the distances theorem

the hybrid distances theorem
Suppose that A and B are points in the disk, and O is the centre of the disk.
Then |AB| = s(AB)/c(OA)c(OB).

proof
First, note that, if z is the complex number representing A, then d(O,A) = 2arctanh(|z|).
Hence |z| = t(OA). If w represents B, then, similarly, |w| = t(OB).
Also, from the definition of d, we get t(AB) = |z-w|/|w*z-1|.

Now observe that tanh²(x)/(1-tanh²(x)) = sinh²(x)/(cosh²(x)-sinh²(x)) = sinh²(x).
Also, in a similar way 1/(1-tanh²(x)) = cosh²(x).

Thus s²(AB) = |z-w|²/(|z-w|²-|w*z-1|²). We now expand the denominator :
|z-w|²-|w*z-1|² = (z-w)(z*-w*)-(w*z-1)(wz*-1) = (1-|z|²)(1-|w|²).
We see that |z-w|² = s²(AB)(1-|z|²)(1-|w|²). Using the formulae for |z| and |w|, and
the hyperbolic identities, we have, on taking (positive) roots, |z-w| = s(AB)/c(OA)c(OB).
But |AB| = |z-w|, so we have the result.