**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.

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!

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.