Proof. Without loss of generality we assume that the set $A$ consists of exactly $d+2$ points which we number . Denote by $a_{i,j}$ the $j$ 'th component of $i$ 'th vector, and write the components in a matrix as \begin{equation*} M=\begin{bmatrix} a_{1,1}&a_{2,1}&\dots&a_{d+2,1}\\ a_{1,2}&a_{2,2}&\dots&a_{d+2,2}\\ \vdots& \vdots&\ddots & \vdots\\ a_{1,d}&a_{2,d}&\dots&a_{d+2,d}\\ 1&1&\dots&1 \end{bmatrix}. \end{equation*}Since $M$ has fewer rows than columns, there is a non-zero column vector $\mathbf{\lambda}$ such that $M \mathbf{\lambda}=0$ , which is equivalent to the existence of a solution to the system
Let $A_1$ be the set of those $a_i$ for which $\lambda_i$ is positive, and $A_2$ be the rest. Set $s$ to be the sum of positive $\lambda_i$ 's. Then by the system (1) above \begin{equation*} \frac{1}{s}\sum_{a_i\in A_1} \lambda_i a_i=\frac{1}{s}\sum_{a_i\in A_2} (-\lambda_i) a_i \end{equation*}is a point of intersection of convex hulls of $A_1$ and $A_2$ .