R. J. Webster's Proof
|The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle.|
This proof appeared in Mathematics Magazine, Vol. 43, No. 4 (Sep., 1970), pp. 209-210.
Let ABC be a triangle with inradius r and circumradius R, and let the adjacent trisectors of angles A, B, C meet in A', B', C' as illustrated below. Morley's theorem states that ΔA'B'C' is equilateral. The proofs of this theorem, which are usually given, do not include calculation of the side of this triangle.
In this note we prove Morley's theorem by showing that the side of this triangle is
In any triangle ABC, we have a = 2R sin(A), etc. Let
|A'B||= 2R sin(3α) sinγ / sin(β + γ)|
|= 2R (4sinα sin(60° + α) sin(60° - α) sinγ) / sin(60° - α)|
|= 8R sinα sinγ sin(60° + α).|
Similarly, BC' = 8R sinα sinγ sin(60° + γ). Consider now triangle DEF shown above, where
Editorial Note: Professor C. N. Mills of Illinois State University at Normal as a tour de force found above expression for a side of the Morley triangle by a straightforward use of elementary Cartesian analysis! His complete proof required some twenty 8½×11 sheets of paper.
Copyright © 1996-2018 Alexander Bogomolny