We know that a point $x$ of a convex set $K$ in a normed linear space $X$ is said to be an extreme point if for any $y,z\in K$ and for any $0<\lambda<1$, $$x=\lambda y+(1-\lambda)z,$$ then $x=y=z$.
I want to show that this is equivalent to show that for any $y,z\in K$, $$x=\frac{y+z}{2}\implies x=y=z.$$
$(\implies)$ is obvious. But how to show the reverse implication? Any suggestion is appreciated.
Let $x=\lambda y+(1-\lambda)z.$
Define $\lambda = \frac{1}{2} (\alpha + 1)$. Then $0 < \alpha <1.$ We have:
$$x = \frac{1}{2} (\alpha + 1) y + (1-\frac{1}{2} (\alpha + 1))z$$ $$x = \frac{1}{2} (\alpha + 1) y + \frac{1}{2} (1-\alpha)z$$ $$x = \frac{1}{2} y + \frac{1}{2} (\alpha y + (1-\alpha)z)$$ Define $z' = \alpha y + (1-\alpha)z.$ By convexity, $z' \in K.$ Therefore, from the assumption, it follows that $x = y = z'.$ But then this implies, $(1-\alpha) y = (1-\alpha)z,$ or $z= y.$