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I see one of the problems of my transformational geometry text book in the topic of reflection, i.e.:

there are two parallell lines g and h and two different points A and B. Both points are inside the area bounded by the lines. Draw the shortest route from A to g then to h then to B!

I assume, of course, we have to apply reflection, i.e. reflect A on g resulting A' and reflect B on h resulting B' then connect A' to B' implying there are two points C and D from the intersection of A'B' and g and A'B' and h respectively. So the shortest route is ACDB.

My question is that is my work correct? if yes, can anybody help with the reason? and if it is incorrect, please show the true work.

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enter image description here

And $\overline {A'B'} = \overline {AC} + \overline {CD} + \overline {DB}$

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  • $\begingroup$ why adding DA to the route? the problem just ask from A to g then to h then to B? $\endgroup$ – wawar05 Jan 26 '17 at 2:35
  • $\begingroup$ Misreading the question. I thougth you wanted to go from A to g to B to h and back to A. The correct path is A to g to h and then to B. I will make a new diagram. $\endgroup$ – Doug M Jan 26 '17 at 2:37
  • $\begingroup$ could you give an explanation why AC + CD + DB is the shortest past? $\endgroup$ – wawar05 Jan 26 '17 at 7:02
  • $\begingroup$ The trip has a component of distance in the direction parallel to lines g,h and a component of distance perpendicular to g,h. If you construct a right triangle with vertexes A', B' and one leg parallel to g, and the other leg perpendicular. The legs of this triangle are the minimal distances you must travel in these tow directions, and the hypotenuse is then the shortest path that travels these minimal distances. $\endgroup$ – Doug M Jan 26 '17 at 18:44

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