The projection of a barycenter onto a plane is the barycenter of the projected triangle?

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When projecting the vertices $A$, $B$ and $C$ of a traingle onto a plane, getting the points $A'$, $B'$ and $C'$, is the projection of $G$ (barycenter of the triangle $ABC$) the same as the barycenter of $A'B'C'$?

I think the way is showing that the projection of the midpoints of the sides of the triangle are the midpoints of the projected sides.

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The answer is yes, and the reason is both deep and straightforward.

A projection onto a plane is an affine map. This means by definition that it respects all affine combinations, and barycentres are a special kind of affine combinations (where all weights are equal).

An easy way to see why such a projection is affine is that if you choose the origin of $\Bbb R^n$ carefully (namely, on the plane onto which you are projecting), then the projection is a linear map. Linear maps are trivially affine.