Let $x_1,\cdots,x_k, x_{k+1}$ be $(k + 1)$ vectors in $\mathbb{R}^n$.
Suppose that $x_{k+1}$ is a convex combination of $x_1,\cdots,x_k$.
Show that $$\text{conv}(x_1,\cdots,x_k, x_{k+1})=\text{conv}(x_1,\cdots,x_k)$$
This result is an analogue to a well-known fact in linear algebra saying that if a subspace spanned by $k$ linearly independent vectors, then every other vectors not equal to them, can be written as the linear combination of those $k$ linearly independent vectors.
Formally:
$x_{k+1}$ is a linear combination of $x_1,\cdots,x_k$ then
$$\text{span}(x_1,\cdots,x_k, x_{k+1})=\text{span}(x_1,\cdots,x_k)$$
If $x \in \text{conv}(x_1,\ldots,x_k,x_{k+1})$, then there are $\lambda_1,\ldots,\lambda_k,\lambda_{k+1} \ge 0$ with $\sum^{k+1}_{i=1}\lambda_i = 1$ such that $$x = \lambda_1 x_1 + \cdots + \lambda_k x_k + \lambda_{k+1} x_{k+1}.$$ However, since $x_{k+1} \in \text{conv}(x_1,\ldots, x_k)$, there are $\mu_1,\ldots, \mu_k \ge 0$ with $\sum^k_{i=1} \mu_i = 1$ such that $$x_{k+1} = \mu_1 x_1 + \cdots + \mu_k x_k.$$ Then \begin{equation} \tag{1}x = (\lambda_1 + \lambda_{k+1} \mu_1) x_1 + \cdots + (\lambda_k + \lambda_{k+1}\mu_k)x_k.\end{equation} Putting $\rho_j = \lambda_j + \lambda_{k+1}\mu_j$ for $j=1,\ldots, k$, we see that $$\sum^k_{j=1} \rho_j = \sum^k_{j=1} \lambda_j + \lambda_{k+1}\sum^k_{j=1} \mu_j = (1-\lambda_{k+1}) + \lambda_{k+1}\cdot 1 = 1. $$Thus $(1)$ shows that $x \in \text{conv}(x_1,\ldots,x_k)$.