MATH325: Lecture 14


Syllabus | Schedule | Grading Keys/Rubrics

This class was mostly review for the sample test, but we did also cover the following:

Lemma: The perpendicular bisector of segment AB is the set of points equidistant from points A and B. (In other words, |PA| = |PB| if and only if P is a point on the perpendicular bisector of segment AB.)

Brief Proof: If P is on the perpendicular bisector and M is the midpoint of AB, then by side-angle-side triangle AMP is congruent to triangle BMP. Therefore |AP| = |BP|.

On the other hand, if |AP| = |BP| then triangle ABP is isosceles. If M is the midpoint of AB then by side-side-side triangle AMP is congruent to triangle BMP. From this one can show that line PM is perpendicular to line AB, and so that line P lies on the perpendicular bisector of segment AB.//

Theorem: Given three non-colinear points A, B and C we can construct a center O and radius OA so that points A, B and C lie on CO(|OA|).

Proof: Construct the perpendicular bisector l of segment AB. For any point P on line l, CP(|PA|) goes through points A and B, because |PA| = |PB| is the radius of the circle.

Similarly, construct the perpendicular bisector m of segment BC.

Let O be the point of intersection of lines l and m. Then |OA| = |OB| = |OC| by the lemma. Therefore points A, B and C lie on CO(|OA|).//

Extra Credit: Suppose two chords AB and CD bisect each other at point M. Prove (or give a counter-example) that M must be the center of the circle.//

We also reviewed the Star Trek Lemma and its application to problem 1 on the sample test and the proof of theorem 1.9.1.