# Triangular geometric construction with 2 medians and 1 side

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?

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

## 2 Answers

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 $$M = m_a \cap m_b$$. 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 $$b$$.