Chapter 4.4, Problem 4PSA

To determine

To Prove: PQ is perpendicular bisector of RS.

Explanation of Solution

Given information: Circles with centers P and Q, RS is common chord.

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

Proof:

  $\begin{array}{l}PR=PS\text{(Each}\text{equal}\text{to}\text{radius)}\\ OR=QS\text{(Each}\text{equal}\text{to}\text{radius)}\\ PQ=PQ\text{(Common)}\\ \therefore \Delta PRQ\cong \Delta PSQ(\text{SSS})\\ \therefore \angle RPQ=\angle SPQ(\text{CPCT})\\ \text{Let}\text{PQ}\text{intersects}\text{RS}\text{at}\text{M}\text{.}\\ PM=PM(\text{Reflection}\text{property)}\\ \therefore \Delta PRM\cong \Delta PSM(\text{SAS})\\ \therefore RM=SMAnd\angle RMP=\angle SMP(\text{CPCT})\\ \angle RMP=\angle SMP={180}^{\circ }(\text{linear pair axiom})\\ \therefore \angle RMP={90}^{\circ }=\angle SMP.\end{array}$