Chapter 4.3, Problem 1PSA

To determine

To Prove: $PS\perp QR$

Explanation of Solution

Given information: A circle with centre P ; S is midpoint of QR.

Formula used: Congruency of triangles, linear pair axiom.

Proof:

  $\begin{array}{l}S\text{is}\text{midpoint}\text{of}QR\text{(Given)}\\ \therefore QS=SR\\ PS=PS(\text{Reflection}\text{property})\\ PQ=PR(\text{Each equal to radius})\\ \therefore \Delta PQS\cong \Delta PSR(\text{SSS})\\ \therefore \angle PSQ=\angle PSR(CPCT)\\ \angle PSQ+\angle PSR={180}^{\circ }(\text{linear}\text{pair}\text{axiom})\\ \therefore \angle PSQ={90}^{\circ }=\angle PSR\\ \therefore PS\perp QR\end{array}$