Chapter 9.4, Problem 36PPS

(a)

To determine

To Prove: The two column proof for the given data.

Explanation of Solution

Given:

The reflection in the line $p$ maps $\overline{BC}$ to $\overline{B^{\prime }{C}^{\prime }}$ and A reflection in the line $q$ maps $\overline{B^{\prime }{C}^{\prime }}$ to $\overline{B^{″}{C}^{″}}$ . Also, $p\left|\right|q,AD=x$ .

The objective is to prove $\overline{B{B}^{″}}\perp p,\overline{B{B}^{″}}\perp q$ .

Calculation:

Consider the reflection in the line $p$ maps $\overline{BC}$ to $\overline{B^{\prime }{C}^{\prime }}$ and the reflection in the line q maps $\overline{B^{\prime }{C}^{\prime }}$ to $\overline{B^{″}{C}^{″}}$ . The lines $p\left|\right|q$ and x is the distance between p and q.

The perpendicular segment is the line that divides the line segment into two intervals. Thus, p is the perpendicular bisector of $\overline{B{B}^{\prime }}$ and q is the perpendicular bisector of the line $\overline{B^{\prime }{B}^{″}}$ and the segment is added as,

  $\overline{B{B}^{\prime }}+\overline{B^{\prime }{B}^{″}}=B{B}^{″}$

Thus, as the line is perpendicular to the portion of the segment and is perpendicular to the whole segment then the $\overline{B{B}^{″}}\perp p,\overline{B{B}^{″}}\perp q$ .

Hence, proved.

(b)

To determine

To Prove: The two column proof for the given data.

Explanation of Solution

Given:

The reflection in the line $p$ maps $\overline{BC}$ to $\overline{B^{\prime }{C}^{\prime }}$ and A reflection in the line $q$ maps $\overline{B^{\prime }{C}^{\prime }}$ to $\overline{B^{″}{C}^{″}}$ . Also, $p\left|\right|q,AD=x$ .

The objective is to prove $B{B}^{″}=2x$ .

Calculation:

Consider by the reflection of the image,

  $\begin{array}{l}\overline{BA}=\overline{A{B}^{\prime }}\\ \overline{B^{\prime }D}=\overline{D{B}^{″}}\end{array}$

As, the congruent segments are the line segments that are equal in the length then,

  $\begin{array}{l}BA=A{B}^{\prime }\\ B^{\prime }D=D{B}^{″}\end{array}$

Then, by the segment addition is,

  $\begin{array}{l}BA+A{B}^{\prime }+B^{\prime }D+D{B}^{″}=B{B}^{″}\\ 2A{B}^{\prime }+2B^{\prime }D=B{B}^{″}\end{array}$

Consider the segment addition as,

  $A{B}^{\prime }+B^{\prime }D=AD$

Then,

  $\begin{array}{l}2A{B}^{\prime }+2B^{\prime }D=B{B}^{″}\\ 2AD=B{B}^{″}\\ 2x=B{B}^{″}\end{array}$

Hence, proved.