# Show [TA] xx [AB]=[TB] xx [BC]=[TC] xx [CA] by vectors?

Let T be the centroid of a triangle $\mathrm{△}ABC$
How can I prove that (using vectors):
$\stackrel{\to }{TA}×\stackrel{\to }{AB}=\stackrel{\to }{TB}×\stackrel{\to }{BC}=\stackrel{\to }{TC}×\stackrel{\to }{CA}$
($×$ is for vector-product).
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Alden Holder
$\begin{array}{rl}& \text{Noting that}\\ \stackrel{\to }{TA}=\phantom{\rule{thickmathspace}{0ex}}& \frac{\stackrel{\to }{A}+\stackrel{\to }{B}+\stackrel{\to }{C}}{3}\\ & \text{we get}\\ \stackrel{\to }{\left[TA\right]}×\stackrel{\to }{\left[AB\right]}=\phantom{\rule{thickmathspace}{0ex}}& \left(\stackrel{\to }{A}-\stackrel{\to }{T}\right)×\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)\\ =\phantom{\rule{thickmathspace}{0ex}}& \left(\stackrel{\to }{A}-\left(\frac{\stackrel{\to }{A}+\stackrel{\to }{B}+\stackrel{\to }{C}}{3}\right)\right)×\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)\\ =\phantom{\rule{thickmathspace}{0ex}}& \left(\frac{1}{3}\right)\left(2\stackrel{\to }{A}-\stackrel{\to }{B}-\stackrel{\to }{C}\right)×\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)\\ =\phantom{\rule{thickmathspace}{0ex}}& \left(\frac{1}{3}\right)\left(\stackrel{\to }{A}×\stackrel{\to }{B}+\stackrel{\to }{B}×\stackrel{\to }{C}+\stackrel{\to }{C}×\stackrel{\to }{A}\right)\\ & \text{Similarly, we get}\\ \stackrel{\to }{\left[TB\right]}×\stackrel{\to }{\left[BC\right]}=\phantom{\rule{thickmathspace}{0ex}}& \left(\frac{1}{3}\right)\left(\stackrel{\to }{A}×\stackrel{\to }{B}+\stackrel{\to }{B}×\stackrel{\to }{C}+\stackrel{\to }{C}×\stackrel{\to }{A}\right)\\ & \text{and}\\ \stackrel{\to }{\left[TC\right]}×\stackrel{\to }{\left[CA\right]}=\phantom{\rule{thickmathspace}{0ex}}& \left(\frac{1}{3}\right)\left(\stackrel{\to }{A}×\stackrel{\to }{B}+\stackrel{\to }{B}×\stackrel{\to }{C}+\stackrel{\to }{C}×\stackrel{\to }{A}\right)\\ & \text{which proves the claim.}\end{array}$