Chapter 4, Problem 21RP

To determine

To Prove: bisector $\angle ABC$ is perpendicular bisect of AC.

Explanation of Solution

Given information: Circle circumscribing $\Delta ABC$ as shown in figure.

  $\angle ABO=\angle CBO.$

  

Formula Used: Properties of circle.

Proof:

  $\begin{array}{l}\angle ABO=\angle CBO\mathrm{....}\left(giving\right)\\ \therefore 2\angle ABO=2\angle CBO\\ \therefore \angle AOM=\angle COM.\end{array}$

  $OA=OC$ (Each equals radius).

OM=OM (common)

  $\therefore \Delta AOM\cong \Delta COM\mathrm{....}(SAS)$

  $\therefore \angle OMA=\angle OMC$ And AM=MC (CPCT)

By linear pair axiom.

  $\begin{array}{l}\angle OMA+\angle OMC={180}^{\circ }\\ \therefore \angle OMA={90}^{\circ }=\angle OMC.\end{array}$