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.