Napoleon's Theorem via Two Rotations

Napoleon's theorem claims that the centers A', B', C', of the equilateral triangles A''BC, AB''C, ABC'', erected on the sides (either all inwardly or all outwardly) of a given triangle ABC form an equilateral triangle. The applet below serves to illustrate a very simple proof of this result.

Use the scroll bar at the bottom of the applet to rotate B'C' 30° clockwise about A and A'C' 30° counterclockwise about B. B'C' is mapped onto SR, with S on AC and R on AC''. This is because ∠B'AC = C'AC'' = 30 °. Similarly, A'C' maps onto PQ, with P on BC and Q on BC''.

In an equilateral triangle the distance from a vertex to the center equals κ = √3/3 of the side, meaning that

It follows that triangles ACC'' and ASR are similar as are triangles BCC'' and BPQ, implying that PQ||CC''||RS and, in addition,

So that B'C' = A'C'. Similarly, A'B' = B'C' = A'C'.

References

1. M. R. F. Smyth, MacCool's Proof of Napoleon's Theorem, Irish Math. Soc. Bulletin 59 (2007), 71-77 (available online)

Napoleon's Theorem

• Napoleon's Theorem
• A proof with complex numbers
• A second proof with complex numbers
• Napoleon's Theorem, Two Simple Proofs
• Napoleon's Theorem via Inscribed Angles
• A Generalization
• Douglas' Generalization
• Napoleon's Propeller
• Napoleon's Theorem by Plane Tessellation
• Fermat point
• Kiepert's theorem
• Lean Napoleon's Triangles
• Napoleon's Theorem by Transformation
• Napoleon's Theorem via Two Rotations
• Napoleon on Hinges
• Napoleon's Relatives
• Napoleon-Barlotti Theorem
• Some Properties of Napoleon's Configuration
• Fermat Points and Concurrent Euler Lines I
• Fermat Points and Concurrent Euler Lines II
• Escher's Theorem

Copyright © 1996-2009 Alexander Bogomolny