceles triangle with double the given angle opposite that side, and the newly constructed vertex of this triangle will be the center of the circumscribed circle you are looking for.
If the given angle is obtuse: Use the supplementary angle of the given angle to construct the isosceles triangle. The  new vertex you construct is the center of the circumscribed triangle in this case.
When completing the construction you need to be mindful of which of these two cases you are in: the side you start with divides the circumcircle into two arcs; the third vertex of your triangle will lie on the same side as the center (i.e. on the longer arc) if the angle was acute, and on the opposite side (i.e. on the shorter arc) if it was obtuse.
If the angle just happened to be a right angle, then the center of the circumscribed circle is the midpoint of the given side (and you will have infinite number of solutions if the median length was half the size of the given side, the third vertex  being anywhere on the circumcircle; or no solution if it was not).
A: 
I found this method experimentally. I am still looking for some theorem to prove following statement:
For a certain measure of BC which is the base of triangle ABC and radius of a circle center O at mid point of BC, with radius equal to measure of median $m_a$ we have:
$\angle BAC - \angle BPO= constant$
In figure AC||PO is the condition of above relation.
We use this fact to construct a triangle it's base, median and angle $\alpha $ opposite to base is given:
Draw a circle center on midpoint O of base BC. Draw PO and PB such that:
$\angle BPO=\frac{\alpha}2$
Draw a line from C parallel with PO, it intersect the circle at A. Connect A to B , The measure of  $\angle BAC$  will be:
$\angle BAC- \angle BPO=d$
Now to get $\angle BAC=\alpha$, construct $\angle BPC = \frac {\alpha}2 + d$ and draw a line from C parallel with PO to get vertex A.
In figure givens are $\alpha=40^o$, $BC=110 mm$ and $m_a=115 mm$
A: See the picture below:


*

*Draw segment AB and its middle point C.

*Draw the perpendicular bisector g and angle ACC'.

*Draw the  line i perpendicular to AC'.

*Let D the intersection of g and i.

*Draw the circle $\Lambda$ with center D and radius AD.

*Draw the circle $\Gamma$ with center C and radius $l$ (the median length).

*Let E and E' the intersections of $\Lambda$ and $\Gamma$.

*Draw triangles ABE or ABE', solutions of the problem.

