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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.

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Is $m_a$ a distance or a line? –  i. m. soloveichik Aug 31 '12 at 19:10
    
@i.m.soloveichik: $m_{a}$ is the measure of the median relative to the vertex $A$. –  spohreis Aug 31 '12 at 19:13
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3 Answers

up vote 7 down vote accepted

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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If $2m$ is bigger than $b+c$, there is no triangle $ABC$. –  spohreis Aug 31 '12 at 19:01
    
@spohreis, yes, and the parallelogram does not exist either. I'm not sure what you are trying to say. –  Will Jagy Aug 31 '12 at 19:21
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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.

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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 $OM$.

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