## 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 | |
---|---|---|

Author: | Den Roussel | |

Date: | 12 Sep 1998 15:05:10 -0400 |

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

Buy this applet What if applet does not run? |

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

Copyright © 1996-2018 Alexander BogomolnyBuy this applet What if applet does not run? |

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 A_{1}, AB' at A_{2}, and so on. The lines A_{1}B_{2}, B_{1}C_{2} and C_{1}A_{2} form ΔA_{0}B_{0}C_{0}, which is equilateral.

An easy proof follows from the observation that lines A_{1}B_{2}, B_{1}C_{2} and C_{1}A_{2} are parallel to the sides of Morley's triangle.

For example, we know that _{1}A_{2} and BB_{1} as P. I am going to show that _{1} = C/3 + 60°,_{1}A_{2}||A'C', and similarly for other two sides.

∠BPC_{1} equals the sum of angular measures of arcs _{1} = 2·C/3_{2}B_{1} = (A + B)/3._{1} = 2C/3 + (A + B)/3 = C/3 + 60°.

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

Copyright © 1996-2018 Alexander Bogomolny63805846 |