Medians of a Triangle

By: Ginger Rhodes

Problem: Prove that the three medians of a triangle are concurrent and that the point of concurrence, the centroid, is two-thirds the distance from each vertex to the opposite side.

For the definition of a median of a triangle click HERE.


Let us begin by constructing two medians of a given triangle. Let the point of intersection of the medians be labeled point G.





So BD = DA and BE = EC.


Now, I will construct a line through B and G that intersects segment AC.


Since D is the midpoint of segment BA and E is the midpoint of BC I can let BD = AD and BE = EC. Using Cevaís Theorem (proved in final assignment, click HERE to take a look),



By substitution,






This implies AH = HC and BH is a median. Since BH was constructed to go through G, I can now say the three medians are concurrent.


Now to letís show the centroid is two-thirds the distance from each vertex to the opposite side.


We can use ∆DEG~∆CAG by AA~. [DE is parallel to AC because D and E are the midpoints of the sides of the triangle, which implies alternate interior angles are congruent]




Since D and E are midpoints of segments BA and BC, respectively, and therefore I can conclude DE=1/2AC. Now, triangle DIG and triangle CGH are similar by Angle-Angle-Similarity, by a ratio 1:2. Using this, I can say IG=1/2GH.


I showed BH is the median of ∆ABC and H is the midpoint of AC earlier; therefore GH and GI are the medians of ∆AGC and ∆DEG, respectively.


Since I is on segment DE and the endpoints of segment DE are the midpoints of the sides of the triangle we can conclude BI = IH.


Now, by segment addition postulate we know BI+IG+GH=BH


BIĖIG+IG+IG+GH =BH, add and subtract IG from the same side of the equation


GH+2IG+GH=BH, since IG+GH=IH and IH=BI


GH+GH+GH=BH, since IG=1/2GH or 2IG=GH




So, letís use segment addition postulate again to get BG+GH=BH.


Now we can make a substitution to get BG+1/3BH=BH. 


Subtract 1/3BH from both sides to get BG=2/3BH.


This shows the centroid, point G, is two-thirds the distance from the vertex B to the opposite side AC. We can use a similar proof to show the other two vertices.