I'm reading a proof of the following lemma
Assume that $K$ is a compact convex set in a Hausdorff locally convex space, and $K$ is metrizable with the induced topology. Then the set $\textrm{ex}(K)$ of extreme points of $K$ is a $G_\delta$ set (countable intersection of open sets).
For "extreme points", the following definition is used:
Let $X$ be a non-empty convex set. A point $x \in X$ is an extreme point of $X$ if the condition $x = tx_1 + (1-t)x_2$ for $x_1, x_2 \in X$ and $0 <t < 1$ implies that $x = x_1 = x_2$. The set of extreme points of $X$ is denoted by $\textrm{ex}(X)$.
One step of the proof says that
Suppose $x\in K\backslash\textrm{ex}(K)$. There exist two distinct points $y,z\in K$ such that $x=(y+z)/2$.
Using the negation of the condition in the definition, one has
$x=ty+(1-t)z$ for some $y,z\in K$ and $0<t<1$ with $y\not=x$ or $z\not=x$.
My question: how can one make $t=1/2$?
[Reference] Here is the whole proof for the lemma
We show that $K\setminus\textrm{ex}(K)$ is a $F_\sigma$ set (countable union of closed sets). Suppose $x\in K\setminus\textrm{ex}(K)$. There exist two distinct points $y,z\in K$ such that $x=(y+z)/2$. Let $d$ be a metric defining the topology of $K$. Define $$ F_n=\{x\in K:\textrm{there exist }\ y,z\in K\textrm{ such that }x=(y+z)/2\textrm{ and }d(x,y)\geq 1/n\}. $$ Note that $F_n$ is closed and $$ K\setminus\textrm{ex}(K)=\bigcup_{n\geq 1} F_n. $$
Let $x \in K \setminus \operatorname{ex} (K)$ and $y_1,z_1\in K$ with $x = ty_1 + (1-t)z_1$ for some $t \in (0,1)$ and $y_1 \neq x \neq z_1$ (if $x = ty_1 + (1-t)z_1$ with $0 < t < 1$, then either $y_1 = z_1 = x$ or $y_1 \neq x \neq z_1$; we assume that $y_1 \neq x$ or $z_1 \neq x$, hence neither of the two points equals $x$).
If perchance we already have $t = \frac{1}{2}$, we're done. Otherwise, by exchanging the names of $y_1$ and $z_1$ we may assume $0 < t < \frac{1}{2}$. Then let $s = 2t$ and $y_2 = sy_1 + (1-s) z_1$. By convexity of $K$ we have $y_2 \in K$, and we have
$$\frac{y_2 + z_1}{2} = \tfrac{1}{2}(2ty_1 + (1-2t)z_1) + \tfrac{1}{2}z_1 = ty_1 + \bigl(\tfrac{1}{2} - t\bigr)z_1 + \tfrac{1}{2}z_1 = ty_1 + (1-t)z_1 = x.$$
Since $s > 0$, we have $y_2 \neq z_1$.