Prove the centroid lies on all three medians

Prove the centroid lies on all three medians

To prove that three medians meet, intersect two of them first. Express their intersection in the two independent side directions. Then show that this very point lies on the third median; a diagram alone cannot establish the concurrence.

A median joins a vertex to the midpoint of its opposite side. In a nondegenerate triangle $OAB$, let $M,N,R$ be the midpoints of $OB,OA,AB$. The intersection $G$ of medians $AM$ and $BN$ has two representations:

$$\overrightarrow{OG}=(1-\lambda)\mathbf a+\frac\lambda2\mathbf b=\frac\mu2\mathbf a+(1-\mu)\mathbf b.$$

Check the third median

The complete Self-Reading Example 8 works through both path expansions, compares independent coefficients, and solves $\lambda=\mu=2/3$. Thus $\overrightarrow{OG}=(\mathbf a+\mathbf b)/3$, $AG:GM=2:1$, and $\overrightarrow{OG}=(2/3)\overrightarrow{OR}$. The last vector equality proves that $G$ lies on the third median segment too.

Relabel $O$ as $C$: the three medians of any nondegenerate triangle $ABC$ meet at the centroid, two thirds of the way from each vertex to the opposite midpoint. The source’s final physical remark applies to a uniform triangular lamina in a uniform gravitational field: its centroid is also its centre of gravity. Equal side-midpoint marks and the $2:1$ division refer to different lengths; do not confuse them.