Chapter 4.4, Problem 3PSA

To determine

To Prove: AD is perpendicular bisector of BC.

Explanation of Solution

Given information: Circle with centre O, AB=AC.

Formula used: congruency of triangles, linear pair axiom.

Proof:

  $\begin{array}{l}AO=OA(\text{Reflection}\text{property)}\\ OB=OC(\text{Each}\text{equals}\text{Radius})\\ AB=AC(\text{Given})\\ \therefore \Delta AOB\cong \Delta AOC(\text{SSS})\\ \therefore \angle OAB=\angle OAC(\text{CPCT})\\ AD=AD(\text{Reflection}\text{property)}\\ \therefore \Delta ADB\cong \Delta ACD(\text{SAS})\\ \therefore DB=DCand\angle ADB=\angle ADC(\text{CPCT})\\ \angle ADB+\angle ADC={180}^{\circ }(\text{linear pair axiom})\\ \therefore \angle ADB={90}^{\circ }=\angle ADC.\end{array}$