How can you halve a segment when the ruler has no markings? You need not guess: circles specify equal distances, and their intersections determine the required line.

A geometric construction is a finite sequence of permitted steps, together with a proof of the result. An unmarked straightedge draws lines; a compass draws circles and transfers a given length as a radius. You will use triangle congruence criteria and geometric loci.

First, agree on the tools

We do not read centimetres from the straightedge or use its two edges as ready-made parallel lines. A line requires two distinct points already constructed. A circle requires a constructed centre and a radius length. In the school model, the compass can retain its opening to transfer a segment. Even the stricter classical model permits length transfer through auxiliary constructions. Euclid I.2, AMSI.

A physical drawing has errors. The proof concerns an ideal model: radii and theorems justify equality, rather than pencil accuracy.

A midpoint and a right angle in one algorithm

Let A≠B. The solution has four parts.

  1. Analysis. We seek the perpendicular bisector: its points are equidistant from A and B. Intersections of equal circles supply two such points.
  2. Construction. Draw circles centred at A and B, each with radius AB. Label their intersections P and Q, and draw PQ. Let M be the intersection of PQ and AB.
  3. Proof. PA=PB and QA=QB, as radii of equal circles. P and Q lie on the perpendicular bisector, so PQ is that line. Thus AM=MB and PQ⊥AB.
  4. Discussion. There are two circle intersections because AB<2AB. They determine one line PQ and one midpoint M. The chosen radius matters: for two equal circles, it must exceed AB/2.

1 · Two circles of radius AB

AB

2 · Two intersections P and Q

ABPQ

3 · Line PQ and midpoint M

ABPQM

Angle bisector: circles through O and Q

OUVQ

SSS: two positions, one shape

ABC₁C₂234
Uniform scale within each panel; lengths in arbitrary units. In the first three AB=2 and both circle radii equal AB. The steps add intersections, then line PQ and M. The bisector example uses 60°: OU=OV=UV, auxiliary circles meet at O and Q, and Q≠O is chosen. Last: base 4, radii 2 and 3; C₁ and C₂ are reflected positions.

The auxiliary circles remain visible so you can trace where P and Q come from. Checking with a tape measure would not replace the third step.

An angle bisector without a protractor

From vertex O of a given nonzero angle smaller than 180°, draw an arc of any radius. It meets the sides at U and V, with OU=OV. Draw circles centred at U and V, both with radius UV. Choose a common point Q inside the angle, distinct from O, and draw ray OQ.

OU = OV, UQ = VQ, OQ = OQ.

Triangles OUQ and OVQ are congruent by SSS, so the angle is bisected. In the illustrated 60° example, one auxiliary intersection is O; choose the other one, Q. Euclid I.9.

Copy the data and construct a triangle

To mark length AB from P along a given ray, draw a circle centred at P with radius AB. Its intersection with the ray gives the other endpoint.

To copy an angle, cut off two points on its sides with an arc of one radius. At the new vertex, repeat that radius and the chord between the two points. The third side of the reconstructed triangle fixes the angle by SSS. Choose the side of the new ray as required. Euclid I.23.

For SSS, lay off base c and intersect circles of radii a and b from its endpoints. For SAS, copy two sides and the angle between them. For ASA, lay off a side and copy the adjacent angles; their rays meet at the third vertex. Here a given angle is a geometric object to copy, not a demand to construct any arbitrary number of degrees.

What counts as a solution?

With base 4 and radii 2 and 3, the third vertex has two mirror positions. These are two constructions relative to the fixed base, but one congruence class of triangles. Specifying a half-plane leaves one position.

With base 5, those circles are tangent. One point lies at the required distances, but all three vertices are collinear: there is no nondegenerate triangle. With base 6, there is no common point at all. Count circle intersections and admissible figures separately. Euclid I.22.

Extra factOptional · you may skip this

These tools cannot divide every angle as requested

You can trisect 90°: construct an equilateral triangle and halve its 60° angle to obtain 30°. But an exact trisection of a given 60° angle into three 20° angles is impossible with unmarked straightedge and compass alone. This is a separate algebraic theorem about constructible numbers, rather than a failed drawing attempt. Commentary on Euclid I.9.

A marked ruler or another tool changes the rules. The basic constructions in this lesson do not require the impossibility theorem.