Plane Geometry - Bruce E. Shapiro

SECTION 19. ANGLE BISECTORS 97Definition 19.4 Let A and B be distinct points. A perpendicular bisectorof AB is a line l such that the midpoint of AB lies on l, and l ⊥ ←→ AB.Theorem 19.5 (Existence and Uniqueness of Perpendicular Bisectors)If A and B are distinct points then there exists a unique perpendicularbisector of AB.Proof. By the midpoint existence theorem, the segment AB has a uniquemidpoint M.By the protractor postulate there exists a point C ∉ ←→ AB such that ∠AMC =90.Let l = ←−→ MC.Since l ⊥ ←→ AB and M ∈ l, l is the perpendicular bisector. This provesexistence.The midpoint is unique by the midpoint existence/uniqueness theorem. Bythe protractor postulate there is only one ray on each side of ←→ AB that isperpendicular to AB and passes through M. These two rays are oppositeand on the same line and hence the bisector is unique.Definition 19.6 Angles ∠BAC and ∠DAE form a vertical pair (are verticalangles) if −→ −→−→ −→AB and AE are opposite and rays AC and AD are oppositeor rays −→ −→−→ −→AB and AD are opposite and AC and AE are opposite.Figure 19.3: Angles α and β are vertical anglesTheorem 19.7 (Vertical Angle Theorem) Vertical angles are congruent.Proof. Assume the geometry shown in figure 19.3. Then −→ −→AB and AE areopposite; and −→ AC and −→ AD are opposite. We need to show that α = β.Angles β = ∠EAD and ∠EAC are a linear pair. Henceβ + ∠EAC = 180Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.