Thread: Prove that P(n) o P(m) = T (ab)

I need help to prove the following proposition.

Proposition: Let m and n be two parallel straight lines. Let AB be a line segment that first intersects m and then n, that is perpendicular to both m and n, and whose length is twice the distance between m and n. Then Pn o Pm = T (AB)

Note: P is reflection and T is translation

Thank you!

Since the problem can be translated and rotated at will, choose m and n so that the question is most easily answered. I choose m as the x=0 line, and n as the x=a line. Let your test point A be (x,0). Then P(m) on A gives (-x,0), and P(n) on (-x,0) gives (x+2a,0). Clearly that's T(ab).

Originally Posted by qmech

Since the problem can be translated and rotated at will, choose m and n so that the question is most easily answered. I choose m as the x=0 line, and n as the x=a line. Let your test point A be (x,0). Then P(m) on A gives (-x,0), and P(n) on (-x,0) gives (x+2a,0). Clearly that's T(ab).

I went to someone for this problem, and that person showed me a different way of proof. Your is good, but it is still unclear why it is T(ab) for the T(ab) is not defined. Thank you.