Similar triangles in stereographic projection I'm reading about the spherical representation and the Riemann sphere, and the projection transformation that takes a point on the sphere to a point on the (extended) complex plane. An exercise is to show that if $Z,Z'$ are the stereographic projections onto the sphere of $z,z'$ lying on the complex plane, then the triangles $NZZ'$ and $Nzz'$ are similar ($N$ is the north pole).
I think that one way would be to calculate the stereographic projections explicitly, using the formula involving coordinates. For such a nice statement, however, I'm sure there must be a way to look at it geometrically. I've tried drawing pictures, but still can't see why the statement must be true. One observation I have is that the longer $NZ$ is, the shorter $Nz$ is. So that means the similarity should be in the order $\triangle NZZ'\sim\triangle Nz'z$. But how can I actually show the similarity?
 A: Triangles are similar if they agree in all their corner angles. And stereographic projection is conformal, i.e. it preserves angles. If you have both of these, you are pretty much done. If you don't have proven conformality yet, I'd tackle that, since it will be useful in other cases as well.
One approach might be using the fact that Möbius transformations are conformal (if you know that), and you can show that a stereographic projection onto the sphere combined with a rotation of the sphere combined with a stereographic projection back to the plane will result in a Möbius transformation. For this reason, it is enough to show conformality in a single point of the sphere, e.g. the one opposite the center of projection where things are very symmetric.
A: An approach without using a conformal argument: indeed, $\bigtriangleup NZZ' \sim \ \bigtriangleup Nz'z$. To see this, notice that side lengths of the triangles obey
$$ \frac{NZ}{Nz'} = \frac{NZ'}{Nz} $$
You can prove this using the usual distance function and the stereographic projection 
$$ z = \frac{x_1 + i x_2}{1-x_3} $$
where $z \in \mathbb{C}$ and $x_1,x_2, x_3$ are the usual coordinates in $\mathbb{R}^3$. It should only take a few lines of work if you write $x_1, x_2, x_3$ in terms of $z, \overline{z}, |z|^2$ in the ways that are normally done when introducing the stereographic projection (happy to share some more steps if this doesn't become clear). 
Aside: you correctly guessed that 
$$ \frac{NZ}{Nz} \neq \frac{NZ'}{Nz'} $$
An argument for why: if this were true $ZZ'$
would have to always be parallel to $zz'$, which is not true. For example, take $Z$ and $Z'$ in different hemispheres of the Riemann sphere. This might have been a good first guess, which when realizing that it doesn't work leads you to the method that I suggest.
A: I noticed a similar triangle image on paper that mapped to the sphere as you are describing.
I found the symmetry to be 12 points on a sphere. It's splits into 6 at the top and mirrors at the equator. You can see it demonstrated in the video.
I'm unsure I understand your question, but this is the code I used to determine the 12 points of a similar triangle when the first two points are fixed at 1 and -1 on the real line.
def similar_points(z):
    x = 4*(z-1)/abs(z-1)**2+1
    y = 4*(z+1)/abs(z+1)**2-1
    xc = x.conjugate()
    yc = y.conjugate()
    zc = z.conjugate()
    return x,y,z,xc,yc,zc,-x,-y,-z,-xc,-yc,-zc

https://youtu.be/MCjOebr14-0
