and

Ravi E. Dattatreya

We show a construction and a general ratio result for triangles that
can be used to segment a line into various parts. The construction has as
a special case the **GLaD
construction**. (*Mathematics* *Teacher*, January 1997).

For any triangle ABC, select an arbitrary point M. Join AM, BM and CM, and extend them to cut BC, AC and AB at D, E and F, respectively. Join DE. Let the point of intersection of CF with DE be N. Join AN and extend it to meet CB at G.

Click **here** for a Geometer's Sketchpad file
for this construction.

Applying **Ceva's
theorem** to triangle ABC and the point M, we obtain:

Also, from **Jim
Wilson's result**, we have:

Now, apply Ceva's theorem to triangle ADC and the point N. We have:

Substituting for AM/MD,

Simplifying,

From this:

Q.E.D.

If D is the midpoint of BC, that is, AD is a median, and M is any point
on the median, then we have:

and, therefore,

In other words,

If we repeat the construction with N as the point, we can obtain the
point H such that

and so on. Thus, we can segment the line BC into integral divisions without
drawing any parallel lines. Just the location of the midpoint D of BC is
required. (**Picture**)

Click **here** to see a Geometer's Sketchpad construction
of these integral divisions.

If we extend the point B to infinity along CB, then the line ME will
become parallel to BC. Also, if we extend the point A to infinity along
BA, DM and GN, which are segments of DA and GA, will become parallel. Then,
our construction begins to resemble a GLaD-like construction applied to
the quadrilateral DMEC which has actually become a parallelogram (see also
Jim Wilson's comments on the GLaD construction.)

In exploring the literally hundreds of various distances, ratios and
areas that result in repeating this construction with the Geometer's Sketchpad,
we found very interesting relationships. Perhaps some of these are worth
exploring.

APPENDIX