Chapter 9.4, Problem 37PPS

(a)

To determine

To Prove: The paragraph proof for the given data.

Explanation of Solution

Given:

The lines l and m intersect at the point p. A is any point that is no on l or m.

The objective is to prove that if you reflect A in m, and the reflect its image $A^{\prime }$ in l, $A^{″}$ is the image of A after a rotation about point P.

Calculation:

By the definition of the reflection the m is the perpendicular bisector of $A{A}^{\prime }$at S and l is the perpendicular bisector of line $\overline{A^{\prime }{A}^{″}}$ . As, the perpendicular bisectors is the line that divides the line segment in equal in equal parts then,

  $\overline{AR}\cong \overline{A^{\prime }R}$

As the two points form one line than the auxiliary segments are,

  $\overline{AP},\overline{A^{\prime }P},\overline{A^{″}P}$

As, the perpendicular bisector is the line segment that divides the two equal parts then,

  $\angle ARP,\angle A^{\prime }RP,\angle A^{\prime }SP,\angle A^{″}RP$

Then, use reflective property as,

  $\begin{array}{l}\overline{RP}\cong \overline{RP}\\ \overline{SP}\cong \overline{SP}\end{array}$

Then, use SAS congruence as,

  $\begin{array}{l}\Delta ARP\cong \Delta A^{\prime }RP\\ \Delta A^{\prime }RP\cong \Delta A^{″}RP\end{array}$

Then, by CPCTC

  $\begin{array}{l}\overline{AP}\cong \overline{A^{\prime }P}\\ \overline{A^{\prime }P}\cong \overline{A^{″}P}\end{array}$

Hence, the rotation moves every point of the pre-image with the specified angle and the direction about the fixed point. Then,

  $\angle APR,\angle A^{\prime }PR,\angle A^{\prime }PS,\angle A^{″}PS$

Hence, proved.

(b)

To determine

To Prove: The paragraph proof for the given data.

Explanation of Solution

Given:

The lines l and m intersect at the point p. A is any point that is no on l or m.

The objective is to prove $m\angle AP{A}^{″}=2\angle \left(m\angle SPR\right)$ .

Calculation:

Consider that the congruent segment are those that are equal in length as,

  $\begin{array}{l}m\angle APR=m\angle A^{\prime }PR\\ m\angle A^{\prime }PS=m\angle A^{″}PS\end{array}$

Then, by addition postulate,

  $\begin{array}{l}m\angle APR+m\angle A^{\prime }PR+m\angle A^{\prime }PS+m\angle A^{″}PS=m\angle AP{A}^{″}\\ m\angle A^{\prime }PR+m\angle A^{\prime }PS=m\angle SPR\end{array}$

Then, by substitution,

  $\begin{array}{l}m\angle A^{\prime }PR+m\angle A^{\prime }PR+m\angle A^{\prime }PS+m\angle A^{\prime }PS=m\angle AP{A}^{″}\\ 2\angle SPR=m\angle AP{A}^{″}\end{array}$

Hence, Proved.