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I am trying to show this.

The most basic way I can think of to show this is to assume that $f$ is a rigid transformation that fixes two points $P,Q \in\mathbb{R}^2$ and let $\alpha \in \mathbb{R}^2$ so $f(P) = P, f(Q) = Q$ and $ f(\alpha) \neq \alpha$.

If we consider the centroid of $PQ\alpha$ and the centroid of $PQf(\alpha)$ given by

$C_{PQ\alpha}$ and $C_{PQf(\alpha)}$, from a basic calculation we can see that they are not equal which implies that the transformation is not Euclidean(I think), let alone an orientation preserving Euclidean transformation(rigid). I am taking Euclidean to mean distance preserving in this case. I think it is more properly called an isometry.

Is this a legitimate proof? Is there a better/more elegant way to show this?

EDIT: I see a flaw in this. If we consider a rotation about a vertex of a triangle in $\mathbb{R}^2$ by $\beta\neq 0$ then the centroid of the triangle has changed. By my logic, a rotation about a point is not distance preserving. So yeah, a big flaw.

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    I think we're missing some assumptions. The nonidentity transformation $(x, y) \mapsto (-x, y)$ fixes the line $x = 0$. [Later] Ah, I see you write "an orientation preserving [...] transformation." So I guess we're only focusing on the orientation-preserving ones. – 2017-01-22
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    @pjs36 Yes, symmetries can be considered as "rigid transformations". – 2017-01-22
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    @JeanMarie Right, but the question only mentions rigid transformations (which, I thought, included reflections). But it seems that orientation-preserving is assumed also, albeit not in the problem statement. – 2017-01-22
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    The definition I have defines a rigid transformation as a orientation preserving isometry. – 2017-01-23

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If your definition of a rigid transformation does exclude orientation-reversing ones, how exactly do you have them defined? What can ypu build on?

In general I'd define a rigid transformation as a length-preserving one, i.e. an isometry. In that definition, a reflection would be considered rigid. Now your proof has a problem with that. Sure, the centroid of the reflected triangle would be different, but so what? Doesn't say anything about the transformation.

In general I'd do the following: assume you have any point in the plane. Then you have the two lengths to your two fixed points. If on the other hand you have those two lengths, then you have two circles around your two fixed points. Two circles intersect in up to two points. These two points will lead to triangles of different orientation. If you want to preserve orientation, then only one of them can be chosen, namely the original one. So that original point has to be fixed as well. This proof relies on the definition with respect to preservation of lengths and orientation, but it also builds on some properties of circles and their intersections which you may not have established yet.

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    I think what your saying boils down to: The distance between the two fixed points is clearly fixed. If the distance between the point (_any_) and one of these fixed points varies then the distance preserving nature of our rigid transformation is violated? – 2017-01-23
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    @HMPARTICLE: I'd not put it like this. It's half the story I think. The fact that varying the distance violates the assumption is one half, but the other half is the fact that withozt varying the distance you cannot help ending up at the original position. – 2017-01-23