# Proof

• Let $A,B,C,D$ be points in of an $n$-dimensional number space $\mathbb R^n$ with $A\neq B$ and $C\neq D,$ and the origin $O\in\mathbb R^n.$
• "$\Rightarrow$": Assume, $L_2:=\{Q\in \mathbb R^n:\; \overrightarrow{OQ}=\overrightarrow{OC}+t\cdot \overrightarrow{CD}\}$ is a straight line, which equals the straight line $L_1:=\{P\in \mathbb R^n:\; \overrightarrow{OP}=\overrightarrow{OA}+s\cdot \overrightarrow{AB}\}.$
• Let $C\in L_2.$
• Since $L_2=L_1$, we have trivially then $C\in L_1.$
• This means that $\overrightarrow{OC}=\overrightarrow{OA}+s_0\cdot \overrightarrow{AB}$ for some $s_0\in\mathbb R.$
• Moreover, there is an $s_1\in\mathbb R$ with $\overrightarrow{OC}+\overrightarrow{CD}=\overrightarrow{OA}+s_1\cdot \overrightarrow{AB}.$
• Subtracting the first equation from the second gives us $\overrightarrow{CD}=(s_1-s_0)\cdot \overrightarrow{AB}.$
• Set $c:=(s_1-s_0).$ We have $c\neq 0$ (otherwise we would have $\overrightarrow{OC}=\overrightarrow{OC}+\overrightarrow{CD},$ which would mean $\overrightarrow {CD}=\overrightarrow{OO},$ or $C=D,$ but this case has been excluded.
• "$\Leftarrow$": Now, assume that $C\in L_1$ and $\overrightarrow{CD}=c\cdot \overrightarrow{AB}$ with $0\neq c\in \mathbb R.$
• Since $C\in L_1,$ then $\overrightarrow{OC}=\overrightarrow{OA}+s_0\cdot \overrightarrow{AB}$ for some $s_0\in\mathbb R.$
• By assumption, $c\cdot \overrightarrow{AB}=\overrightarrow{CD}$ for a $c\in \mathbb R$ with $c\neq 0.$
• Any point $Q\in L_2$ with $\overrightarrow{OQ}=\overrightarrow{OC}+t\cdot \overrightarrow{CD}$ for $t\in\mathbb R$ can be, therefore, written as $\overrightarrow{OQ}=\overrightarrow{OA}+s_0 \overrightarrow{AB}+tc\cdot \overrightarrow{AB}=\overrightarrow{OA}+(s_0+tc)\cdot \overrightarrow{AB}.$
• Since $c\neq 0$, the real number $s_0+tc$ runs through all real numbers $\mathbb R$ as $t$ does.
• Therefore, $Q$ runs through all points of the straight line $L_1,$ as $t$ runs through all real numbers $\mathbb R.$
• Therefore, $L_2=L_1$.

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