In the Extended Euclidean Plane, H, let A=(0,0), B=(5,0), and C=(4,0). Show that there exists a point D=(x,0) for some real number x, such that Rx(A,B;C,D)=pi.
I have the formulas for these things. However, I am having trouble applying them correctly.
I have that epsilon= |[A,B] intersect {C,D}| and Rx= (-1)^Epsilon (dE(A,C)/dE (A,D) ) (dE(B,D)/dE (B,C) ) Where dE= Euclidean distance function.
Is dE just the distance between the points?
Any help is appreciated.
Yes, $d_E$ here means the usual (Euclidean) distance.
First note that if nonzero vectors $u, v$ are parallel ($v=\lambda u$) then we can divide them: $\frac vu:=\lambda$ which is a signed version of the more generally writable $\frac{|v|}{|u|}$.
To clearly see what happens here, you should prove that the cross ratio as defined in the post, coincides with the following $$\frac{\vec{AC}}{\vec{CB}}\, /\, \frac{\vec{AD}}{\vec{DB}}$$ based on the above notation of parallel vectors.
For that, observe that $\epsilon$ is odd iff exactly one of $C,D$ lies between $A$ and $B$.