# Morley's Theorem

G. Zsolt Kiss' proof

This is the latest proof that was sent (September 02, 2006) to me by G. Kiss Zsolt from Hungary. Like Conway's and Newman's, this one, too, employs the cheating (backward) strategy. Also, one can't miss its relation to Conway's proof. It uses the same add-ons but in a more natural and symmetric manner.

Assume angles a, b, c are given such that a + b + c = 60°. Pick up an equilateral triangle and, on its sides form isosceles trapezoids with base angles 2a, 2b, 2c, respectively, and the sides equal to the top base.

In such a trapezoid B_{0}B_{1}C_{2}C_{0}, the diagonals serve as the bisectors of the base angles. Indeed, assuming _{0}B_{1}C_{2} = 2a,_{1}B_{0}C_{0} = 180° - 2a._{1}B_{0}C_{0} is isosceles, its base angles equal

Now we form ΔABC by passing three straight lines through the dangling vertices of the trapezoids:

We get three small isosceles triangles, ΔB_{0}B_{1}B_{2} is one of them. Its apex angle B_{1}B_{0}B_{2} is found from

∠B_{1}B_{0}B_{2}
| = 360° - 60° - (180° - 2a) - (180° - 2c) |

= 2a + 2c - 60° | |

= 2(60° - b) - 60° | |

= 60° - 2b. |

Hence its base angles equal

while the supplementary angle _{0}B_{1}A = 120° - b.

_{0}C

_{2}A = 120° - c.

It follows that ∠C_{0}B_{1}A = 120° - a - b so that the two sum up to 180°:

_{0}C

_{2}A + ∠C

_{0}B

_{1}A = 180°.

This means that quadrilateral AC_{2}C_{0}B_{1} is cyclic. In the same manner, quadrilateral AC_{0}B_{0}B_{1} is also cyclic. Hence so is the pentagon AC_{2}C_{0}B_{0}B_{1}. Since, in the circumcircle, the three chords B_{0}B_{1}, B_{0}C_{0}, and C_{0}C_{2} are equal, lines AB_{0} and AC_{0} trisect angle at A, which, as easily seen is 3a.

Now trying to reverse the steps, assume in ΔA'B'C' angles equal 3a, 3b, 3c so that

Compared to Conway's or Newman's, it may be just a little longer but does not come even close to using trigonometry.

Hubert Shutrick sent me the following comment (with the entire proof on a separate page):

The third diagram in the proof shows how the vertices of ΔABC can be constructed using the points A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2} although A_{2} and C_{1} are not shown. Here is another motivation for these six points. Note that it is a straight edge and compass construction. Given the equilateral triangle A_{0}B_{0}C_{0}, the required vertex A has to be such that the angle B_{0}AC_{0} is a and the locus of points making this angle is the arc of a circle through B_{0} and C_{0}. Any other chord of the same length will also subtend the angle a for points on the circle. That is why B_{1} is an interesting point. The angle B_{0}B_{1}C_{0} is a so B_{1} is on the circle and, since the length B_{0}B_{1} is 1, the angle B_{0}AB_{1} will be a for points A on the circular
arc. Hence, if the theorem is true, B_{1} must be on the side AC of the required triangle.

By symmetry, the six points mentioned above must be on the sides of the triangle ABC. The proof then consists of showing that it's angles are indeed 3a, 3b and 3c, simple angle calculations. There are two special cases that should be mentioned. There is the case

I found this proof long ago, in the 1970s, while teaching Euclidean geometry to prospective mathematics' teachers at Karlstad's university college.

Amic has observed that a correction is in order:

I read

"There are two special cases that should be mentioned. There is the case |

There is a mistake : the other case is when a or b or

And I don't see why the case

So in my opinion this proof is wrong for a triangle ABC with a right angle.

My comment is that the proof can be saved by drawing the necessary line not "through two points" but at a certain angle at the "double point". For example, if _{1} and B_{2} coincide and AC should be drawn as to form angles _{1}C_{2} and, respectively, B_{2}A_{1}.

Hubert Shutrick has later added the following:

If B' is the midpoint of B_{1}B_{2}, then B_{0}B' is orthogonal to AC and ∠C_{0}B_{0}B' is _{0}C_{0}C' is _{0}B', C_{0}C' and B_{0}C_{0} is indeed 3a. As for the special case that I bungled, if _{1} = B_{2} = B',_{0}B'.

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