assume that the parallel lines l and l’ have a common perpendicular segment mm’
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Let the perpendicular between L and L' meet L' at point N. M is a point on L and M' is a point on L', so that MM' is a transverse.

Now we have a right triangle MM'N where N is the right angle. The triangle rule says that MN+NM'≥MM'. The minimum value of MM' is when either MN or NM' is zero. They cannot both be zero because that would make MM' zero, since it can’t have negative length. But MN is the distance between the parallel lines and is not zero; so NM'=0 and MM'≤MN. Therefore MM' lies on the perpendicular MN and to reach L' it cannot be shorter than MN. The inequality says that it can’t be greater, so MM' must be the same length as MN. So MN, the common perpendicular, must be the shortest distance between the parallel lines.

 

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