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Theorem HA4
Suppose that K,L are complete hyperbolic lines.
If K,L meet in D then K,L have two axes, otherwise they have one.
proof
We have already discussed the case where they meet in D.
Otherwise, they meet in X, a point of C, or are ultraparallel.
In the first case, suppose that K = AX, L = BX.
By the existence theorem, there is a unique hyperbolic line H through X and
perpendicular to the hyperbolic line AB. Inversion in H swaps fixes X and swaps
A and B. It therefore swaps K = AX and L = BX, so is an axis.
Now suppose that inversion in a hyperbolic line H' swaps K and L. Since inversion
maps C to itself, it must fix X. It must also swap A and B. It follows that H' must
be perpendicular to the hypebolic line AB. Since it passes through X, it must be H.
In the second case, the common perpendicular theorem shows that there is
a unique hyperbolic line M perpendicular to both K and L. Suppose that M meets
K at U and L at V. Let H be the hyperbolic perpendicular bisector of the hyperbolic
segment UV. Then inversion in H swaps U and V and maps UV to itself. Since K and
L are the (unique) perpendiculars to UV through U and V, the inversion must swap
K and L. Conversely, an inversion which swaps K and L must map UV to itself, as
the unique common perpendicuar. It therefore swaps U and V, so is inversion in
the hyperbolic line H already defined.
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