Vector Sum of Rotated Triangle is Zero

Theorem

Let $T = \triangle ABC$ be embedded in the Cartesian plane $\CC$.

Let the sides of $T$ be directed line segments, that is, vectors.

Let $\CC$ be rotated anticlockwise about the origin by an angle $\theta$.

Let the triangle $T$, as defined in the rotated coordinates, be T'.

Let $\mathbf{0}$ be the zero vector.

Let $+$ be the vector sum.

Then:

$T' = \mathbf{AB}' + \mathbf{BC}' + \mathbf{CA}' = \mathbf{0}$

Proof

Let $\mathbf{AB}$ have the same magnitude as $AB$ and direction from $A$ to $B$.

Let similar definitions hold for $\mathbf{BC}$ and $\mathbf{CA}$.

By definition of vectors:

$\mathbf{AB} = \paren { B_x - A_x, B_y - A_y }$

Let $\mathbf{AB}$ in the rotated coordinates be designated $\mathbf{AB}'$.

By definition of vectors:

$\mathbf{AB}' = \paren { B_x' - A_x', B_y' - A_y' }$

By Equations defining Plane Rotation the $x$-component of $\mathbf{AB}'$ is:

$\paren { B_x \cos \theta - B_y \sin \theta - A_x \cos \theta + A_y \sin \theta }$

By Vector Magnitude is Invariant Under Rotation the above expression is equal to:

$\paren { B_x - A_x }$

and the y-component of $\mathbf{AB}'$ is:

$\paren { B_x \sin \theta + B_y \cos \theta - A_x \sin \theta - A_y \cos \theta }$

By Vector Magnitude is Invariant Under Rotation the above expression is equal to:

$\paren { B_y - A_y }$

That is, the sides of $T$ are equal to the sides of $T'$:

$\norm {AB} = \norm {AB'}$

$\norm {BC} = \norm {BC'}$

$\norm {CA} = \norm {CA'}$

By Triangle Side-Side-Side Congruence:

$T \cong T'$

$T'$ consists of the vectors $\mathbf{AB}' + \mathbf{BC}' + \mathbf{CA}'$.

By definition, the terminal point of $\mathbf{CA}$ is the initial point of $\mathbf{AB}$.

It follows that: $T = \mathbf{AB} + \mathbf{BC} + \mathbf{CA} = \mathbf{0}$

By definition, the terminal point of $\mathbf{CA}'$ is the initial point of $\mathbf{AB}'$.

It follows that: $T' = \mathbf{AB}' + \mathbf{BC}' + \mathbf{CA}' = \mathbf{0}$

$\blacksquare$