Using only a straight edge and compass, geometrically construct a triangle from the three given segments representing the base (side $c$), and the medians to the other two sides ($m_a$ and $m_b$)

I've looked at numerous websites trying to get help with this construction, but all I could find were constructions of triangles given 2 sides and a median or 2 medians and an altitude. I know I'm supposed to start with side $c$ and draw circles using the medians, but that's about it. How do I go about doing this?

  • 4
    $\begingroup$ Are you aware that medians cut each other into parts with the ratio $1:2$? $\endgroup$
    – Blue
    Commented Mar 6, 2019 at 16:53

2 Answers 2


So let's try to understand the problem first, so we can construct the figure.

Let's draw a triangle $\triangle ABC$, with $AB$ the horizontal side. Then the median from $A$ onto $BC$ will intersect $BC$ at $D$, and similarly $E$ is the intersection of the median from $B$ with $AC$. Due to similarity, you have $ED||AB$. You also have $ED=\frac 12 AB$. Now extend $AB$ past the $B$ point (away from $A$) to $B'$, suc that $B'B=\frac 12 AB=ED$. Note that $EBB'D$ is a parallelogram, so $B'D=BE$. In the triangle $\triangle ADB'$ you know that the lengths are $AB'=1.5c$, $AD=m_a$ and $DB'=EB=m_b$.

So now to construct the figure:

  1. draw a long straight line.
  2. select point $A$ at the beginning of this line
  3. use the compass to construct $B$ at distance $c$ from $A$
  4. use compass to get the center of $AB$, say $M$
  5. use compass to go $BM$ away from $A$ on the $AB$ line, to $B'$
  6. use compass to calculate $D$ at $m_a$ away from $A$ and $m_b$ away from $B'$
  7. construct $E$ in the same way, using $A'$ away from $B$

The last step can alternatively be calculated by drawing a parallel to $AB$ through $D$ and then use the intersection of this line with the center centered on $B$, and radius $m_b$


Let $M = m_a \cap m_b$. enter image description here Then $$|AM| = \frac 23m_a,\quad |BM| = \frac 23m_b$$ so we may construct $\triangle ABM$ from $3$ known sides.

Then we may extend medians to their full lengths, which gives us the midpoints of sides $a, b$.

Connecting these midpoints with points $A, B$ we obtain sides $a$ and $b$.


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