Triangle Median Explorations with
By Donna Greenwood
Given line segments j, k,
m.If these are the medians of a
triangle, construct the triangle.
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
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.
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
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.
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.
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.
up the points, take measurements and it works!
and Final Notes
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
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.
second observation/question is: why
does creating parallel line segments “work”? See
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.