I have to prove the following:
Let $B=A+t\vec{AC}$. Let $t:=(A, B, C)=\frac{\vec{AB}}{\vec{AC}}$. Prove that $(B, A, C)=\frac{t}{t-1}$.
I've been trying by two different ways but I always obtain the same result. Here's my attempt:
We have $(A, B, C)=t$. $$(A, B, C) \implies B=A+t\vec{AC},$$
so
$$(B, A, C) \implies A=B+t\vec{BC}.$$
From here:
$A=B+t\vec{BC} \implies A=A+\vec{AB}+t\vec{BC} \implies \vec{0}=\vec{AB}+t\vec{BC} \implies \vec{BA}=t\vec{BC} \implies t=\frac{\vec{BA}}{\vec{BC}}=(B, A, C).$
But this is not what I wanted. What am I doing wrong?
Thanks.
So, if I deduce from your question correctly, $s = (B, A, C)$ is the number defined by $$ A = B + s\def\v{\overrightarrow}\v{BC} \tag 1$$ or $$ \v{BA} = s\v{BC}\tag{1'} $$ We now that $$ B = A + t\v{AC} \iff \v{AB} = t\v{AC} \tag 2 $$ Starting with (1'), we have \begin{align*} s\v{BC} &= \v{BA}\\ \iff s\v{BA} + s\v{AC} &= \v{BA}\\ \iff (1-s)\v{BA} &= s\v{AC}\\ \iff (s-1)\v{AB} &= s\v{AC}\\ \stackrel{(2)}\iff (s-1)t\v{AC} &= s\v{AC}\\ \stackrel{\v{AC} \ne 0}\iff (s-1)t &= s\\ \iff s = \frac{t}{t-1} \end{align*} So if $\v{AC} \ne 0$, we must have $s = \frac{t}{t-1}$.