Erik D. Jacobson

Centroid Proof, 4#14 | Home


Consider the three medians of a triangle.  It is a fact that these segments are concurrent, and further, that the point of intersection is two-thirds the distance from each vertex to the opposite side.  The following steps demonstrate how GSP might be used to help high school students understand these relationships and develop a sense of proof.





The first step is to draw two medial lines and label their point of intersection point C.  A successful proof will show that (1) point C is one the third medial line (the one from vertex I) and in addition will (2) show that segment IC is twice the length of CM3. 


It is important to point out that it is not sufficient to draw a line from I to M3 and observe whether or not if goes through C; the argument that C is on the third median must follow from what is known about the median.


The important question to ask when proving (1) is, �What does it mean that C is on the median IM3?� The answer is that points I, C, and M3 are collinear.



Next, draw an auxiliary line through C that is parallel to side IJ. Since M2 is the midpoint of IJ and since triangle K1K2K is similar to triangle IJK by AA similarity, we know that C is the midpoint of segment K1K2.



By an analogous argument, when an auxiliary line is drawn through C parallel to side IK we find that C is the midpoint of segment J1J2. By SAS congruence, triangle CJ2K2 is congruent to triangle CJ1K1.  Further, triangle IK1J1 is similar to triangle IJK.  It follows that IK1CJ1 is a parallelogram and that triangle IK1J1 is congruent to triangle CJ1K1.  Furthermore, segment IC bisects (and is bisected by) segment J1K1.


To prove (1) and (2), it remains to show that M3 is the midpoint of segment K2J2 because the congruence of triangles IJ1K1, CK1J1, and CJ2K2 implies that A, C, and M3 are colinear—satisfying (1)—and that the medians are congruent, satisfying (2). 



Why is M3 the midpoint of segment K2J2?  Now, KJ2 is congruent to JK2 because JJ1K1K2 and KK1J1J2 are parallelograms. The midpoint M3 of segment JK is therefore the midpoint of J2K2.


It remains only to prove (3). Now, C is the midpoint of K1K2 and J1J2 by triangle similarity and so triangle CJ1K1 is congruent to CJ2K2. Since these triangles are congruent, they must have congruent medians. By the previous observation that J1K1 bisects IC, we see that (3) is in fact true.