Classical Geometry: Euclidean, Transformational, Inversive, and Projective

CHAPTER 2

CONCURRENCY

2.1 Perpendicular Bisectors

This chapter is concerned with concurrent lines associated with a triangle. A family of lines is concurrent at a point P if all members of the family pass through P.

In preparation, we need a few additional facts about parallel lines.

Theorem 2.1.1. Let l₁ and l₂ be parallel lines, and suppose that m₁ and m₂ are lines with m₁ ⊥ l₁ and m₂ ⊥ l₂. Then m₁ and m₂ are parallel.

Proof. Let P be the point l₁ ∩ m₁ and let Q be the point l₂ ∩ m₂. By the parallel postulate, l₁ is the only line through P parallel to l₂, and so m₁ is not parallel to l₂ and consequently must meet l₂ at some point S. In other words, m₁ is a transversal for the parallel lines l₁ and l₂. Since the sum of the adjacent interior angles is 180°, it follows that l₂ must be perpendicular to m₁.

But l₂ is also perpendicular to m₂ and so m₁ and m₂ must be parallel.

In the above proof, if we interchange l₁ and m₁ and l₂ and m₂, we have:

Corollary 2.1.2. Suppose that l₁ ⊥ m₁ and l₂ ⊥ m₂. Then l₁ and l₂ are parallel if and only if m₁ and m₂ are parallel.

One of the consequences of this corollary is that if two line segments intersect, then their perpendicular bisectors must also intersect. This fact is crucial in the following theorem, the proof of which also uses the fact that the right bisector of a segment can be characterized as being the set of all points that are equidistant ...