Solution:

We are given a line, a pair of distinct points $A$ and $A_{1}$ that are reflections of each other across that line and a third point $B$ waiting to be reflected across the line. We can assume that the point labeled $A$ is on the same side of the line as $B$.

Given line $l$ and two points $A$ and $B$ on the same side of the line, there are three possible ways line $l$ and line $\overleftrightarrow{A B}$ can align on the plane:
(1) line $l$ and line $\overleftrightarrow{A B}$ intersect at an oblique (non-right) angle
(2) line $l$ is parallel to line $\overleftrightarrow{A B}$
(3) line $l$ is perpendicular to line $\overleftrightarrow{A B}$

These three possibilities set up three cases. A solution to the first case creates a straightedge procedure that can be easily used to solve the other two cases.

## Case 1:

There are two key ideas to this construction. First, the reflection of any point on line $l$ is the point itself. Second, the intersection of two reflection lines is the reflection of the intersection of the two original lines. The general plan is to reflect two lines that intersect at $B$. We do this by having lines that go through $B$ and contain two points whose reflections are known.

Setup

Complete diagram

- $C$ is where line $l$ meets $\overleftrightarrow{A B}$
- $D$ is where line $l$ meets $\overline{A_{1} B}$
- $B$ is where $\overrightarrow{A D}$ meets $\overrightarrow{A_{1} C}$

Procedure

$\overleftrightarrow{A C}$ is the first line containing $B$, and $\overleftrightarrow{A_{1} C}$ is its reflection. $\overleftrightarrow{A_{1} D}$ contains $B$ and has reflection $\overleftrightarrow{A D}$. Since $B$ is the intersection of $\overleftrightarrow{A C}$ and $\overleftrightarrow{A_{1} D}$, the reflection of $B$ is the intersection of $\overleftrightarrow{A_{1} C}$ and $\overleftrightarrow{A D}$.

Note that we now have a straightedge procedure that allows us to reflect a point (the target) across a line if we have a pair of points that we use as guide points of a reflection across the line. This works when the target point ($B$) and the same side example point ($A$) are on a line that intersects the line of reflection ($l$).

Case 2:

In this case, line $\overleftrightarrow{A B}$ is parallel to line $l$. No point $C$ as in case 1 is available. We can use straightedge operation (c) and select a point $H$ that is not on line $\overleftrightarrow{A B}$ and not on line $\overleftrightarrow{A A_{1}}$.

Select any appropriate point $H$.

Use case 1 on $H, A, A_{1}$ to get $H_{1}$.

- select $H$ so that $\overleftrightarrow{H A}$ is neither parallel nor perpendicular to line $l$
- $I =$ line $l \cap \overleftrightarrow{A H}$
- $J =$ line $l \cap \overline{A_{1} H}$
- $H_{1} = \overrightarrow{A J} \cap \overrightarrow{I A_{1}}$

Procedure to reflect $H$.

Now we are set to use our straightedge procedure to reflect $H$ across line $l$ using points $A$ and $A_{1}$ as the guide points. This is shown in figure 5. Finally we can use $H$ and $H_{1}$ as guide points for reflecting $B$ across line $l$ as shown in figure 6.

Use $H$ and $H_{1}$ to construct $B_{1}$

- $K$ is where line $l$ meets $\overleftrightarrow{B H}$
- $L$ is where line $l$ meets $\overline{H_{1} B}$
- $B_{1}$ is where $\overrightarrow{H L}$ meets $\overline{H_{1} K}$

Procedure using $H$ and $H_{1}$ to construct $B_{1}$

## Case 3:

When $\overleftrightarrow{B A}$ is perpendicular to line $l$, $B$ is on $\overleftrightarrow{A A_{1}}$ and so will be $B_{1}$. As in case 2, we need to pick a point $S$ so that line $\overleftrightarrow{A S}$ is neither parallel nor perpendicular to line $l$. Again we use $A$ and $A_{1}$ as guide points to reflect $S$. Now using $S$ and $S_{1}$, we can get $B_{1}$ as shown below.

With $B$ on line $A A_{1}$, select $S$.

Get $S_{1}$; use $S$ and $S_{1}$ to construct $B_{1}$.

- Select $H$ so that $\overleftrightarrow{H A}$ is neither parallel nor perpendicular to line $l$
- $T$ is where line $l$ meets $\overleftrightarrow{A S}$
- $U$ is where line $l$ meets $\overline{A_{1} S}$
- $S_{1}$ is where $\overrightarrow{A U}$ meets $\overrightarrow{T A_{1}}$
- $V$ is where line $l$ meets $\overleftrightarrow{B S}$
- $B_{1}$ is where $\overrightarrow{V S_{1}}$ meets $\overrightarrow{B A_{1}}$

Procedure