The Fundamental Theorem of Similarity Geometry If L = (A,B) and L' = (A',B') are lists of distinct points of E, then there is a unique element of S+(2) which maps L to L'.

Proof Let A, B, A', B' have complex coordinates α, β, α', β' respectively. An element s of S+(2) has the form s(z) = γz+δ, with γ ≠ 0. This maps L to L' if and only if γα + β = α', and γβ + δ = β'. As A, B are distinct α ≠ β, and as A', B' are distinct, α' ≠ β'. As (α - β) ≠ 0, the equations have the unique solution γ = (α' - β')/(α - β) and δ = (αβ' - α'β)/(α-β). As α' ≠ β', γ ≠ 0, so s(z) = γz + δ is in S+(2). Thus we have a unique element of S+(2) mapping L to L'. This is a purely algebraic proof, and gives no idea how the map can be constructed. We can show that such a map exists in a more geometrical way: We begin with a dilation d about A mapping B to C, where |AC| = |A'B'|. We then apply a translation t mapping A to A'. Suppose that t(C) = C'. Since the translation preserves distance, |A'C'| = |AC|, and hence |A'C'| = |A'B'|. Finally, we apply a rotation r about A' to map C' onto B'. Then s = rotod maps L to L'. Each of our maps is direct, so s is in S+(2). Unfortunately, the uniqueness of s is not at all clear!

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