Let's assume that perpendiculars can be drawn to any segment using a straightedge only. Given a circle, it becomes a trivial matter in locating the center of that circle, by first drawing two arbitrary chords, and then finding the midpoints of the chords (which is possible using a straightedge only and not using the circle), and then drawing the perpendiculars through them. But this is an impossibility as per Jacob Steiner's Theorem. Hence construction of a perpendicular to any line segment using a straightedge only is an impossibility.

However, by cheating, using a double edged straightedge, such as a common ruler, one can easily draw a rhombus, which diagonals are perpendicular.

As a note, given a line segment and any point not on it, a line through this point and parallel to the line segment can be constructed using a straightedge only. Unfortunately, one still cannot construct a true rhombus with this means alone.

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TopNewestLet's assume that perpendiculars can be drawn to any segment using a straightedge only. Given a circle, it becomes a trivial matter in locating the center of that circle, by first drawing two arbitrary chords, and then finding the midpoints of the chords (which is possible using a straightedge only and not using the circle), and then drawing the perpendiculars through them. But this is an impossibility as per Jacob Steiner's Theorem. Hence construction of a perpendicular to any line segment using a straightedge only is an impossibility.

However, by cheating, using a double edged straightedge, such as a common ruler, one can easily draw a rhombus, which diagonals are perpendicular.

As a note, given a line segment and any point not on it, a line through this point and parallel to the line segment can be constructed using a straightedge only. Unfortunately, one still cannot construct a true rhombus with this means alone.

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