Chapter 5 Section 1: Perpendicular & Angle Bisectors

Chapter 5 Section 1: Perpendicular & Angle Bisectors GOAL: Prove and apply theorems about perpendicular and angle bisectors

A Perpendicular ( ┴) Bisector is a segment, line or ray that “cuts” another segment, line or ray AT ITS MIDPOINT and FORMS A 90o INTERSECTION.

Properties of a Perpendicular Bisector…

1. ANY point on the ( ┴ ) Bisector is EQUIDISTANT from the endpoints of the segment being “cut”. 2. The ( ┴ ) Bisector “cuts” any segment EXACTLY in half, revealing the Mid-Point. 3. The ( ┴ ) Bisector forms not one but four (4) 90o angles where it “cuts” a segment.

PROOF of Theorem 5-1-1 (Perpendicular Bisector Theorem)

● C Since segment CP is the Perpendicular Bisector, then it cuts segment AB exactly in half. That means that AP  BP by definition of Bisection. It is also known that the A ● │ ● │ ●B P angles formed at Point P are Right Angles by definition of Perpendicular. Since CP  CP by the Reflexive Property then it can be verified that APC  BPC by (SEE Page 300) SAS Postulate. Since AC  BC are the segments connecting to point C which is a point ON the Perpendicular Bisector, they are congruent by CPCTC.

PROOF of Theorem 5-1-3 (Angle Bisector Theorem)

The shortest distance between two points…is a straight line. More specifically the shortest distance between two points is a straight line ALONG THE PERPENDICULAR segment from point to point or point to line. The distance from a point is defined as the length of the ( ┴ ) segment from the given point to the given line.

B Let’s say that < A has been bisected by AD. That ● indicates that the two smaller angles created are equal in measure and are  . Since the distance from a point to a line is along the ( ┴ ) then there A D o ● ● are 90 angles formed at the intersection. Because AD is a shared segment then it is  to

C itself. Therefore, ADB  ADC by AAS and the ● distance of DB is the same as DC by CPCTC

(SEE Page 301)