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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply