Ch 5.2

We will look at the properties of a perpendicular bisector to solve algebraic problems.

Look at the point of concurrency created by the perpendicular bisectors and explore the relationship it creates.

A perpendicular bisector is a perpendicular line that passes through the midpoint of a segment.

There are 3 in a triangle (one for each side) that cross at a point of concurrency.

Points on the perpendicular bisector are equidistant from the endpoints of the line segment.

A point of concurrency is the point where three or more lines, rays, or segments intersect.

Concurrent figures are the actual 3 or more figures that intersect.

The point of concurrency of the perpendicular bisectors, called the circumcenter, creates congruent sides.

..\Perpendicular Bisector.gsp

The segments from the vertices to the point of concurrency are congruent to each other.

Just what I need, more stuff to set

equal to each other!

G

F

ED

B

AC

16

G

F

ED

B

AC

3x + 8

5x - 4

11x - 40

G

F

ED

B

AC

7x + 3 4x + 24

8x - 4

Because CP is the perpendicular bisector of AB, we know that AP = PB by the definition of a perpendicular bisector. We also know that <APC is congruent to <CPB, because they are both 90o angles created by the perpendicular line. Then because CP is a shared side by the reflexive property CP is congruent to itself. Finally by the Side-Angle-Side congruence theorem, we know that ∆APC ∆BPC. Since the two triangles are congruent, congruent parts of congruent triangles are congruent, so CA = CB.

We are given than CA = CB, so we also know that CA CB, by the definition of congruent segments. Then we also know that Line PC is perpendicular to segment AB. Then by the reflexive property we can say that CP is congruent to itself. By Hypotenuse leg ∆PAC ∆PBC. Therefore, we know that AP = PB because congruent parts of congruent triangles are congruent. Then by the definition of a perpendicular bisector, we can conclude that C is on the perpendicular bisector.

eWorkbook 5.2