# By Donna Greenwood

Given line segments j, k, m.If these are the medians of a triangle, construct the triangle

Click here for a sketch to manipulate. You can manipulate the medians, j, k and m to see how it impacts the constructed triangle ABC. This is the “backwards” version of creating a triangle of medians for any given triangle, so that’s the approach I took for the construction.

# Steps to a Triangle Construction

1.  Replicate one of the medians (m in this sketch) by drawing a parallel line to segment m through an arbitrary point (C). Construct a circle with radius m at C. One of the intersections of the line with the circle and point C determine the endpoints of m.

2. Create a triangle out of the given segments. At the endpoints of the new segment m, create two circles of radius j and k respectively. One of the intersections of the two circles is the third vertex of the triangle, so pick one.

3. Now we have a triangle whose leg lengths are j, k and m. From now on, references to j, k and m are to those segments, rather than the givens. Let one of the segments of the triangle be one of the medians of the new triangle (again, m in this sketch). The vertex is at point C. Note that if we could use measurement in constructions (which we can’t, of course), then we could simply measure 2/3 of the way along this median to find out where the medians intersect. Instead, construct the centroid of this triangle of medians.

4. Create a parallel at L to segment k in the sketch and replicate the segment (BL in this sketch). The endpoint is at point B in the “new” triangle.

5. By the definition of a median, segment CL is 1/2 the length of the side of the triangle. To get vertex A, construct the ray CL and a circle at L with radius CL. The intersection is the third vertex of the triangle, point A.

6. Connect up the points, take measurements and it works!

# Observations and Final Notes

Why does this work? There are a couple of observations made and used in the construction that haven’t been proven. The first: Is point L, the centroid of the triangle of medians, really the midpoint of side AC of DABC? This was the foundation of the construction. Here’s a sketch:

Point L was constructed as the centroid of the triangle of medians, DGEC.LB was constructed as the line parallel to segment k (FG in this sketch). We let m be a median of the constructed, giving point G as the midpoint of side AB of the constructed triangle. So, by triangle mid-segment theorem, DALB is similar to DAFG, and the similarity ratio is 1/2, so AF = 1/2 AL, or AF = FL. Since L is the centroid of DGEC, FL (=AF) = 1/3 FC, and LC = 2/3 of FC, and L is the midpoint of AC.
The second observation/question is: why does creating parallel line segments “work”? See the sketch:

The answer is basically the same as for the first question. Constructing the parallels creates similar triangles. In this sketch, look at DGEC and DOHC. The measure of angle EGC = the measure of angle HOC.  The other angles aren’t marked, but you can visually follow the sets of similar triangles to see the congruent angles.