MISSION · Geometry
Constructions with Compass and Straightedge
Two tools, no measuring, and every arc carries a proof with it
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The idea in one sentence
A compass draws every point at one fixed distance from a centre, so an arc is not a decoration — it is a statement about distance, and two arcs crossing pin down a point that satisfies two statements at once.
The rules of the game
You get exactly two tools. A compass draws circles and arcs. A straightedge draws lines through two points, and has no marks on it. You may not measure a length or an angle. You may not use the ruler’s numbers even if it has some.
This looks like a pointless restriction until you notice what it buys. A measured bisector is right to within the width of a pencil line. A constructed one is exactly right, and the construction itself is the proof of why.
Perpendicular bisector of a segment
To cut \(AB\) in half at a right angle:
- Open the compass wider than half of \(AB\) — any width past halfway works.
- With the point on \(A\), draw an arc above and below the line.
- Without changing the width, do the same from \(B\).
- Draw the line through the two crossing points.
Why it works. Both crossing points were drawn at the same radius from \(A\) and from \(B\), so each is equally distant from the two ends. The set of all points equidistant from \(A\) and \(B\) is exactly the perpendicular bisector, so two such points determine it. Join the crossings to \(A\) and \(B\) and you get two triangles congruent by SSS, which forces the right angle.
Keeping the compass width fixed in step 3 is the whole construction. Change it, and the crossings are no longer equidistant, and the line is no longer a bisector.
Bisecting an angle
To cut angle \(B\) in half:
- Centre on the vertex and draw an arc across both arms, marking a point on each.
- From each of those two points, draw arcs of equal radius that cross inside the angle.
- Draw the line from the vertex through the crossing.
Why it works. Step 1 puts two points the same distance from the vertex. Step 2 puts the crossing the same distance from each of those. The triangle on one side of your new line has three sides matching the triangle on the other, so SSS again, and the two angles at the vertex must be equal.
Copying a segment and an angle
These two are the building blocks everything else assembles from.
Copying a segment is one compass setting: open it to \(AB\), move to your new starting point, mark it off.
Copying an angle takes the SSS idea once more. Draw an arc across both arms of the original and the same arc from the new vertex. Measure the gap between the two crossings on the original with your compass, then step that gap off on the new arc. Both triangles now have three matching sides, so the angles match.
What can and cannot be built
The classical questions stood open for two thousand years and were all settled in the 1800s, all in the negative. You cannot trisect a general angle, double a cube, or square a circle with these two tools. A regular 17-gon, surprisingly, you can — Gauss was 18 when he found it.
The reason is algebraic rather than geometric. Each new intersection you construct solves a linear or quadratic equation, so every constructible length is built from whole numbers by \(+, -, \times, \div\) and square roots. Trisecting an angle needs a cube root, which never appears on that list.
Common mix-ups
- Changing the compass width mid-construction. Steps say “without changing the width” because the equal radii are what carries the proof.
- Arcs too short to cross. Open past halfway and draw generously; a construction with no intersection has nothing to join.
- Erasing the arcs. They are the working. Leave them showing.
- Measuring instead of constructing. A protractor answer scores nothing on a construction question, however accurate.
- Confusing a perpendicular bisector with a median. Both hit the midpoint; only the bisector is guaranteed perpendicular, and a median goes to the opposite vertex instead.
Check yourself
What is the set of all points equidistant from \(A\) and \(B\), which congruence rule justifies the perpendicular bisector, and what do you get by bisecting a right angle twice?
Equidistant from both ends means the perpendicular bisector of \(AB\). The equal compass radii give three matching sides, so SSS. And \(90 \to 45 \to 22.5\) degrees.
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