## Proof of Theorem E1
Suppose that r maps B to C. Since the map is an isometry, |OC| = |AB| (= |OX| as above)._{L}Let M be the internal bisector of <XOC. Then r maps C to X (since |OC| = |OX|)._{M}
Consider the composite r°_{L}t.Considering the effect of each component in turn, we see that s fixes O and X.By Lemma 1, s is the identity or is r, reflection in the real axis._{x}Using the fact that reflections are self-inverse, we have either t = r°_{L}r, or_{M}t = r°_{L}r°_{M}r._{x}Thus, each isometry is the composite of (at most three) reflections. |

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