Chapter 3.4, Problem 9PSB

To determine

To prove:In a circle with given center, the given ray is the bisector of a given angle

Explanation of Solution

Given information:

In a circle with center O,

  $\angle \text{ NOG }\cong \angle \text{ POG}$

Formula used:

Congruent Parts of Congruent Triangles are Congruent (CPCTC)

SAS criteria of congruency

Same shape and same size object are always equal, by definition of congruency

Proof:

It is given that

  $\angle \text{ NOG }\cong \angle \text{ POG}$

  $\begin{array}{l}\text{In }\Delta \text{ AEC and }\Delta \text{ DEB, }\\ \text{⇒}\angle \text{ NOG = }\angle \text{ POG}\mathrm{........}\text{by definition of congruency}\mathrm{.....}\left(a\right)\\ \text{⇒}\angle \text{ NOG + }\angle \text{NOR = }\angle \text{ POG +}\angle \text{POR=180}°\\ \text{⇒}\angle \text{NOR = }\angle \text{POR}\mathrm{......}\left(i\right)\\ \text{since, }\angle \text{ NOG = }\angle \text{ POG, from }\left(a\right)\\ \text{In }\Delta \text{NOR and }\Delta \text{POR, }\\ \angle \text{NOR = }\angle \text{POR}\mathrm{......}\left(i\right)\\ \overline{\text{ON}}=\overline{\text{OP}}\mathrm{..........}\text{radii of given circle}\mathrm{.........}\left(ii\right)\\ \overline{\text{OR}}=\overline{\text{OR}}\mathrm{.........}\text{common side }\mathrm{.............}\left(iii\right)\\ \text{From }\left(i\right),\left(ii\right)\text{ and }\left(iii\right)\\ \Delta \text{NOR }\cong \Delta \text{POR}\mathrm{......}\text{SAS criteria is satisfied}\\ \angle \text{NRO = }\angle \text{PRO}\mathrm{.......}\text{CPCTC}\\ \therefore \overline{\text{RO}}\text{ will also bisect }\angle \text{NRO}\\ \text{Hence proved}\end{array}$