Final Project Part B

(Posted in the server on Nov.30 & Dec.10, 2009)

Chen Tian

In Assignment 4 I showed my own (1st) proof of the concurrence of the three medians of any triangle.

In this additional investigation, I will first briefly introduce 3 other ways to prove the concurrence of the three medians of any triangle. The first two of the three are pretty well known. But in the first one using Ceva's Theorem, I would leave the proof of Ceva's Theorem to students, or else, the proof would be too trival to students.

Here are other two proofs offered by Dr. Roy Smith for our geometry class.

One using vectors operation is essentially the same as the last one of three other proofs, but clearer stated and with some interesting generalization.

Now, I am going to show another proof (2nd one) I got by myself in a flash on Oct. 9 2009, which Dr. Roy Smith has shared with my Geometry classmates (MATH7200 UGA Fall 2009). I think this proof is much better than my previous proof, and I think it is a pretty neat proof.

Ok, here we go with the quick proof as follows.

(1) Given any triangle ABC, let D, E, and F be, respectively, midpoints of AB, AC, and BC. The medians are the segments connecting any vertex of the triangle and the midpoint on its opposite side. It is obvious that any two medians do meet inside the triangle ABC.

(2) We connect two of the three midpoints of the three sides so we get a medial triangle DFE. Claim: Every median of the original triangle is also a median of the medial triangle. Easily we can prove DF||AC, EF||AB, DE||BC (by either similarity or the properties of parallelograms and congruent triangles, which was shown in the Brief proof (2) among the 3 other ways above.). Medians BE, CD, and AF meet, respectively, DF, EF, and DE at G, H, and J. Thus |DG|=(1/2)|AE|, |GF|=(1/2)|EC|. But by construction |AE|=|EC|, so |DG|=|GF|; similarly, |FH|=|HE|, |DJ|=|JE|, as desired. Or, for example, we can consider BE, DF as the diagonals of parallelogram DBFE, so they bisect each other, i.e., |DG|=|GF|.

(3) Thus any two medians of the original triangle meet inside the medial triangle.

(4) We repeat constructing medial triangle of the previous one. We see any two medians of the original triangle must meet inside all medial triangles.

(5) Hence, we claim: all three pairs of medians meet exactly at the same point. Since the diameter of each medial triangle is 1/2 that of the previous one, by Archimedean Axiom there is only one point common to all the triangles constructed, “the limit of the shrinking triangles”.

Done with this cool proof.

Furthermore, in the last two of three other proofs I briefly stated above and also in the first proof provided by Dr. Smith, one common mathematical fact is used, which is that the centroid on any median is 2/3 of the way from the vertex which the median is through to the corresponding middle point on the opposite side of the triangle.

It is a very useful fact, which can be used to prove some other geometry theorems, for instance, the excenter (O), the centroid (G), and the orthocenter(H) of any trianlge lie on a line (called the Euler line of the triangle) and |GH| is always double of |GO|, i.e., |GH|=2|GO|. Here is the beautiful GSP document I created for demonstration. Any pair of parallel lines (one through a vertex and H, the other through the opposite midpoint and O) and the corresponding median through the vertex and the midpoint form two similar triangles. Therefore, by similarity, |GH|=2|GO| is always true. More generally, if we draw a line through G parallel to the pair of lines defined above, then any transversal (median is the special case) through G cutted between the pair of lines is divided into segments with constant ratio 2:1 (or the other way around).

Illustration figure:

You may want to check out the complete proof (stated differently from mine) about Euler line from Prop.5.7 on Page 55 in Hartshorne's Geometry: Euclid and Beyond (2005).

A little bit more explanation about the denotation G for centroid.