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My homework question is what is the product of rotations through opposite angles α,−α about two distinct points. The answer is clearly a translation, but I'm not sure how to prove it. My idea on how to prove this is to draw a triangle, ABC, and its two rotations, A'B'C' and A"B"C", and then show AA", BB", CC" are parallel, however I don't know how to get to this point.

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Is this in $\mathbb{R}^3$? – ncmathsadist Oct 10 '13 at 1:48
What level of math class is this? High school geometry, or something in college? (Or, said another way: how rigorous of a proof are they looking for?) – apnorton Oct 10 '13 at 1:48
@ncmathsadist this is in R2 – David Grinberg Oct 10 '13 at 1:54
@anorton This is a college level geometry cource – David Grinberg Oct 10 '13 at 1:54

Let $A$ and $B$ be the centers of the two rotations, and let $R$ be the rotation about $A$ by the angle $\alpha$. Of course then $R^{-1}$ is the rotation about $A$ by $-\alpha$, but we want the rotation by $-\alpha$ about $B$ instead. Let $T$ be the translation that sends $B$ to $A$. Then the rotation about $B$ by $-\alpha$ is $T^{-1}R^{-1}T$, i.e., first move $B$ to $A$ by $T$, then do the rotation about $A$, and finally move $A$ back to $B$. (The composite operation sends $B$ to itself and rotates all directions by $-\alpha$, since $T$ doesn't affect directions. So the composite is the desired rotation around $B$.) So the combination "first rotate around $A$ by $\alpha$ and then rotate around $B$ by $-\alpha$" is the composition $T^{-1}R^{-1}TR$. Now $R^{-1}TR$, being a conjugate of a translation, is itself a translation $T'$, so we have the composite of two translations, $T^{-1}T'$, and this is a translation.

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