Homeomorphism of Interior of Convex Polygon to Open Unit Disk 
Show that the interior of a non-degenerate convex polygon in $\mathbb{R}^2$ is homeomorphic to the open unit disk in $\mathbb{R}^2$. 

My attempt: Let $P$ be the set of points in the polygon, let $c \in \text{int}(P)$. Consider arbitrary $x \in \text{int}(P)$, $x \neq c$. Let $R := \{c+t(x-c): t > 0\} \cap P$. Let $b := \sup_{y \in R} d(y, c)$. Define \begin{eqnarray*}
f(x) := \frac{d(x,c)}{b}(x-c).
\end{eqnarray*}
Finally let $f(c) = (0,0)$.
The problem: while I am pretty sure this is the correct map, I am a hard time proving continuity as I cannot see how to get explicit formulas without knowing which polygon I am mapping to the open unit disk.
P.S. I've seen questions very close to this one, but all answers stopped short of explaining how to show continuity of the map.
 A: First, your quantity $b$ is not a constant, it depends on the ray $R$, and so ostensibly $b$ depends on $x$. But it should be clear that $b=b(\theta)$ really only depends on the angle $\theta$ of the ray $R$. Let me denote $R(\theta)$ as the ray of angle $\theta$ emanating from $c$, and $p(\theta)$ as the point on $R(\theta) \cap P$ furthest from $c$, so $b(\theta) = |p(\theta)-c|$. 
The key is to prove that $b(\theta)$ is continuous at each $\theta_0$. Continuity of $f(x)$ away from $c$ then follows, because $\theta = \theta(x)$ has an obviously continuous formula near each $x \ne c$, involving the arctangent function.
By convexity, there exists a closed half-plane $H$ with boundary line $L$ containing $p(\theta_0)$ such that $P \cap H \subset L$. Also, $c \not\in H$ by non degeneracy. Let $\overline H$ be the opposite half-plane to $H$ with the same boundary line $L$. It follows that for angles $\theta'$ near $\theta_0$ we have $p(\theta') \in \overline H$. From this it follows that $b(\theta)$ is upper semicontinuous.
Consider any angle $\theta'$ near $\theta_0$ (any $\theta'$ not pointing opposite $\theta_0$ will do). By convexity, for each $q \in \overline{p(\theta_0) p(\theta')}$ the interior of the segment $\overline{cq}$ must be contained in $P$. From this it follows that $b(\theta)$ is lower semicontinuous. 
