Sum of decomposition coefficients of collinear points is $1$, but here seems to be a violation. Below is a proposition in a text on analytic geometry:

Given two points $A$ and $B$, $A\ne B$, $O$ an arbitrary point, if point $C$ is collinear with $A$ and $B$, then vector $\overrightarrow{OC}$ can be decomposed by $\overrightarrow{OA}$ and $\overrightarrow{OB}$, and the sum of decomposition coefficients is $1$. That is, $\overrightarrow{OC}=a\overrightarrow{OA}+b\overrightarrow{OB}$ and $a+b=1$.

The text further states that the conclusion holds even if $O$ is collinear with $A$ and $B$ (I checked the proof and could find no problem), only the decomposition coefficients are not unique (but this seems to have nothing to do with my question).
Now, I put $O$ on line $\ell=\overline{AB}$. Take two points $C_1$ and $C_2$ on $\ell$, so we have $$\overrightarrow{OC_1}=a_1\overrightarrow{OA}+b_1\overrightarrow{OB} \qquad \qquad\overrightarrow{OC_2}=a_2\overrightarrow{OA}+b_2\overrightarrow{OB}$$ with $a_1+b_1=a_2+b_2=1$ by the proposition.
Next, let's consider $\overrightarrow{OC_1}+\overrightarrow{OC_2}$. Since $O$, $C_1$, $C_2$ are all on line $\ell$, vectors $\overrightarrow{OC_1}$ and $\overrightarrow{OC_2}$ are collinear with $A$ and $B$, so is their sum. We have $$\begin{align}
\overrightarrow{OC_1}+\overrightarrow{OC_2} &=a_1\overrightarrow{OA}+b_1\overrightarrow{OB}+a_2\overrightarrow{OA}+b_2\overrightarrow{OB} \\
&=(a_1+a_2)\overrightarrow{OA}+(b_1+b_2)\overrightarrow{OB}
\end{align}$$
However, the sum of the decomposition coefficients of $\overrightarrow{OC_1}+\overrightarrow{OC_2}$ is $a_1+a_2+b_1+b_2=1+1=2$, not $1$ any more, a violation of the proposition.

The proposition of the text should be correct, so where was I wrong in applying the proposition? How to reconcile this contradiction?

Thank you for your help.
 A: My bad. In the case where $O,A,B,C_1,C_2$ are collinear, the decomposition rule should be "there exists a decomposition where the sum of decomposition coefficients is $1$" instead of the sum always being $1$. I believe the ambiguity of the subject of "can be decomposed" is at fault here.
To illustrate this, consider the case where all these points are on the real line, so they can be represented as real directed numbers. Let $O = 0, A = 2, B = 3$. Then $\overrightarrow {OA} = 2, \overrightarrow {OB}=3$.
Take the case $C_1 = A =2$. Then a decomposition could be $\overrightarrow {OC_1} = 1\cdot\overrightarrow {OA} + 0 \cdot\overrightarrow {OB}$, with sum of decomposition coefficients $1$. Another decomposition could be $\overrightarrow {OC_1} = 0\cdot\overrightarrow {OA} + \frac23\cdot\overrightarrow {OB}$, with sum of decomposition coefficients $\frac23$. We can even decompose it as $\overrightarrow {OC_1} = -2\cdot\overrightarrow {OA} + 2\cdot\overrightarrow {OB}$, with sum of decomposition coefficients $0$. This shows that the decomposition is not unique. Most decompositions violate the "rule", but there still exists a decomposition with sum of decomposition coefficients $1$.
For example, if we have $C_2 = B = 3$, we can say:
$$\overrightarrow {OC_1} = 1\cdot\overrightarrow {OA} + 0 \cdot\overrightarrow {OB}$$
$$\overrightarrow {OC_2} = 0\cdot\overrightarrow {OA} + 1 \cdot\overrightarrow {OB}$$
$$5=\overrightarrow {OC_1}+\overrightarrow {OC_2}  = 1\cdot\overrightarrow {OA} + 1 \cdot\overrightarrow {OB} = -2\cdot \overrightarrow {OA} + 3 \cdot\overrightarrow {OB}$$
where the latter decomposition satisfies the rule but not the former.
