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- Thread starter hypermorphism
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Place the rectangle along the line so that the edge of the rectangle is at the terminating point. Draw a line perpendicular to the segment. Do the same for the other side, and close the resulting rectangle. Draw the two diagonals. They will intersect at the halfway point of the line segment. Repeat this again on the other side of the line so that you have two rectangles of equal length. The two sets of diagonals are at the halfway point along the axis of the line segment, so connecting the two intersection points of the diagonals will bisect the line exactly.

Correct?

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That'll do it! There's another way as well, without having to have a right-angle present initially. For those who are still reading, restrict the problem as stated further in this manner: The straight edge is "infinitely long", so you cannot access all four sides, only two parallel sides. It is of adjustable width. You can rotate the straight edge, but of course you cannot tell from sight whether you have rotated it exactly 90 degrees.

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I'm not sure if you mean to use the 'rectangular' nature of the stright edge or not. In classic straight edge and compass geometry this is illegal.

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jimmysnyder said:

I'm not sure if you mean to use the 'rectangular' nature of the stright edge or not. In classic straight edge and compass geometry this is illegal.

I knew I was forgetting something. Since I didn't forbid it, it's a valid answer. An additional restriction is that you cannot fold the surface that the line segment is on.

Also, the reason I allow the straight edge to be 2-dimensional as opposed to its stricter cousin is in order to compensate for something the compass would normally be doing.

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Good job! Curiously, you cannot create a perpendicular if you just stick to plane rotations.

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Simple if I have a rectangle with diagonal longer than the straight line. Say the rectangle is designated by ABCD and the line by EF. Coincide A and E. Let BC be cut any where by F and mark B. Do it on the other side of the line and mark B'. BB' bisects the original line. (PS: Two lines originating from two edges of diameter form a right angle triangle on the circumference.

I got the clue of reading hidden posts after submitting my reply. One of the guys already answered it.

I got the clue of reading hidden posts after submitting my reply. One of the guys already answered it.

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