Problem
GEO-B3-M05-P020 Generalized Napoleon
On the sides of triangle \(ABC\), external isosceles triangles with outer vertices \(A_1,B_1,C_1\) are constructed on \(BC,CA,AB\). The apex angles at \(A_1,B_1,C_1\) are \(2\alpha,2\beta,2\gamma\), where \(\alpha+\beta+\gamma=180^\circ\). Prove that the angles of triangle \(A_1B_1C_1\) are \(\alpha,\beta,\gamma\).
C. Hint 1. Construct an auxiliary triangle congruent to one of the built triangles.
D. Hint 2. Reduce the angle \(A_1B_1C_1\) to an angle in the original hexagon.
Consider hexagon \(A B_1 C A_1 B C_1\). The sum of angles at \(A_1,B_1,C_1\) is \(2(\alpha+\beta+\gamma)=360^\circ\), so the sum of the remaining three angles is also \(360^\circ\).
Construct an auxiliary triangle congruent to one of the external isosceles triangles so that one side lies on a neighboring side of the hexagon. Then two pairs of triangles become congruent by two sides and the included angle. It follows that twice one angle of triangle \(A_1B_1C_1\) equals the corresponding angle of the hexagon, namely \(2\alpha\). Hence that angle is \(\alpha\). Cyclically, the other angles are \(\beta\) and \(\gamma\).
This is a stronger Napoleon version; for \(\alpha=\beta=\gamma=60^\circ\), the result is an equilateral triangle.