Centroid Theorem

Consider a triangle with vertices $A,$ $B,$ and $C$ as well as two of its medians. Let $G$ be the point of intersection of the medians.

Let $Q$ be a point on $\overline{AC}$ such that $\overline{QE}$ is parallel to $\overline{BD}.$

In the diagram, $\angle EQC$ and $\angle BDC$ are corresponding angles. Since $\overline{EQ}$ and $\overline{BD}$ are parallel, these two angles are congruent by the Corresponding Angles Theorem. The same is true for $\angle QEC$ and $\angle DBC.$

Therefore, $△ECQ$ and $△BCD$ have two pairs of congruent angles and are similar by the Angle-Angle Similarity Theorem. Similar reasoning can be used to show that $△AGD$ and $△AEQ$ are also similar. By the definition of a median, $E$ is the midpoint of $\overline{BC},$ and therefore, $E$ divides $\overline{BC}$ into two congruent segments. Note that congruent segments have equal lengths. This information and the Segment Addition Postulate imply that the length of $\overline{BC}$ is two times the length of $\overline{EC}.$ Therefore, the scale factor of the similar triangles is $\frac{1}{2}.$ That means $DC=2QC.$ Furthermore, by the Segment Addition Postulate, $DC=DQ+QC.$ Then, using the Transitive Property of Equality and the Subtraction Property of Equality the following is obtained. Since $QC$ and $DQ$ are equal, $Q$ is the midpoint of $\overline{DC}.$ Remembering that $AD=DC,$ the ratio of $DQ$ to $AD$ can be calculated. Note that corresponding parts of similar triangles are proportional. Therefore, since $△AGD$ and $△AEQ$ are similar, the ratio of $DQ$ to $AD$ is equal to the ratio of $GE$ to $AG.$ This information can be used to express $AG$ in terms of $AE.$ This means that $AG$ is two-thirds of $AE.$ Now, consider $△ABC$ and its medians $\overline{CF}$ and $\overline{AE}.$ Let $K$ be the point of intersection of these medians.

Let $R$ be a point on $\overline{AB}$ such that $\overline{ER}$ is parallel to $\overline{CF}.$

By following the same reasoning as before, it can be proved that $AK$ is two-thirds of $AE.$ Therefore, $G$ and $K$ are the same points. That means the medians are concurrent — they meet at one point.

Before it was shown that $AG=\frac{2}{3}AE.$ By using similar arguments, it can be also shown that $BG=\frac{2}{3}BD$ and $CG=\frac{2}{3}CF.$