Interior point equivalence I try to proof the following statements which I have problems with

Let $C\subset R^n$ be a convex set and $x \in C$ an interior point of the convex set $C$.
  
  Prove: $x\in \text{int}(C) \iff$ for all $z\in C$ exists a point $y \in C$ and $\alpha \in (0,1)$ such that $x=(1-\alpha)z+y\alpha$

I would like to show $\impliedby$ first.
Since  $C$ is convex, the convex combination $x=(1-\alpha)z+y\alpha \in C$.
How do I show that $x\in \text{int}(C)$.
Would appreciate any help for either $\implies$ or the other direction.
 A: Let $C$ be an open segment in the plane $\mathbb{R}^2$, for example, $C=(0,1)\times\{0\}$. Then $C$ is convex and every point $x\in C$ has the property that for all $z\in C$ there exist $y\in C$ and $\alpha\in (0,1)$ such that $x=(1-\alpha)z+\alpha y$. But $\operatorname{int}(C)=\emptyset$, so the implication $\Longleftarrow$ does not hold.
To prove $\Longrightarrow$, let us take an arbitrary convex set $C\subseteq\mathbb{R}^n$ and points $x\in\operatorname{int}(C)$, $z\in C$. If $z=x$ then we can take $y=x$ and arbitrary $\alpha\in (0,1)$. So we may assume that $z\neq x$. Let us consider an open half-line $H=\{(1-\beta)z+\beta x:\beta\in(1,\infty)\}$. Then $x\in\operatorname{cl}(H)$, hence there exists $y\in\operatorname{int}(C)\cap H$. If $y=(1-\beta)z+\beta x$ for some $\beta\in(1,\infty)$ then for $\alpha=1/\beta$ we have $x=(1-\alpha)z+\alpha y$, $\alpha\in(0,1)$.
Edit: As it was pointed out by Stephen Pietromonaco, it has a sense to assume that $C$ has nonempty interior. In that case the implication $\Longleftarrow$ is valid.
Let us prove it.
Assume that $x\in C$ has the property that for every $z\in C$ there exist $y\in C$ and $\alpha\in (0,1)$ such that $x=(1-\alpha)z+\alpha y$. We have to show that $x\in\operatorname{int}(C)$, provided that $\operatorname{int}(C)\neq\emptyset$. Let $B$ be an arbitrary open ball contained in $C$. We may assume that $x\notin B$, otherwise we are done. By a translation of the coordinate system we may also ensure that $x=0$. Let us pick linearly independent vectors $z_1,\dots,z_n\in B$. By our assumption on $x$, there are $y_1,\dots,y_n\in C$ and $\alpha_1,\dots,\alpha_n\in (0,1)$ such that for all $i\in\{1,\dots,n\}$, $x=(1-\alpha_i)z_i-\alpha_i y_i$. Since $C$ is convex, we can find linearly independent vectors $v_1,\dots,v_n\in C$ such that $-v_1,\dots,-v_n\in C$; $v_i$ is a suitable multiple of $z_i$.
Let us prove that $0$ is an interior point of the polytope $P$ defined as a convex hull of the set $\{v_1,\dots,v_n,-v_1,\dots,-v_n\}$. We prove that $0$ cannot be expressed as a convex combination of $v_1,\dots,v_n,-v_1,\dots,-v_n$ with some coefficients equal to zero. Assume that $0=\beta_1 v_1+\dots+\beta_n v_n-(\gamma_1 v_1+\dots+\gamma_n v_n)$, where $\beta_i,\gamma_i\in [0,1]$ and $\beta_1+\dots+\beta_n+\gamma_1+\dots+\gamma_n=1$. Since $v_1,\dots,v_n$ are linearly independent, we must have $\beta_i-\gamma_i=0$ for every $i$. But this implies that $\beta_i,\gamma_i\in(0,1)$ for every $i$ (this is immediate if $n>1$; for $n=1$ we use the fact that $v_1\neq 0$).
So $x=0$ must be an interior point of $P$, and hence also of $C$.
The given proof is probably not the simplest one. I will appreciate if someone offers a more straightforward argument. Also, I am using an unproven statement that the algebraically defined interior of an $n$-dimensional polytope in $\mathbb{R}^n$ is topologically open, which is possibly only a reformulation of the problem, not its solution.
