Angle Trisectors on Circumcircle
The problem below has been taken from a discussion on the geometry.research newsgroup.
| Subject: | | A Theorem concerning the Trisectors of a Triangle |
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| Author: | | Den Roussel |
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| Date: | | 12 Sep 1998 15:05:10 -0400 |
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While investigating the Morley triangle, I came upon this theorem which some might find interesting.
If the Angular trisectors of a triangle are produced to the Circumcircle, then the Chords of adjacent
trisectors form an Equilateral triangle.
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
Copyright © 1996-2009 Alexander Bogomolny
What is meant here is this. In  ABC, angle trisectors are drawn that form Morley's triangle A'B'C'. Extend the trisectors till their intersection with the circumcircle of ABC. AC' cuts it at A1, AB' at A2, and so on. The lines A1B2, B1C2 and C1A2 form A0B0C0, 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 + 60o. Denote the point of intersection of C1A2 and BB1 as P. I am going to show that BPC1 = C/3 + 60o, 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 + 60o. Q.E.D.
Copyright © 1996-2009 Alexander Bogomolny
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