#Prove 2 triangles have the same centroid (centre of gravity)

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tribal remnant
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Here is what the exercise wants from us:

Let $t$ be a positive real number. On the sides $AB$, $AC$, $BC$ of triangle $ABC$, consider points $C_t$, $B_t$, and $A_t$, respectively, such that $\frac{AC_t}{C_tB} = \frac{BA_t}{A_tC} = \frac{CB_t}{B_tA} = t$.

\begin{enumerate}[a)]
\item Prove that for any $t$, triangles $ABC$ and $A_tB_tC_t$ have the same centroid.
\end{enumerate}

\

The problem

I need to prove that they have the same centroid. I was thinking of a vectorial approach but could not come up with anything.

The solution for this problem begins like so:
From the hypothesis, we immediately deduce that $\overrightarrow{CA_t} = \frac{1}{t+1} \overrightarrow{CB}$ and $\overrightarrow{BC_t} = \frac{1}{t+1} \overrightarrow{BA}$ respectively, and $\overrightarrow{AB_t} = \frac{1}{t+1} \overrightarrow{AC}$.

But how did they deduce that?

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thin abyssBOT
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pale vortexBOT
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Frieren
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chrome tulip
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Well, if a a vector AB has a point C on it such that |AC|/|CB| = k, then:
AC = (k/(k + 1))AB
CB = (1/(k + 1))AB

tribal remnant
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oh word? is that like a property

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i had no clue

chrome tulip
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It basically amounts to a system:
x/y = a
x + y = b

tribal remnant
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Thanks a lot @chrome tulip I managed to prove it!

unborn zealotBOT
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@tribal remnant has given 1 rep to @chrome tulip

chrome tulip
tribal remnant
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