[proofplan]
We compare the two displayed points by expressing the second origin $o'$ as a translate of the first origin $o$. If $u=o'-o$, then each displacement from $o'$ differs from the corresponding displacement from $o$ by subtracting $u$. The coefficient condition $\sum_i \lambda_i=1$ makes the total translation error exactly cancel.
[/proofplan]
[step:Express the change of origin as a vector in the model space]
Define the vector $u \in V$ by
\begin{align*}
u=o'-o.
\end{align*}
By the defining torsor identities of an [affine space](/page/Affine%20Space) modeled on $V$, this means
\begin{align*}
o'=o+u.
\end{align*}
For each index $i \in \{1,\ldots,m\}$, the displacement from $o'$ to $p_i$ satisfies
\begin{align*}
p_i-o'=(p_i-o)-u.
\end{align*}
Indeed, both sides are vectors in $V$, and adding the right-hand side to $o'$ gives
\begin{align*}
o'+((p_i-o)-u)=o+u+(p_i-o-u)=o+(p_i-o)=p_i.
\end{align*}
By uniqueness of the vector carrying $o'$ to $p_i$, the identity follows.
[/step]
[step:Compute the weighted displacement from the new origin]
Define the vectors $s,s' \in V$ by
\begin{align*}
s=\sum_{i=1}^m \lambda_i(p_i-o)
\end{align*}
and
\begin{align*}
s'=\sum_{i=1}^m \lambda_i(p_i-o').
\end{align*}
Using the identity from the previous step and linearity of finite sums in the [vector space](/page/Vector%20Space) $V$, we obtain
\begin{align*}
s'=\sum_{i=1}^m \lambda_i((p_i-o)-u).
\end{align*}
Therefore
\begin{align*}
s'=\sum_{i=1}^m \lambda_i(p_i-o)-\left(\sum_{i=1}^m \lambda_i\right)u.
\end{align*}
Since $\sum_{i=1}^m \lambda_i=1$, this becomes
\begin{align*}
s'=s-u.
\end{align*}
[guided]
We now track exactly how the vector part of the affine combination changes when the origin changes. Define
\begin{align*}
s=\sum_{i=1}^m \lambda_i(p_i-o)
\end{align*}
and
\begin{align*}
s'=\sum_{i=1}^m \lambda_i(p_i-o').
\end{align*}
Both $s$ and $s'$ are vectors in $V$, because each displacement $p_i-o$ and $p_i-o'$ lies in $V$, and $V$ is closed under scalar multiplication and finite sums.
From the previous step, for every $i \in \{1,\ldots,m\}$,
\begin{align*}
p_i-o'=(p_i-o)-u.
\end{align*}
Substituting this identity into the definition of $s'$ gives
\begin{align*}
s'=\sum_{i=1}^m \lambda_i((p_i-o)-u).
\end{align*}
Now we use distributivity and additivity in the vector space $V$: scalar multiplication distributes over vector subtraction, and finite sums may be regrouped. Hence
\begin{align*}
s'=\sum_{i=1}^m \lambda_i(p_i-o)-\sum_{i=1}^m \lambda_i u.
\end{align*}
Since $u$ is independent of $i$, the second sum is
\begin{align*}
\sum_{i=1}^m \lambda_i u=\left(\sum_{i=1}^m \lambda_i\right)u.
\end{align*}
The coefficient hypothesis gives $\sum_{i=1}^m \lambda_i=1$, so
\begin{align*}
s'=s-u.
\end{align*}
This is the cancellation mechanism: changing the origin from $o$ to $o'=o+u$ subtracts exactly $u$ from the weighted displacement because the weights sum to one.
[/guided]
[/step]
[step:Translate back to the affine space and cancel the origin shift]
Using $o'=o+u$ and $s'=s-u$, we compute in the affine action:
\begin{align*}
o'+s'=(o+u)+(s-u).
\end{align*}
By associativity of the affine action of $V$ on $A$,
\begin{align*}
(o+u)+(s-u)=o+(u+s-u).
\end{align*}
The vector-space cancellation in $V$ gives
\begin{align*}
u+s-u=s.
\end{align*}
Therefore
\begin{align*}
o'+s'=o+s.
\end{align*}
Substituting the definitions of $s$ and $s'$ gives
\begin{align*}
o'+\sum_{i=1}^m \lambda_i(p_i-o')=o+\sum_{i=1}^m \lambda_i(p_i-o).
\end{align*}
This is the desired equality.
[/step]
