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?
 A: 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.
A: Let $A$ lie on $NW$ such that $XA \parallel PV,$
and $B$ lie on the extended line $NY$ such that $XB \parallel PZ,$
as in the diagram below.

Then $$\angle WAX = \angle WVP = \frac 12 \angle NOV = \angle AWX,$$
so $\Delta AXW$ is isoceles with $WX = AX.$
Also, $$\angle YBX = \angle NZP = \frac 12 \angle NOZ = \angle NYS = \angle BYX,$$ 
so $\Delta BXY$ is isoceles with $XY = BX.$
But $AX : PV = NX : NP = BX : PZ,$ and from $PV = PZ$ it then follows
that $AX = BX$ and therefore $WX = XY.$
No coordinates, no trigonometric functions, no inversions, just classical high school geometry.
A: Let
$$
\alpha = \angle SNY
\\\beta = \angle SNX
\\\gamma = \angle SNW = \angle OVN
\\\theta = \angle VOP = \angle ZOP
\\\phi= \angle NPO
$$
By considering the angle subtended by the chord VZ at N and O,
$$
\theta = \gamma - \alpha
$$
and by considering the triangle NPO,
$$
\phi = \pi - ((\pi - 2\gamma) + \theta + \beta) = \gamma + \alpha - \beta
$$
If the circle has radius $R$,
$$
OP = R\sec{\theta}
$$
Applying the sine rule to triangle NPO gives
$$
\frac{\sin{\beta}}{R\sec{\theta}} = \frac{\sin{\phi}}{R}
$$
and therefore
$$
\sin{\beta}\cos{(\gamma - \alpha)} = \sin{( \gamma + \alpha - \beta)}
$$
On expanding and simplifying this gives
$$
\tan{\beta} = \tfrac{1}{2}(\tan{\alpha} + \tan{\gamma})
$$
which is the required result, since
$$
SY = 2R\tan{\alpha}
\\SX = 2R\tan{\beta}
\\SW = 2R\tan{\gamma}
$$
A: First, consider the inversion with pole $N$ that exchanges the circle and the line $SYXW$. This moves $Z,V$ to $Y,W$. Hence, $Z,Y,W,V$ are concyclic and the triangles $NZV$ and $NWY$ are similar.
Now look at the triangle $NZV$; let $M$ be the midpoint of $ZV$. $P$ is the intersection of the tangents drawn to the circumcircle at the vertices $Z,V$. It is well-known that the line $NP$ is the symmedian in the triangle $NVZ$ that starts from the vertex $N$, so the lines $NM$ and $NX$ are symmetrical about the bisector of bisector of $\angle ZNV$.
Let $\varphi$ be the similitude that transforms the triangle $NZV$ to $NWY$. That transform can be obtained by applying a homothety with center $N$ and a reflection about the angle bisector of $\angle ZNV$. Then 
$$
\varphi(M) = \varphi(AM\cap ZV) = AP \cap YW = X.
$$
Since $M$ is the midpoint of $ZV$, the point $X=\varphi(M)$ is the midpoint of $\varphi(Z)\varphi(V)=WY$.
