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) .