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In a high school book the author said these paths have all the same distance. Is that true? how to convince myself (and my students as well) they all have the same length?

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    $\begingroup$ If all of the lines that appear to be parallel are indeed parallel then we have a "taxicab metric" for distance. en.wikipedia.org/wiki/Taxicab_geometry $\endgroup$ – Doug M Jan 26 '18 at 20:31
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    $\begingroup$ @DougM That sounds like the makings of a great answer. $\endgroup$ – jpmc26 Jan 26 '18 at 21:08
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    $\begingroup$ You should have asked why those paths have the same length. $\endgroup$ – steven gregory Jan 27 '18 at 16:39
  • $\begingroup$ @stevengregory edited. Thank you! $\endgroup$ – user42912 Jan 28 '18 at 16:19
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By a sequence of reflections of setions of each path ... each path can be shown to the same length as the red path.

enter image description here

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    $\begingroup$ Could you please just draw the lines more right? $\endgroup$ – user42912 Jan 26 '18 at 22:35
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    $\begingroup$ The original black lines aren't even parallel/perpendicular exactly (though that's fine since it's only a diagram). The additional lines are fine for making the point whether or not they align perfectly too. $\endgroup$ – Nij Jan 26 '18 at 23:23
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    $\begingroup$ @Nij Why do they have to be perpendicular anyway? Being parallel is necessary, though. $\endgroup$ – Weijun Zhou Jan 27 '18 at 7:18
  • $\begingroup$ If the original is perpendicular, so does the generic path. But there's not a term so far as I'm aware of express "there are two sets of parallel lines and the angle between any representatives of the sets are at the same angle as all others". $\endgroup$ – Nij Jan 27 '18 at 8:00
  • $\begingroup$ @Nij "Two sets of parallel lines" is enough. The latter part goes without saying. $\endgroup$ – Weijun Zhou Jan 27 '18 at 9:13
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Turn the picture by about $55^\circ$ clockwise, so that the paths are made of horizontal and vertical segments.

enter image description here

Then you will see that any of those paths is as long as it takes to move from $A$ to the 'right', all the way to the place 'right above $B$' and then move 'down' all the way to $B$ (darker black line added on the picture).

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When I was 5th grade student or something like that, I couldn't understand how they could be equal in a similar question and one of my brothers showed me a figure as I drawed and I completely understand it when I saw that figure. The figure I attached is as same as that one and since I didn't want to mess up the original one, I showed it on path $I$ and $II$:

enter image description here

Note that same colored segments have the same length.

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Translating vectors doesn't change their magnitude.

Or:

taxicab

The small red (green) vectors sum to the big red(green) vector.

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All the angles appear to be right angles, and paths go in only two directions: let's call them N (up-and-right in the diagram) and E (down-and-right). Now imagine, for each path, moving all the E segments down the bottom, and all the N segments over to the right. You'll get the same path from A to B for all 4, and rearranging segments doesn't change the length.

Edit: Donald Splutterwit has posted a picture of exactly what I was describing.

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  • $\begingroup$ Aha! Thanks for your answer. I was staring at @DonaldSplutterwit's answer, still not seeing how that helped explain anything, as to me it's basically just an extra line he added. But matching it with your explaination makes it obvious! $\endgroup$ – BruceWayne Jan 28 '18 at 8:28
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I hope you defined that every path had endpoints A and B.

You may like to play 'architect' with your students. Define a 'step' in terms of rise and run. (where both are > 0 which you may elect not to do in the beginning). Then have them design their flight of stairs where individual steps may have variable rises and/or runs with the object of connecting the first and second floors of a building with parallel floors. Hopefully you or someone may illustrate a 1 or 2 step solution. Label bottom and top A & B. Generalize by comparing the sum of Rises,,, Sum of Runs.
You may elect to investigate the case where Rises and Runs ---> 0

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