= Solution
An <extreme point> $p\in P$ cannot be written as a nontrivial <convex combination> of two distinct <points> of $P$: if $p=\lambda u+(1-\lambda)v$ with $u,v\in P$ and $0<\lambda<1$, then $u=v=p$.
A <vertex of a linear polyhedron> is a <feasible point> at which the <active constraints> include $n$ <linearly independent> row normals. Equivalently, some $n$ independent bounding <hyperplanes> intersect uniquely at that <point>. This definition does not require $P$ to be bounded or full-dimensional. For a <linear polyhedron>, it is also equivalent to the <point> uniquely maximizing a suitable <linear function>: sum the selected active inequalities to construct one such <linear function>.
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