Napoleon's Theorem/Lemma 1

Lemma for Napoleon's Theorem

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

Let the sides of $\triangle ABC$ be the vectors $\mathbf u$, $\mathbf v$, and $\mathbf w$.

Let the interior of $T$ lie to the left of the vector path.

Let $\mathbf v$ be rotated by an angle of $60 \degrees$ anticlockwise.

Let the rotated vector be $\mathbf v'$.

Then:

$\mathbf u + \mathbf v' = \mathbf 0$

where $\mathbf 0$ denotes the zero vector.

Let: $v = \norm {\mathbf v}$, $v' = \norm {\mathbf v'}$, and $u = \norm {\mathbf u}$.

$v = v'$

$v = u$

By definition of equilateral triangle the angle between $\mathbf u$ and $\mathbf v$ is $120 \degrees$.

By definition of rotation the angle between $\mathbf u$ and $\mathbf v'$ is $180 \degrees$.

That is, the direction of $\mathbf v'$ is opposite to that of $\mathbf u$.

It follows that:

$\ds \mathbf u$

$=$

$\ds -\mathbf v'$

$\ds \leadsto \ \ $

$\ds \mathbf u + \mathbf v'$

$=$

$\ds \mathbf 0$

$\blacksquare$