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.