# Finding the glide reflection using a compass or straightedge

Given two congruent triangles that are not a rotation, translation or reflection of each other; how can I find the glide reflection (the last remaining option) using only compass and straightedge.

Find the Glide Line and Glide Vector gives an algebraic solution but I would like a solution with a compass or straightedge.

In this diagram, We want to glide the triangle $\triangle ABC$ to $\triangle A'B'C'$. They must meet your requirements. Of course, under those requirements they will also have opposite orientations.

Under your requirements, at least one pair of corresponding sides of the triangles will not be parallel. In the diagram, I used $\overline{BC}$ and $\overline{B'C'}$. Extend those sides (to lines $h$ and $j$) until they intersect, at point $D$.

Those intersecting lines at $D$ have two angle bisectors. Construct the one (called $l$) such that the direction of vector $\overrightarrow{BC}$ relative to point $D$ is opposite that of vector $\overrightarrow{B'C'}$ relative to point $D$. I.e. the orientation of vector $\overrightarrow{BC}$ after reflection in $l$ would match the orientation of vector $\overrightarrow{B'C'}$.

(If somehow my analysis that there would be a pair of intersecting extended sides is false, just choose a pair of sides that are parallel [and not anti-parallel] and extend one of those sides into line $l$.)

Now construct the line $m$ through point $B'$ and perpendicular to line $l$, then line $n$ through point $B$ parallel to line $l$. Lines $m$ and $n$ are perpendicular: call their intersection point $B'_1$. Construct point $F$ as the midpoint of segment $\overline{B'B'_1}$.

Finally, construct line $p$ through point $F$ perpendicular to line $m$. This is the line of reflection for your glide transformation. Construct vector $\overrightarrow{BB'_1}$, the translation vector for your glide transformation.

To show this is correct, I have shown the translation of $\triangle ABC$ under vector $\overrightarrow{BB'_1}$. We see that $\triangle A'B'C'$ is indeed the reflection of that triangle in line $p$.