Show $[TA] \times [AB] =[TB] \times [BC]=[TC] \times [CA]$ by vectors?

Question

Let $T$ be the centroid of a triangle $\triangle ABC$.

How can I prove that (using vectors):

$$\overrightarrow{TA} \times \overrightarrow{AB} =\overrightarrow{TB} \times \overrightarrow{BC}=\overrightarrow{TC} \times \overrightarrow{CA} $$ ($\times$ is for vector-product).

Answer 1

\begin{align*} &\text{Noting that}\\[4pt] \overrightarrow{TA}=\;&\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3} \\[6pt] &\text{we get}\\[6pt] \overrightarrow{[TA]} \times \overrightarrow{[AB]} =\;&(\overrightarrow{A}-\overrightarrow{T})\times(\overrightarrow{B}-\overrightarrow{A})\\[4pt] =\;&\left(\overrightarrow{A}- \left( \frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3} \right)\right)\times(\overrightarrow{B}-\overrightarrow{A})\\[4pt] =\;&\bigl({\small{\frac{1}{3}}}\bigr)( 2\overrightarrow{A}-\overrightarrow{B}-\overrightarrow{C}) \times(\overrightarrow{B}-\overrightarrow{A})\\[4pt] =\;&\bigl({\small{\frac{1}{3}}}\bigr)( \overrightarrow{A}{\times}\overrightarrow{B} + \overrightarrow{B}{\times}\overrightarrow{C} + \overrightarrow{C}{\times}\overrightarrow{A} ) \\[6pt] &\text{Similarly, we get}\\[6pt] \overrightarrow{[TB]} \times \overrightarrow{[BC]} =\;& \bigl({\small{\frac{1}{3}}}\bigr)( \overrightarrow{A}{\times}\overrightarrow{B} + \overrightarrow{B}{\times}\overrightarrow{C} + \overrightarrow{C}{\times}\overrightarrow{A} )\\[6pt] &\text{and}\\[6pt] \overrightarrow{[TC]} \times \overrightarrow{[CA]} =\;& \bigl({\small{\frac{1}{3}}}\bigr)( \overrightarrow{A}{\times}\overrightarrow{B} + \overrightarrow{B}{\times}\overrightarrow{C} + \overrightarrow{C}{\times}\overrightarrow{A} )\\[4pt] &\text{which proves the claim.}\\[4pt] \end{align*}