# Exterior of a circle in Hilbert geometry is segment connected.

Assuming the parallel axiom (P), show that if E, C are two points outside a circle $\Gamma$, then there exists a third point D such that the segments ED and DC are entirely outside $\Gamma$.

Draw a line EC if EC does not intersect $\Gamma$ then pick a point D between E and C which can be done by axiom and the segment ED and DC lie entirely outside the circle as desired.

If line EC intersects the circle. i want to say denote the points of intersection with F and G find the midpoint of segment FG call it $\alpha$ draw the perpendicular passing through $\alpha$ of the line FG. where the perpendicular intersects the circle deonte it H draw line $\alpha$H using the axioms extend the segment to a point D s.t $\alpha$ * H * D then draw segment ED and DC i claim this lay entirely outside of the cirle $\Gamma$ but im not sure... i believe my choice of the midpoint of the segment is the correct one for this construction but im not really sure it guarantees the result despite thinking it does.

i am using Hilberts axioms and other result proved from them.

Anyone have an insight why this construction works or perhaps a better method using Hilberts axioms?

This is questions 11.1B in Geometry by Hartshorne.