Chapter 4.3, Problem 9PSB

To determine

To Verify: $AO\perp BC.$

Explanation of Solution

Given information: circles with centers O, LB=LC.

Formula used: Properties of circle, congruency of triangles, linear pair axiom.

Proof:

  $\begin{array}{l}\angle B=\angle C(\text{given})\\ \therefore AB=AC\text{(Sides opposite to equal angles are equal)}\\ OC=OC\text{(Each equal to redius)}\\ AO=AO(\text{Reflective}\text{property)}\\ \therefore \Delta ABO\cong \Delta ACO(\text{SSS})\\ \therefore \angle BAE=\angle CAE(CPCT)\\ AE=AE\text{(Reflective}\text{property)}\\ \therefore \Delta ABE\cong \Delta ACE(\text{SAS})\\ \therefore \angle AEB=\angle AEC(\text{CPCT})\\ \angle AEB+\angle ACE={180}^{\circ }(\text{Linear pair axiom})\\ \therefore \angle AEB={90}^{\circ }=\angle AEC\\ \therefore AO\perp BC.\end{array}$