<s>[INST] Stereographic projection proof that is geometrical. Given tangents $VP$ and $ZP$ on circle intersecting at $P$.
Prove: $YX=XW$

Heres what I do know. Obviously segments $ON, OV, OS$ are all congruent since they are just radii. I also know that $VP$ and $ZP$ are congruent since they are tangent lines intersecting. There is also an isosceles triangle $ONV$ since two of the sides are just radii which means angles $ONV$ and $NVO$ are congruent. I also know $OVZ$ is an isosceles triangle. Based off of all the information I have figured out I am not sure how to use it to show that $YX = XW$. Thoughts? Ideas? Solutions?
 [/INST] This is a coordinate-based proof. Choose (w.l.o.g.) the following coordinates for your points:
\begin{align*}
N&=\begin{pmatrix}0\\1\end{pmatrix} &
V&=\frac{1}{a^2+1}\begin{pmatrix}2a\\a^2-1\end{pmatrix}\\
S&=\begin{pmatrix}0\\-1\end{pmatrix}&
Z&=\frac{1}{b^2+1}\begin{pmatrix}2b\\b^2-1\end{pmatrix}
\end{align*}
This is using a rational parametrization of the unit circle, i.e. using the tangens of half the angle as the parameter. Then you have the tangents
$$2ax + (a^2-1)y = a^2+1\qquad 2bx + (b^2-1)y = b^2+1$$
which intersect in
$$P=\frac1{ab+1}\begin{pmatrix}a+b\\ab-1\end{pmatrix}$$
The stereographic projections of these points are
\begin{align*}
W&=\begin{pmatrix}2a\\-1\end{pmatrix} &
Y&=\begin{pmatrix}2b\\-1\end{pmatrix} &
X&=\begin{pmatrix}a+b\\-1\end{pmatrix} &
\end{align*}
So you can see that they are equidistant, as claimed.
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