Napoleon's Theorem/Lemma 2

Lemma

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

Let $\mathbf{v}$ be a vector in $\CC$ with magnitude $\dfrac 1 3 \norm {AB}$ and direction $\vec {BA}$.

Let a unit rotation be anticlockwise by $60^{\circ}$, and denote vector $\mathbf{v}$ after this rotation as $\mathbf{v}'$.

Then the vector path from $B$ to the incenter $O$ of $T$ is $\mathbf{v} + -\mathbf{v}' '$ and the vector path from the incenter $O$ to $A$ is $\mathbf{v}' + \mathbf{v}$.

Proof

The angle between $\mathbf{v}$ and $\mathbf{v}'$ is $60^{\circ}$.

If $\mathbf{v}$ is rotated again:

the angle between $\mathbf{v}$ and $\mathbf{v}' '$ is $120^{\circ}$.

By the definition of vector, for an arbitrary vector $\mathbf{w}$:

$\mathbf{w} + (-\mathbf{w}) = \mathbf{0}$, where $\mathbf{0}$ is the zero vector.

This accounts for all of the vectors in the figure.

By these definitions:

$-\mathbf{v} = \mathbf{v}$

We define vector paths from the vertices of an equilateral triangle to its incenter.

By Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral:

the incenter of $T$ is also the centroid.

By Medians of Triangle Meet at Centroid:

the centroid divides the altitude in two parts: the part nearest the base is $\dfrac 1 3$ of the whole.

By construction:

it is evident that the black dotted point in the figure is the centroid of $T$.

To see this, note that:

the point lies on the intersection of medians of the triangle

also:

the distance from the base to the point is $1/3$ the total height.

Also by construction:

the correctness of the paths is evident.

$\mathbf{v} + (- \mathbf{v}' ')$ is a vector path from $B$ to $O$.

$\mathbf{v}' + \mathbf{v}$ is a vector path from $O$ to $A$.

$\blacksquare$