Theorem: Napoleon's Theorem

Let $\triangle ABC$ be an arbitrary triangle.
Let $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$ be equilateral triangles constructed on $AB$, $BC$ and $AC$ respectively on the exterior of $\triangle ABC$.
Let $O_1$, $O_2$ and $O_3$ be the incenters of $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$.
Then $\triangle O_1 O_2 O_3$ is an equilateral triangle.
:400px

Proof:
=== {{Lemma|Napoleon's Theorem|1}} ===
{{:Napoleon's Theorem/Lemma 1}}{{qed|lemma}}
=== {{Lemma|Napoleon's Theorem|2}} ===
{{:Napoleon's Theorem/Lemma 2}}{{qed|lemma}}
==== Vectors ====
Side $a = BC$ lies opposite vertex $A$ of $\triangle ABC$.
Let the vector $\mathbf{a}$ have magnitude $\dfrac 1 3 a$ in the direction of $\vec {BC}$.
$O_1$ is the incenter of the $\triangle BCD$ with side $a$ from $\triangle ABC$.
Side $b = CA$ lies opposite vertex $B$ of $\triangle ABC$.
Let the vector $\mathbf{b}$ have magnitude $\dfrac 1 3 b$ in the direction of $\vec {CA}$.
$O_2$ is the incenter of the $\triangle CAE$ with side $b$ from $\triangle ABC$.
Side $c = AB$ lies opposite vertex $C$ of $\triangle ABC$.
Let the vector $\mathbf{c}$ have magnitude $\dfrac 1 3 c$ in the direction of $\vec {AB}$.
$O_3$ is the incenter of the $\triangle ABF$ with side $c$ from of $\triangle ABC$.
==== Construction of Vector Paths ====
Let $\mathbf{p}$ be the vector path from $O_1$ to $O_2$ through vertex $C$.
By {{Lemma|Napoleon's Theorem|2}}:
:The first part of $\mathbf{p}$ from $O_1$ to $C$ is $\mathbf{a}' + \mathbf{a}$.
Also by {{Lemma|Napoleon's Theorem|2}}:
:The second part of $\mathbf{p}$ from $C$ to $O_2$ is $\mathbf{b} + - \mathbf{b}' '$.
By addition:
:$\mathbf{p} = \mathbf{a}' + \mathbf{a} + \mathbf{b} + - \mathbf{b}' '$
400px
Let $\mathbf{q}$ be the vector path from $O_2$ to $O_3$ through vertex $A$.
By {{Lemma|Napoleon's Theorem|2}}:
:The first part of $\mathbf{q}$ from $O_2$ to $A$ is $\mathbf{b}' + \mathbf{b}$.
By {{Lemma|Napoleon's Theorem|2}}:
:The second part of $\mathbf{q}$ from $A$ to $O_3$ is $\mathbf{c} + - \mathbf{c}' '$.
By addition:
:$\mathbf{q} = \mathbf{b}' + \mathbf{b} + \mathbf{c} + - \mathbf{c}' '$
==== Test for Equilateral Triangle ====
By the test from {{Lemma|Napoleon's Theorem|1}}, we construct $\mathbf{p} + \mathbf{q}'$: 
:If the result is $\mathbf{p} + \mathbf{q}' = \mathbf{0}$, the two vectors $\mathbf{p}$ and $\mathbf{q}$ are sides of an equilateral triangle.
:$\mathbf{q}' = \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' - \mathbf{c}' ' '$
By definition of plane rotation and that $\theta = \frac 1 6$ of a complete rotation:
:$- \mathbf{c}' ' ' = \mathbf{c}$
Substituting:
::$\mathbf{q}' = \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' + \mathbf{c}$
Then:
{{begin-eqn}}
{{eqn | l = \mathbf{p} + \mathbf{q}'
      | r = \mathbf{a}' + \mathbf{a} + \mathbf{b} - \mathbf{b}' ' + \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' + \mathbf{c}
      | c = Addition
}}
{{eqn | l =
      | r = \mathbf{a}' + \mathbf{a} + \mathbf{b} + \mathbf{b}' + \mathbf{c}' + \mathbf{c}
      | c = Cancel terms
}}
{{eqn | l =
      | r = \mathbf{a} + \mathbf{b} + \mathbf{c}
      | c = Vector Sum of Rotated Triangle is Zero
}}
{{eqn | l =
      | r = \mathbf{0}
      | c = Vector Sum of Triangle is Zero
}}
{{end-eqn}}
The result follows.
{{qed}}