Confusion constructing the triangle $ABC$ given the lengths $AB$, $AC$ and the length of the median $m_a$?

Question

I'm trying to answer the following problem:

I did in a way and went to look for the solution, this is it:

I tried to follow it using Geogebra, I think it goes this way:

1. Given the setup of the problem, I think we have an arbitrary location for the point $A$ and the given lengths.

2. So we pick any point $M$, construct the ray from $A$ to $M$ and using the compass, we mark $A'$.

3. How do I show that $A'MC \equiv AMB$? I tried the following: Construct circles with the radius we were given. If this is valid, I can easily prove it using SSS.

Here are my further questions:

• I am interpreting the question as saying that we have the measures of the sites and the measure of the median, instead of interpreting it as saying we have the position of each point. Is that correct?
• Is the procedure for $3$ valid?
• Why construct the triangle $AA'C$? Shouldn't we just construct $AC$, since we have constructed $ABM$ and $A'CM$?
• After that, it says to construct the mid point of $AA'$ but isn't it already constructed? What I used in $2$ already seems to show that $M$ is the midpoint of $AA'$.

Answer 1

The point is that if you consider $A'$ as the symmetric of $A$ with respect to the midpoint of $BC$, $ABA'C$ is a parallelogram and $AA'=2m_a$. By reverse-engineering this approach you may start by constructing the triangle $ABA'$, where $AB=AB,BA'=AC,AA'=2m_a$, then take $C$ as the symmetric of $B$ with respect to the midpoint of $AA'$.

Of course there are many alternatives. For instance, by drawing $AB$ and a circle $\Gamma$ centered at $A$ with radius $AC$ we have that for any $P\in\Gamma$ the midpoint of $PB$ lies on a circle centered at the midpoint of $AB$. By intersecting this circle with the one centered at $A$ with radius $m_a$ we find the position(s) of the actual midpoint of $BC$, then $C$.