Angle Trisectors on Circumcircle

The problem below has been taken from a discussion on the geometry.research newsgroup.

While investigating the Morley triangle, I came upon this theorem which some might find interesting.

It has been suggested that this result might follow from Morleys theorem. So far, I have been unable to make this connection. Any thoughts?

Den Roussel

Discussion

|Activities| |Contact| |Front page| |Contents| |Geometry|

What is meant here is this. In ΔABC, angle trisectors are drawn that form Morley's triangle A'B'C'. Extend the trisectors till their intersection with the circumcircle of ΔABC. AC' cuts it at A₁, AB' at A₂, and so on. The lines A₁B₂, B₁C₂ and C₁A₂ form ΔA₀B₀C₀, which is equilateral.

An easy proof follows from the observation that lines A₁B₂, B₁C₂ and C₁A₂ are parallel to the sides of Morley's triangle.

For example, we know that ∠BA'C' = C/3 + 60°. Denote the point of intersection of C₁A₂ and BB₁ as P. I am going to show that ∠BPC₁ = C/3 + 60°, from which C₁A₂||A'C', and similarly for other two sides.

∠BPC₁ equals the sum of angular measures of arcs BC₁ = 2·C/3 and A₂B₁ = (A + B)/3. Whence, ∠BPC₁ = 2C/3 + (A + B)/3 = C/3 + 60°. Q.E.D.

|Activities| |Contact| |Front page| |Contents| |Geometry|