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In a previous problem I asked about the notation used here.. I'm still not sure how to show it even though I now get what it's asking.

The following is an exercise from The Four Pillars of Geometry.

3.6.5. Show that the reflections in lines $L$, $M$, and $N$ (in that order) have the same outcome as reflections in lines $L'$, $M'$, and $N$, where $M'$ is perpendicular to $N$

I've had to ask around to clairify parts of the problem; it should be noted that the lines $L'$ and $M'$ are essentially arbitrary with no relation to $L$ and $M$ except that they appear in the same order when you do the reflections. Also, no picture is given as the lines are general though I've drawn a few pictures to help reason it out.

My main issue might be that I'm not sure what tools to use, so please be as specific as possible in your answer.

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    (see comments to answer as well) Yeah, I tried to trackback a bit to show that I did look in the book and was in fact, regrettably, still confused. :P It's possible I'm over thinking these things.2011-07-25

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I understand the question as in Mark's comment: $L$, $M$ and $N$ are given lines, and we want to show that there are lines L' and M' with M'\perp N such that reflections in L', M' and $N$ (in that order) have the same effect as reflections in $L$, $M$ and $N$ (in that order).

First, since the reflection in $N$ is last in both cases, we can cancel it, so the task is to show that there are lines L' and M' such that reflections in L' and M' have the same effect as reflections in $L$ and $M$, subject to M'\perp N.

This is false. If $L$ and $M$ are parallel, reflections in $L$ and $M$ amount to a translation perpendicular to them by twice their distance, and this can only be reproduced by L' and M' if they, too, are parallel to $L$ and $M$; hence we cannot choose M' perpendicular to $N$ unless $M$ already happens to be.

It's true, however, if $L$ and $M$ aren't parallel. In this case, it follows almost immediately from the hint given in the book right above the exercise:

Reflections in any two lines meeting at the same angle $\theta/2$ at the same point $P$ give the same outcome.

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    I wanted "initial point" to be some point in the plane. Then "final" to be the point arrived at after the reflections. Then, I claimed that you could place the lines so that the "intersection" was the center of a circle with "final" and "initial" at some radius. The angle was between the rays coming out of the center going towards "final" and "initial" You're right it's simpler but now I'm wondering if I got it - even if there is a little more justification to be done.2011-07-25