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. Ofcourse 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 haven't yet able to do this, also I haven't able to prove that it is impossible. Do anybody know something about this topic?
Thanks in advance.
Yes, this is possible.
It is possible to construct a pair of perpendicular lines somewhere on plane. 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 $(y,−x)$ is perpendicular for the line line joining $(0,0)$ and $(x, y)$.