Napoleon's Theorem

A proof via rotations that the centroids of equilateral triangles built externally on the sides of any triangle form an equilateral triangle.

Keywords: Napoleon's theorem, equilateral triangle, centroid, rotation

Difficulty: advanced

Napoleon's Theorem

On each side of triangle $ABC$, build an equilateral triangle externally, and let $G_A$, $G_B$, $G_C$ be their centroids. Then $\triangle G_A G_B G_C$ — Napoleon's triangle — is always equilateral.

Proof via rotations

Let $R_A$, $R_B$, $R_C$ be the $120°$ rotations about $G_A$, $G_B$, $G_C$, all in the same direction. Since $G_X$ is the center of the equilateral triangle on the side opposite $X$, each $R_X$ cycles that triangle's three vertices. With the direction chosen consistently:

R_A\colon B \to C, \quad
    R_B\colon C \to A, \quad
    R_C\colon A \to B.
    

Tracking the vertex $B$ through the three rotations in turn:

B \;\xrightarrow{R_A}\; C
      \;\xrightarrow{R_B}\; A
      \;\xrightarrow{R_C}\; B.
    

So $R_C \circ R_B \circ R_A$ fixes $B$. The total rotation angle is $3 \cdot 120° = 360°$, which makes the composition a translation; a translation with a fixed point is the identity:

R_C \circ R_B \circ R_A = \mathrm{id}.
    

Rearranging gives $R_B \circ R_A = R_C^{-1}$, a $240°$ rotation about $G_C$.

Rotation-composition lemma. A rotation by $2\alpha$ about $P$ followed by a rotation by $2\beta$ about $Q$ equals a rotation by $2(\alpha + \beta)$ about some point $R$, where $\triangle PQR$ has angle $\alpha$ at $P$ and angle $\beta$ at $Q$.

Apply the lemma with $2\alpha = 2\beta = 120°$ and center $G_C$: $\triangle G_A G_B G_C$ has $60°$ angles at $G_A$ and $G_B$, so the third angle is also $60°$. Napoleon's triangle is equilateral. $\blacksquare$

The Proof

Step 1: Start with Any Triangle

Take an arbitrary triangle $ABC$.

Step 2: Equilateral Triangles Externally

On each side, build an equilateral triangle pointing outward, with apexes $P_A$, $P_B$, $P_C$ opposite to $A$, $B$, $C$.

Step 3: Napoleon's Triangle

Mark the centroids $G_A$, $G_B$, $G_C$ of the three equilateral triangles. Connecting them gives Napoleon's triangle, which is always equilateral; see the introduction notes for the proof.

Conclusion

Three $120°$ rotations about $G_A$, $G_B$, $G_C$ compose to the identity. The rotation-composition lemma then forces the angles of $\triangle G_A G_B G_C$ to each be $60°$ — equilateral, for any starting triangle $ABC$.