# How to prove two triangles have the same centroid?

Suppose you have a $\triangle ABC$ and three similar exterior triangles $\triangle BCX$, $\triangle CAY$ and $\triangle ABZ$. How can I prove that the centroids of $\triangle ABC$ and $\triangle XYZ$ are the same point?

• Could you indicate which sides are in proportion and/or which angles are equal? (i.e. similar in what specific orientation, as it's not totally clear from the diagram)? – James Harrison Sep 30 '14 at 6:19

Consider the associated vector plane $V$, and the direct similarity transformation $\varphi$ of $V$ that takes $\overrightarrow{BC}$ to $\overrightarrow{BX}$. Then $\varphi$ is linear and, letting $G$ and $H$ be the respective centroids of $ABC$ and $XYZ$, we have $$3\overrightarrow{GH} = \overrightarrow{BX} + \overrightarrow{CY} + \overrightarrow{AZ} = \varphi(\overrightarrow{BC}) + \varphi(\overrightarrow{CA}) + \varphi(\overrightarrow{AB}) = \varphi(\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB}) = \varphi(0) = 0.$$