Let $S:=\{a_0,..., a_k\}$ be a set of geometrically independent points in $\mathbb{R}^n$. The definition of an $n$-plane $P$ spanned by these points is the set of all points $x$ in $\mathbb{R}^n$ such that $x=\sum_{i=0}^k t_ia_i$ for some scalars $t_i$ with $\sum_{i=0}^k t_i = 1$. Munkres states in his book $\textit{Elements of Algebraic Topology}$ the following
Note that $P$ can also be described as the set of all points $x$ such that $$x=a_0+\sum_{i=1}^k t_i(a_i-a_0)$$ for $\textit{some}$ scalars $t_1,...,t_k$
I managed to show that every point written in the form of the first definition can be written in the form of the second definition. I have difficulties to show the other direction. Here is what I have found, the problem lies in Case 2.2:
Let $x=a_0+\sum_{i=1}^k t_i(a_i-a_0)=a_0+\sum_{i=1}^k t_ia_i-(\sum_{i=1}^k t_i)a_0$ for $\textit{some}$ scalars $t_1,...,t_k$. We want to find scalars $s_0,...,s_k$ that add up to $1$ such that $x=\sum_{i=0}^k s_ia_i$.
$\textbf{Case 1}$: If the $t_i$ are all zero, then chose $s_0=1$ and $s_i=0$ for all $i=1,...,k$.
From now on, suppose that at least one of the $t_i$ is different from zero.
$\textbf{Case 2.1}$: If $\sum_{i=1}^k t_i=0$, then necessarily $\sum_{i=1}^k t_ia_i$ is different from the zero vector, since any subset of $S$ is also geometrically independent (this is true, right?), and then take the contraposition. So we have $x=a_0+\sum_{i=1}^k t_ia_i=1\cdot a_0+\sum_{i=1}^k t_ia_i$. Hence, we chose $s_0=1, s_1=t_1,..., s_k=t_k$, which satisfies $\sum_{i=0}^k s_i=s_0+\underbrace{\sum_{i=1}^k t_i}_\text{$=0$ by assumption}=1$
$\textbf{Case 2.2}$: $\underline{\textbf{(Here is the "difficulty".)}}$ Similarily, if $\sum_{i=1}^k t_ia_i=0$, then $\sum_{i=1}^k t_i$ is different from zero by the same argument as above. This implies that $x=a_0-(\sum_{i=1}^k t_i)a_0=(1-\sum_{i=1}^k t_i)a_0$. So we would like to chose $s_0=1-\sum_{i=1}^k t_i$ and all the other $s_i$ equal to zero. But then the $s_i$ do not add up to 1. On the other hand, if we chose for instance $s_1=\sum_{i=1}^k t_i$, then the $s_i$ do add up to 1, but we do not get $x$, since $s_1$ is different from zero by the first line in Case 2.2. Also, there does not seem to be a contradiction when assuming Case 2.2, so what do I miss?
This is immediate:
$$a_0 + \sum _{i=1} ^k t_i (a_i - a_0) = \underbrace {\left( 1 - \sum _{i=1} ^k t_i \right)} _{s_0} a_0 + \sum _{i=1} ^k \underbrace {t_i} _{s_i} a_i $$
and
$$s_0 + s_1 + \dots + s_k = \left( 1 - \sum _{i=1} ^k t_i \right) + t_1 + \dots + t_k = 1 .$$