Thread: Transformations

Show that a translation TPQ is equivalent to the composition of two reflections in parallel lines l1 and l2, where l1 and l2 are perpendicular to PQ. If d is the distance between l1 and l2, show that PQ = 2d.

This is what I have so far. I feel like I am missing some inportant parts.

Proof. Given translation TPQ, pick an arbitrary point, A, on PQ. Construct a line to PQ through A. We shall call this line l1. Reflect point P in l1 to get P'. By the property of reflections, PA = P'A. Let PA = x. Therefore, P'A = x also. Now construct the bisector of P'Q. We shall call this new line, l2. Since l1 is perpendicular to PQ and l2 is perpendicular to PQ, l1 || l2. Let the intersection of l2 and PQ be at point B. Now reflect P' in l2 to get P''. By the property of reflections, P'' will be on Q. Also by the properties of reflections, P'B = P''B = QB. Let the P'B = y. Therefore, P''B = y and QB = y also. Therefore, PQ = x + x + y + y = 2x + 2y and AB (the distance between l1 and l2) = x + y. Let d = AB (the distance between l1 and l2). Therefore, 2d = 2(x + y). Therefore, PQ = x + x + y + y = 2x + 2y = 2(x + y) = 2d. Therefore PQ = 2d.
Given l1 || l2, and an arbitrary point P. Reflect P over l1 to get P'. Construct line segment PP' where A is the intersection of l1 and PP'. By properties of reflections, PA = P'A. Let PA = x. Therefore, P'A = x also. Next, reflect P' in l2. Construct line segment P'P" where B is the intersection of l2 and P'P". By properties of reflections, P'B = P"B. Let the P'B = y. Therefore, P''B = y also. Therefore, PP" = x + x + y + y = 2x + 2y and AB (the distance between l1 and l2) = x + y. Let d = AB (the distance between l1 and l2). Therefore, 2d = 2(x + y). Therefore, PQ = x + x + y + y = 2x + 2y = 2(x + y) = 2d. Therefore PQ = 2d.