Chapter 12.3, Problem 31E

To determine

To prove:

The reflections in intersecting lines theorem.

Explanation of Solution

Given:

  $\overline{PA},\overline{P{A}^{\prime }}and\overline{P{A}^{″}}$

Concept used:

Two geometric figures are congruent if a rigid motion or a composition of rigid motions maps one of the figures onto other.

Calculation:

If two lines $kandm$ intersecting at the point $P$ .

Reflecting the point $A$ in the line $k$ maps to $A^{\prime }$

Reflecting the point $A^{\prime }$ in the line $m$ maps to $A^{″}$

  $A{A}^{\prime }$ that intersects $k$ in the point $l$

  $A^{\prime }{A}^{″}$ that intersects $m$ in the point $n$

Connecting the segment

  $\overline{PA},\overline{P{A}^{\prime }}and\overline{P{A}^{″}}$

Definition of reflection

  $k$ is perpendicular bisector of $\overline{A^{\prime }{A}^{″}}$ .

The pairs of congruent corresponding angles

  $\begin{array}{l}m\angle AlP\cong m\angle A^{\prime }lP={90}^0\\ \angle APl\cong \angle A^{\prime }Pl\end{array}$

The angle between the two lines of reflection is

  $m\angle lPn=m\angle 2+m\angle 1^{\prime }$

The rotation angle

  $\begin{array}{l}m\angle AP{A}^{″}=2\times m\angle 2+2\times m\angle 1^{\prime }\\ m\angle AP{A}^{″}=2\left(m\angle 2+m\angle 1^{\prime }\right)\\ m\angle AP{A}^{″}=2\times m\angle lPn\end{array}$

  

Therefore,

  $m\angle lPn=m\angle 2+m\angle 1^{\prime }$

  $m\angle AP{A}^{″}=2\times m\angle lPn$