$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$ For any vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, if $$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$$ where $\mid{\mathbf{a}\times\mathbf{b}}\mid\neq0$
Show that $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, i.e. a triangle
What I have done so far:
Subtract each term from another, change the sign of cross product when we switch elements s.t. $$\mathbf{a}\times\mathbf{b}+\mathbf{c}\times\mathbf{b}=0, \mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}=0,\mathbf{b}\times\mathbf{c}+\mathbf{a}\times\mathbf{c}=0$$
Which gives $$\mathbf{a}\times\mathbf{(b+c)}=\mathbf{b}\times\mathbf{(a+c)}=\mathbf{c}\times\mathbf{(a+b)}=0$$
Any hints for what to do next? And I think there should be a geometric proof somewhere as well. 

An alternative argument: as $a\times b$ is nonzero, $a$ and $b$ are
nonzero and non-parallel. Also as $(a+b+c)\times a=0$ then $a+b+c$
is a multiple of $a$. Similarly $a+b+c$ is a multiple of $b$. A vector
can only be a multiple of $a$ and of $b$ if it is zero.
