0
$\begingroup$

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?

enter image description here

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – James Harrison Sep 30 '14 at 6:19
1
$\begingroup$

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 $$ 3\overrightarrow{GH} = \overrightarrow{BX} + \overrightarrow{CY} + \overrightarrow{AZ} = \varphi(\overrightarrow{BC}) + \varphi(\overrightarrow{CA}) + \varphi(\overrightarrow{AB}) = \varphi(\overrightarrow{BC} + \overrightarrow{CA} + \overrightarrow{AB}) = \varphi(0) = 0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.