$ \ \vec a × \vec b = \vec b × \vec c = \vec c × \vec a $ implies existance of a triangle Let$\ \vec a, \vec b, \vec c \ $  be noncollinear vectors. Show that the necesssary and sufficient condition for the existence of a triangle $ABC$ with the properties
$\vec {BC}=\vec a ,\vec {CA}= \vec b , \vec {AB}=\vec c \ \ $is$ \ \vec a × \vec b =
\vec b ×
\vec c =
\vec
c ×
\vec 
a $.
I got the left to right implication by multiplying the relation $\vec a+\vec b + \vec c = \vec 0$ with $\vec b $ and $\vec c$, but for the right to left implication I get that $(\vec a + \vec b + \vec c)\times \vec b = \vec 0$ and I don't know how to prove that $\vec a + \vec b + \vec c=\vec 0$ from here.
 A: I will only prove the necessary part.
Let $\vec{s} = \vec{a}\times\vec{b} = \vec{b}\times\vec{c} = \vec{c}\times\vec{a}$ and $\vec{t} = \vec{a} + \vec{b} + \vec{c}$.
We are given $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, so $\vec{b}-\vec{a} \not\parallel \vec{c} - \vec{a}$. This leads to
$$3\vec{s} = \vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}
= (\vec{b}-\vec{a})\times(\vec{c}-\vec{a}) \ne \vec{0}$$
Notice $\vec{a}\cdot \vec{s} = \vec{a} \cdot (\vec{a}\times \vec{b}) = 0$ and similarly, $\vec{b} \cdot \vec{s} = \vec{c} \cdot \vec{s} = 0$. We have
$$\vec{t}\cdot \vec{s} = (\vec{a} + \vec{b} + \vec{c})\cdot \vec{s} = 0$$
Recall the Jacobi identity for vector triple product,
$$\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c}\times\vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}$$
We have
$$\vec{t} \times \vec{s} = \vec{a} \times \vec{s} + \vec{b} \times \vec{s} + \vec{c} \times \vec{s} = \vec{0}$$
This leads to
$$|\vec{t}|^2||\vec{s}|^2 = (\vec{t}\cdot \vec{s})^2 + | \vec{t} \times \vec{s}|^2 = 0$$
Since $\vec{s} \ne \vec{0}$, $|\vec{s}| \ne 0$ and this forces $|\vec{t}| = 0$.
As a result, $\vec{a} + \vec{b} + \vec{c} = \vec{t} = \vec{0}$.
