This chapter is concerned with concurrent lines associated with a triangle. A family of lines is ** concurrent** at a point

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

**Theorem 2.1.1.** *Let l*_{1} *and l*_{2} *be parallel lines, and suppose that m*_{1} *and m*_{2} *are lines with m*_{1} ⊥ *l*_{1} *and m*_{2} ⊥ *l*_{2}. *Then m*_{1} *and m*_{2} *are parallel*.

**Proof.** Let *P* be the point *l*_{1} ∩ *m*_{1} and let *Q* be the point *l*_{2} ∩ *m*_{2}. By the parallel postulate, *l*_{1} is the only line through *P* parallel to *l*_{2}, and so *m*_{1} is not parallel to *l*_{2} and consequently must meet *l*_{2} at some point *S*. In other words, *m*_{1} is a transversal for the parallel lines *l*_{1} and *l*_{2}. Since the sum of the adjacent interior angles is 180°, it follows that *l*_{2} must be perpendicular to *m*_{1}.

But *l*_{2} is also perpendicular to *m*_{2} and so *m*_{1} and *m*_{2} must be parallel.

In the above proof, if we interchange *l*_{1} and *m*_{1} and *l*_{2} and *m*_{2}, we have:

**Corollary 2.1.2.** *Suppose that l*_{1} ⊥ *m*_{1} *and l*_{2} ⊥ *m*_{2}. *Then l*_{1} *and l*_{2} *are parallel if and only if m*_{1} *and m*_{2} *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 ...

Start Free Trial

No credit card required