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It is well known that on Euclidean plane one can construct an isosceles triangle on given straight line by using a ruler and a pair of compasses.

Also it is possible to construct straight line containing given point and parallel to given straight line by using only a ruler, with the condition that we can measure out a segment equal to any given segment.

But in this conditions there are difficulties with constructing perpendicular for given straight line. Of course it is equivalent to constructing of an isosceles triangle on given straight line by using only a ruler, with the condition that we can measure out a segment equal to any given segment.

I'm unable to do this, also I'm unable to prove that it is impossible. Does anybody know something about this topic?

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  • $\begingroup$ Does 'ruler' mean 'straightedge'? It seems like the answer is trivially 'yes' if you are allowed to measure distances (or even just with a straightedge with one marking, as long as the base is sufficiently short). $\endgroup$ – LSpice May 15 '19 at 23:22
  • $\begingroup$ @LSpice Do you see a simpler way then in the answer of Dave Witte Morris? If yes please write it down. $\endgroup$ – Evgeny Kuznetsov May 16 '19 at 9:05
  • $\begingroup$ I think that, despite referring to the base, I was nonetheless thinking of the operation in the title ("construct an isosceles triangle"), which really is trivial with a ruler, rather than the one in the body ("… on [a] given straight line"). Just to make sure, you do mean a ruler, in the sense of neusis? $\endgroup$ – LSpice May 17 '19 at 17:12
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Yes, this is possible.

First, note that it is possible to construct a pair of perpendicular lines. To do this, make a rhombus. (On two lines that meet at a point $A$, construct segments of the same length starting from $A$. Then constructing parallels yields a parallelogram with two adjacent sides of equal length. This is a rhombus.) The diagonals of the rhombus are perpendicular.

Think of the two perpendicular lines as the $x$-axis and the $y$-axis, so we may draw horizontal and vertical lines through any point. This allows us to assume that the origin $(0,0)$ is on the given line. Choose some other point $(x,y)$ on the line. The horizontal and the vertical lines through $(x,y)$ meet the two axes at $(x,0)$ and $(0,y)$. So we may construct the points $(y,0)$ and $(0,-x)$. The vertical line through $(y,0)$ meets the horizontal line through $(0,-x)$ at the point $(y,-x)$. The line joining $(0,0)$ and $(y,-x)$ has slope $-x/y$, and is therefore perpendicular to the original line, which has slope $y/x$.

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  • $\begingroup$ #Dave Witte Morris Yes of course this way we get the perpendicular. Thanks a lot for nice and clear explanation. $\endgroup$ – Evgeny Kuznetsov Sep 12 '15 at 9:52

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