**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
Ceva’s Theorem (proved in final assignment, click HERE to take a look), **

** **

** **

**By substitution,**

** **

** **

**Therefore,**

** **

** **

**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 let’s 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 **

** **

**BI–IG+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**

** **

**GH=1/3BH**

** **

**So, let’s 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. **

** **

** **

** **