# Reflected Triangles

Alignments to Content Standards: G-CO.A.5 G-CO.D.12

The triangle in the upper left of the figure below has been reflected across a line into the triangle in the lower right of the figure. Use a straightedge and compass to construct the line across which the triangle was reflected. ## IM Commentary

This task is a reasonably straight-forward application of rigid motion geometry, with emphasis on ruler and straightedge constructions, and would be suitable for assessment purposes.

## Solution

The line in question is the perpendicular bisector of any pair of corresponding points. We choose the two corresponding points labelled $A$ and $B$ for the illustration below, but either of the other two pairs of points works just as well. To summarize the construction: We construct the circle of radius $\overline{AB}$ cenetered at $A$ and the circle of radius $\overline{AB}$ cenetered at $B.$ These intersect at two points, which we label $P$ and $Q$. Then $P$ and $Q$ are both equidistant from $A$ and $B$, and so lie on the perpendicular bisector of $\overline{AB}$. We conclude that line $PQ$ is precisely the perpendicular bisector, which is the line over which the first triangle was reflected to arrive at the second.

We remark that the above construction is valid even from the strict Euclidean perspective on straight-edge and compass constructions. In a more modern setting, in which one typically allows the use of a compass with memory, one has a little more flexibility. In particular, we could replace the two circles in the above construction with any pair of circles of equal radius at least $\frac{1}{2}\overline{AB}$, centered at $A$ and $B$.