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14.2: Constructions

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    Let us consider geometric constructions with a ruler and a parallel tool; the latter makes it possible to draw a line thru a given point parallel to a given line. By Exerciser 14.1.1, any construction with these two tools is invariant with respect to affine transformations. For example, to solve the following exercise, it is sufficient to prove that the midpoint of a given segment can be constructed with a ruler and a parallel tool.

    Exercise \(\PageIndex{1}\)

    Let \(M\) be the midpoint of segment \([AB]\) in the Euclidean plane. Assume that an affine transformation sends the points \(A\), \(B\), and \(M\) to \(A'\), \(B'\), and \(M'\) respectively. Show that \(M'\) is the midpoint of \([A'B']\).

    The following exercise will be used in the proof of Proposition 14.3.1.


    According to the remark before the exercise, it is sufficient to construct the midpoint of \([AB]\) with a ruler and a parallel tool. Guess a construction from the diagram.

    截屏2021-02-25 上午10.32.50.png

    Exercise \(\PageIndex{2}\)

    Assume that the points with coordinates \((0,0)\), \((1,0)\), \((a,0)\), and \((b,0)\) are given. Using a ruler and a parallel tool, construct points with coordinates \((a\cdot b,0)\) and \((a+b,0)\).


    Let \(O, E, A\), and \(B\) denote the points with the coordinates (0, 0), (1, 0), (\(a\), 0), and (\(b\), 0) respectively.

    To construct a point \(W\) with the coordinates \((0, a + b)\), try to construct two parallelograms \(OAPQ\) and \(BWPQ\).

    To construct \(Z\) with coordinates \((0, a \cdot b)\) choose a line \((OE') \ne (OE)\) and try to construct the points \(A' \in (OE')\) and \(Z \in (OE)\) so that \(\triangle OEE' \sim \triangle OAA'\) and \(\triangle OE'B \sim \triangle OA'Z\).

    Exercise \(\PageIndex{3}\)

    Use ruler and parallel tool to construct the center of a given circle.


    Draw two parallel chords \([XX']\) and \([YY']\). Set \(Z = (XY) \cap (X'Y')\) and \(Z' = (XY') \cap (X'Y)\). Note that \((ZZ')\) passes thru the center.

    Repeat the same construction for another pair of parallel chords. The center lies in the intersection of the obtained lines.

    Note that the shear map (described in Exercise 14.1.2(a)) can change angles between lines almost arbitrarily. This observation can be used to prove the impossibility of some constructions; here is one example:

    Exercise \(\PageIndex{4}\)

    Show that with a ruler and a parallel tool one cannot construct a line perpendicular to a given line.


    Assume a construction produces two perpendicular lines. Apply a shear map that changes the angle between the lines (see Exercise 14.1.2(a)).

    Note that it transforms the construction to the same construction for other free choices points. Therefore, this construction does not produce perpendicular lines in general. (It might produce a perpendicular line only by a coincidence.)

    This page titled 14.2: Constructions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.