×

# Straightedge Only!

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

Note by John Ashley Capellan
2 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 2 years, 11 months ago