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Denote by $G_{XY}$ a glide reflection which reflects around the $XY$-axis and then takes the point $X$ to $Y$. I would like to prove that the composition of two different glide reflections $G_{XY} \circ G_{ZX}$ is a rotation (about some point) by twice the angle between the vectors $XY$ and $XZ$.

I read here that the composition of a glide reflection $G$ with itself is a translation parallel to the axis of reflection of $G$, but am unsure of how to use such result here. Is there a way to prove this without finding exactly the rotation pivot?

I am trying to show this in order to prove a more general statement, namely that a sufficient condition for the composition $G_{DA} \circ G_{CD} \circ G_{BC} \circ G_{AB}$ to be the identity map is that the quadrilateral $ABCD$ is cyclic, i.e. can be inscribed in a single circle.

Note that $G_{BC} \circ G_{AB}$ means that we first perform $G_{AB}$, then do $G_{BC}$ using the original axis of reflection $BC$, not the new one obtain after the first isometry.

Any help would be greatly appreciated.

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Hubble
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  • Picture is of great help. Once you draw it, I am quite sure that you will see a proof. – Ma Ming Jun 02 '14 at 22:32
  • Unfortunately, I'm not a big fan of seeing proof. I'd rather use a formal argument involving geometry or linear algebra if that is possible.

    I have proved the result for a rectangle $ABCD$, but fail to generalize this.

     – Hubble Jun 02 '14 at 22:36

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Got it! This follows from the fact that two distinct intersecting mirrors have a single point in common, which remains fixed. All other points rotate around it by twice the angle between the mirrors. Since you glide along axis which correspond to the sides of your quadrilateral, the composite of a pair of glides is a rotation by twice the angle between them.

Hubble
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