Stolz region - stuck with the proof Given two points $z_1,z_2$ such that $ \lvert z_i\rvert<1$, show that for every point $z\ne 1$ in the closed triangle with vertices $z_1,z_2,1$ following holds: $$ \frac{\lvert 1-z\rvert}{1-\lvert z\rvert}\le K,$$ where $K$ is a constant that depends only on $z_1, z_2.$
Determine the smallest value of $K$ for $z_1= \frac{1+i}{2}, z_2=\frac{1-i}{2}$.
What I tried, it's to write $z=re^{i\theta}$, then $r<1$, I'll prove the result but for  $\left(\dfrac{\lvert 1-z\rvert}{1-\lvert z\rvert}\right)^2$ $= \dfrac{1-2r\cos\theta+r^2}{1-2r+r^2}$, and $\theta$ is bounded by the angles of $z_i$ but I can't see, what I can do now.
 A: Hint: It's better to write $z = 1 - r e^{i\theta}$.  Then $|1 - z| = r$, while $|z|$ can be obtained using the Law of Cosines.
EDIT.  Well, I might as well summarize the whole thing now.  Use polar coordinates centred on $1$ rather than $0$, so $z = 1 - r e^{i\theta}$.  The two points $z_1$ and $z_2$, and therefore also the triangle, are contained inside some sector $-\pi/2 < -\theta_1 \le \theta \le \theta_1 < \pi/2$, and also inside some smaller circle $r = 2 p \cos \theta$ of radius $p < 1$ with centre on the real axis and passing through $1$, as in the picture below:

For all $z\ne 1$ in the shaded region, and in particular for points in the triangle,
$$
\frac{|1-z|}{1-|z|} = \frac{r}{1-\sqrt{1 + r^2 - 2 r \cos \theta}} = \frac{1}{2  \cos \theta - r} \left(1 + \sqrt{1+r^2 - 2 r \cos \theta}\right)
$$
Now $2 \cos \theta - r = (2 - 2p) \cos \theta + 2 p \cos \theta - r \ge (2 - 2 p) \cos \theta_1 > 0$, while $1 + \sqrt{1 + r^2 - 2 r \cos \theta} = 1 + |z| \le 2$, so we get
$$ \frac{|1-z|}{1-|z|} \le \frac{1}{(1-p) \cos \theta_1}$$
