Napoleon

Napoleon triangle of a triangle

Consider a triangle ABC and the equilaterals erected on its sides outwardly. The centers of these equilaterals form another equilateral triangle A'B'C'. This is the Napoleon triangle of ABC.

Triangles AA*C, A*BC are equal and triangle A'CB' is similar to the former two by a similarity with fixed ratio. This leads to a trivial proof. Notice that ABC and A'B'C' are perspective, the intersection point of the lines joining opposite vertices of them lying on the Kiepert hyperbola of ABC.
Notice that AA*, BB*, CC* are the Fermat lines of a triangle (with angles less than 120 degrees). The subject is related to the Fermat point(s) of a triangle.
Notice also that the Napoleon triangle persists even when the triangle is degenerate (its vertices are collinear).

The three images above show the triangles formed by the centers of equilaterals erected on the sides of a basic triangle (ABC) but with orientations different from the previous. The first has two outside and one inside. The second has two inside and one outside and the third has all three inside (inside meaning: the third vertices of the two triangles, ABC and equilateral lying on the same side of their common side). In this third case we have an analogous inner Napoleon equilateral triangle. Supply the arguments to prove it.

Napoleon triangle of a triangle

See Also