Chapter 10.4, Problem 8PSA

To determine

To Prove : $\overline{PQ}\perp \overline{RS}$

Answer to Problem 8PSA

1

Explanation of Solution

Given information : $\overline{QR}\perp \overline{QS}\text{ are tangent to circle with centre P at R and S}\text{.}$

  

  $\begin{array}{l}\text{Join P to R and S}\text{.}\\ \text{In }\Delta \text{PRQ and }\Delta \text{PSQ}\\ \text{PQ}\cong \text{PQ}\text{(Common)}\\ \text{PR}\cong \text{PS}\text{(Equal Radii)}\\ \text{PR}\cong \text{QS}\text{(Length of tangents)}\\ \text{By SSS congruencey}\\ \Delta \text{PRQ}\cong \Delta \text{PSQ}\\ \Rightarrow \angle \text{RPQ}\cong \angle \text{SPQ}\text{(by CPCT)}\\ \text{Let O be the point of intersection of PQ and RS}\text{.}\\ \text{In }\Delta \text{PRO and }\Delta \text{PSO}\\ \text{PO}\cong \text{PO}\text{(Common)}\\ \angle \text{RPO}\cong \angle \text{SPO}\text{(Prove above)}\\ \text{PR}\cong \text{PS}\text{(Equal radii)}\\ \Rightarrow \Delta \text{PRO}\cong \Delta PSO\text{(by SAS)}\\ \Rightarrow \angle \text{POR}\cong \angle \text{POS}\\ \text{But }\angle \text{POR + }\angle \text{POS = 180° (Linear pair)}\\ \Rightarrow \text{2}\angle \text{POR = 180°}\\ \Rightarrow \angle \text{POR = 90°}\\ \Rightarrow \overline{PQ}\perp \overline{RS}\end{array}$