### Trisect?

Another idea for trisecting an angle. Take an arbritrary angle *AOB* and place it in a circle with center at *O.*
Extend lines, bisect angles, connect points, etc., to get this figure.

Now extend *CK* to meet the circle at *N,* and extend *DL* to meet the circle at *N,* and draw some more lines.

**Claim**: the angle *AOB* is trisected by *OM* and *ON.* That is, the three angles *AON, NOM,* and *MOB* are equal. They certainly appear to be equal, even for angles near 180 degrees.

The proposed proof starts with the claim that angle *NDA* equals angle *MOF.* The rest of the proof follows from that statement. By Euclid's
III.20, "In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base," angle *NDA* is half of angle *NOA.* By symmetry, you can see that angle *MOF* is half of angle *MON.* Therefore, *NOA* = *MON,* and, of course, by symmetry the two angles *NOA* and *MOB* are equal, so angle *AOB* is trisected.

Note that two sides of the angles *NDA* and *MOF* are parallel, namely, side *DA* is parallel to side *OF.* Thus, the claim that the two angles are equal is logically equivalent to the statement that the other two sides, namely, side *DN* and side *OM,* are parallel. They look like they're parallel in the figure. You can see that they're not, though, when the original angle *AOB* is very close to 180 degrees.

Thus, this is not an angle trisection. It's close enough, though, so that it fools the eye.

It was proved in the 19th century that angles cannot be trisected with the Euclidean tools of
straightedge and compass.

This page:
Nov, 2011

David E. Joyce

Department of Mathematics and Computer Science

Clark University

Worcester, MA 01610

Homepage