Some optimal solution is an extreme point

Let $A$ be a real $m \times n$ matrix, $b \in \mathbb{R}^m$, $c \in \mathbb{R}^n$, and write $P = \{x \mid Ax \ge b\}$. Assume: (i) $P$ has at least one extreme point (in Mathlib's sense: a point of $P$ not expressible as a strict convex combination $t x_1 + (1-t)x_2$, $0<t<1$, of two points of $P$ other than itself), and (ii) there exists an optimal point, i.e. some $x \in P$ with $c \cdot x \le c \cdot y$ for all $y \in P$ (minimization). The conclusion: there exists a point $x$ that is simultaneously an extreme point of $P$ and optimal ($x \in P$ and $c \cdot x \le c \cdot y$ for every $y \in P$). This transfers attainment to an extreme point; it does not assert uniqueness.