### 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 C_{O}(|OA|).

**Proof:** Construct the perpendicular bisector l of segment AB. For any point P on line l,
C_{P}(|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 C_{O}(|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.