Proof:

We leave (i) as an exercise for the reader. To prove (ii) we use induction on $p$ setting aside the cases $p = 1, 2$ for the reader to investigate. Denoting by ${\bf b}$ the barycenter of $\sigma$, the reader may check that $\Vert{\bf b} - {\bf x}\Vert \leq p(p+1)^{-1}($diam$\;\sigma)$ for any point ${\bf x}$ of $\sigma$. Let $\tau$ be one of the simplicies appearing in the chain $B\sigma$. Then the diameter of $\tau$ equals $\Vert w - z\Vert$ where $w$ and $z$ are two vertices of $\tau$. If one of these is ${\bf b}$ then the result follows from the assertion in the previous sentence. If neither $w$ nor $z$ is ${\bf b}$ then they are both vertices of a face $\tau^{\prime}$ of $\tau$ lying on a face $\sigma^{\prime}$ of $\sigma$. But $\tau^{\prime}$ is then a constituent $(p-1)$ simplex of $B(\sigma^{\prime})$ and by induction hypothesis, the result follows (how?). $\square$