A question about pyramids (polytopes) Let's use the following definition of a face:
A nonempty convex subset $F$ of a convex set $C$ is called a face of $C$ if $\alpha x + (1-\alpha) y \in F$ with $x, y \in C$ and $0 < \alpha < 1$ imply $x, y \in F$.
Let's focus on polytopes. In this context, say that a face $F$ of a $d$-polytope $P$ is a facet if F is a $(d-1)$-face. Moreover, say that a $d$-polytope $P$ is a pyramid with base $B$ if there is a $(d-1)$-polytope $B$ such that $P = \text{conv}(B \cup \{a\})$ for some point $a$.
My question is the following: Let $P$ be a pyramid with the above notation. Let $\widehat B$ be a facet of the base $B$. How can I prove that $\text{conv}(\widehat B \cup \{a\})$ is a facet of $P$?
 A: I think this is easier with the definition of faces as the intersection with supporting hyperplanes. But let's give it a shot with this definition.
Obviously, the set $\DeclareMathOperator{\conv}{conv}F = \conv(\widehat B \cup \{a\})$ is a subset of $\conv(B \cup \{a\}) = P$.
Since $P$ is a $d$-polytope, $a$ is not in the $d-1$-space affinely spanned by $B$.
That is, assuming we are in $d$-space, $a$ is not in the hyperplane $H$ containing $B$.
It follows that each point $x \in P$ (except the apex $a$) can be written as
$\alpha x' + (1 - \alpha)a$ for a unique $x' \in B$ and a unique $\alpha \in [0,1]$. Why? The line $L$ from $a$ to $x$, together with $H$,
spans the whole $d$-space, and the intersection of a transverse codimension-$(d-1)$ subspace and a codimension-$1$ subspace has codimension $d$, i.e., is a point.
Now suppose $x, y \in P$ and $\alpha \in (0,1)$ such that the point $z = \alpha x + (1 - \alpha) y$ is in $F$. Since $x, y \in P$,
we can write $x = \beta x' + (1 - \beta) a$ and $y = \gamma y' + (1-\gamma)a$ for some $x', y' \in B$ and some $\beta, \gamma \in [0,1]$.
So
\begin{align}
z &= \alpha (\beta x' + (1 - \beta) a) + (1 - \alpha)(\gamma y' + (1-\gamma)a) \\
&= \alpha \beta x' + \gamma y' - \alpha \gamma y' + (1 - \gamma - \alpha \beta + \alpha \gamma) a.
\end{align}
Let $\delta = \alpha \beta + \gamma - \alpha \gamma$. (You can check that $\delta = 0$ only if $x = y = z = a$.)
\begin{align}
z &= \delta \left( \frac{\alpha \beta}{\delta} x' + \frac{\gamma - \alpha \gamma}{\delta} y' \right) + (1 - \delta) a.
\end{align}
But $\frac{\alpha \beta}{\delta} x' + \frac{\gamma - \alpha \gamma}{\delta} y'$ is a convex combination of points in $B$, so it is a point in $B$.
Since $z \in F$, $z = \kappa z' + (1 - \kappa) a$ for some $z' \in \widehat B$.
So $\kappa = \delta$ and $z' = \frac{\alpha \beta}{\delta} x' + \frac{\gamma - \alpha \gamma}{\delta} y'$.
But since $\widehat B$ is a face, we must have $x', y' \in \widehat B$.
Therefore, $x, y \in F$.
This shows that $F$ is a face.
