# The Fundamental Proof of Morley's Trisector Theorem

### Nolan L Aljaddou

April 26, 2009

All triangle endpoints are formed by the intersection of three distinct lines. That is, a triangle with three side lines A, B, and C is generated as follows: ∪{(A∩B), (B∩C), (C∩A)} (a single line may be defined by only two distinct points which belong to the same set, and therefore the union of two distinct points can define a line; in this case the union of each point forms each side line of the triangle). The endpoints of the triangle are generated not only by the intersection of each pair of lines, but equivalently by the intersection of the lines of each of their angle’s projections - from the angles they form upon intersection - thus each may be equally replaced by those angle projections, defined by their angles, in the set to form ∪{α, β, γ}; this set includes the side lines of the triangles as well as the endpoints, so to reduce the set to its endpoints alone the intersection is taken, as the mutual intersection of the three angle projections only occurs at the endpoints of the triangle: ∩{∪[α, β, γ]}.

The trisection of a triangle's angles divides them each into three equal values, and one third of each separate original angle respectively may be denoted as a, b and c. Since their original total sum is 180° [3a + 3b + 3c = 180°] the sum of one of each of their trisections (being a third of each original angle) is collectively equal to one third of the total of 180° [a + b + c = (3a + 3b + 3c)/3 = 180°/3] which is 60°.

Since each angle of an equilateral triangle is 60°, each of its angles can be divided into three parts which are equal to a, b, and c - which then form a union to form the angle. Therefore, the equilateral triangle's endpoints themselves are generated by the intersection of the union of these three angles, three times. That is: ∩{∪[∪(a, b, c), ∪(a, b, c), ∪(a, b, c)]} generates the endpoints of the equilateral triangle. The union of the three angles is geometrically equivalent to the addition of the three angles; that is, ∪(a, b, c) = (a + b + c), therefore the previous intersection is equivalent to: ∩{∪[(a + b + c), (a + b + c), (a + b + c)]}, which itself is geometrically equivalent to ∩{(a + b + c) + (a + b + c) + (a + b + c)}, which reduces to ∩{3(a + b + c)} or ∩{(3a + 3b + 3c)} = ∩{∪[3a, 3b, 3c]}, which is the generator of the endpoints of the
original triangle. That is, the intersection of the three original angles' projections not only generates the original triangle's endpoints, but the intersection of their trisection also generates the endpoints of an equilateral triangle - which only occurs for the intersection of adjacent central angle trisections, as the bordering trisections intersect with each other bordering trisection at an entire line - the side of the triangle - and not a single point.

QED

**Note by AlexB**: the above is a version of the original message which can be found - along with a record of the intervening correspondence - on a separate page.

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