Chapter 10.6, Problem 19PSB

To determine

Give the explanation for the given proof.

Explanation of Solution

Given:

The given diagram is

  

  $\angle APR$ and $\angle AQR$ are right. $R$ is the midpoint of $arc\left(AB\right)$ . The given proof is $\overline{PR}\equiv \overline{RQ}$ .

Calculation:

  $R$ is the midpoint of the $arc\left(AB\right)$ .

  $marc\left(AR\right)=marc\left(BR\right)$

Let $marc\left(AR\right)=marc\left(BR\right)=a$

So,

  $\begin{array}{l}marc\left(AB\right)={360}^{°}-\left(marc\left(AR\right)+marc\left(BR\right)\right)\\ ={360}^{°}-2x\end{array}$

Angle formed by tangent and secant $=\frac{1}{2}$

  $=$ difference of intercepted arcs

  $\begin{array}{l}m\angle Q=\frac{1}{2}\left(ArcAB-ArcAR\right)\\ {90}^{°}=\frac{1}{2}\left({360}^{°}-2x-x\right)\\ \Rightarrow {180}^{°}={360}^{°}-3x\\ \Rightarrow {180}^{°}-{360}^{°}={360}^{°}-3x-{360}^{°}\\ \Rightarrow -{180}^{°}=-3x\\ \Rightarrow \frac{-{180}^{°}}{-3}=\frac{-3}{-3}x\\ \Rightarrow x={60}^{°}\end{array}$

Tangent chord angle $=\frac{1}{2}$ intercepted arc

  $\begin{array}{l}m\angle PAQ=\frac{1}{2}marc\left(BA\right)\\ =\frac{1}{2}\left[marc\left(AR\right)+marc\left(BR\right)\right]\\ =\frac{1}{2}\left[{60}^{°}+{60}^{°}\right]\\ ={60}^{°}\end{array}$

Now,

  $\begin{array}{l}m\angle PRQ={360}^{°}-\left(m\angle PAQ+m\angle APR+m\angle RQA\right)\\ ={360}^{°}-\left[{60}^{°}+{90}^{°}+{90}^{°}\right]\\ ={120}^{°}\end{array}$

The angle inscribed on the circle by two intersecting chord $=\frac{1}{2}$ intercepted arc

  $\begin{array}{l}m\angle PAR=\frac{1}{2}marc\left(BR\right)\\ =\frac{1}{2}\cdot {60}^{°}\\ ={30}^{°}\\ m\angle QAR=m\angle PAQ-m\angle PAR={60}^{°}-{30}^{°}={30}^{°}\end{array}$

In $\Delta APR$ and $\Delta AQR$ .

  $\begin{array}{l}\angle APR=\angle AQR\\ \angle PAR=\angle QAR\\ AR=AR\end{array}$

From $ASA$ , $\Delta APR\cong \Delta AQR$

Hence $\overline{PR}\equiv \overline{RQ}$ .