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

On each side of triangle $A B C$ , build an equilateral triangle externally, and let $G_{A}$ , $G_{B}$ , $G_{C}$ be their centroids. Then $△ G_{A} G_{B} G_{C}$ — Napoleon's triangle — is always equilateral.

Strategy: compose the three $120°$ rotations about $G_{A}$ , $G_{B}$ , $G_{C}$ . They cancel to the identity, and a rotation-composition lemma then forces each angle of Napoleon's triangle to be $60°$ .

Step 1: Start with Any Triangle

Take an arbitrary triangle $A B C$ .

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.

Three $120°$ rotations about $G_{A}$ , $G_{B}$ , $G_{C}$ compose to the identity. The rotation-composition lemma then forces the angles of $△ G_{A} G_{B} G_{C}$ to each be $60°$ — equilateral, for any starting triangle $A B C$ . ∎

Notes

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} : B \to C, R_{B} : C \to A, R_{C} : A \to B .

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

B R_{A} C R_{B} A 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} = id .

Rearranging gives $R_{B} \circ R_{A} = R_{C}^{- 1}$ , a $240°$ rotation about $G_{C}$ .

Rotation-composition lemma. A rotation by $2 α$ about $P$ followed by a rotation by $2 β$ about $Q$ equals a rotation by $2 (α + β)$ about some point $R$ , where $△ P QR$ has angle $α$ at $P$ and angle $β$ at $Q$ .

Apply the lemma with $2 α = 2 β = 120°$ and center $G_{C}$ : $△ G_{A} G_{B} G_{C}$ has $60°$ angles at $G_{A}$ and $G_{B}$ , so the third angle is also $60°$ . Napoleon's triangle is equilateral. $■$