My own proof of the concurrence of the three medians of any triangle

(It was my 8th finished write-up. It took me continual days to finially complete it on (Oct.26 & Nov.30, 2009)

Chen Tian


Here is the script tool I created to construct the centroid of a triangle.

The following is the proof I came up with completely by myself, though the mathematical idea was pretty easy.

(You may want to check out my another proof with some other proofs given by others)



Notation: In the following &(***) means angle (***).


Given any triangle ABC, then D, E are, respectively, midpoints of AB, AC.

Join CD, BE, which intersect at point F. Produce AF to R on BC. We want to prove R is the midpoint of BC.

Through F, draw line n parallel to AB and line m parallel to AC. Denote the new intersection points by G, H, J, K, shown in the figure. Join KG.

Now AKFG is a parallelogram. So the diagonals KG, AF bisects each other at point P (1). By vertical angle theorem , &(KFP)=&(JFR), &(GFP)=&(HFR), &(KFG)=&(JFH) (2).

By similarity n||AB implies GF:AD=CF:FD=FH:DB, and so AD=BD implies GF=HF (3).

For similar reason, m||AC and AE=CE implies KF=JF (4).

Thus, by (SAS) , Triangle(FGK) is congruent to Triangle(FGJ), since (3) (2) (4). So &(FGP)=&(FHR), &(FKP)=&(FJR) (5).

Hence, KG||HJ, i.e., KG||BC.

Thus, KBHG, KJCK are all parallelograms, and so KG=BH=JC.

By (ASA) Triangle(FKP) is congruent to Triangle(GJR) and Triangle(FGP) is congruent to Triangle(FRJ), since (2) (3) (5). So KP=JR, JP=HR.

But by (1) KP=JP, so JR=HR. Therefore, BH+HR=CJ+JR,i.e., BR=CR, i.e., R bisects BC, i.e., R is the midpoint of BC, as desired.



Note that afterwards Dr. Jim Wilson showed me some other ways to construct "critical parallelograms" about the given triangle, which was cool. I enjoyed. I would like to ask students to try them out as exercises.


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