Straightedge Only!

Given a segment, construct a perpendicular line to the segment using only the straightedge.

#Geometry #Construction

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Michael Mendrin - 6 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$