Basic Ruler and Compass Constructions Revision Notes | GCSE AQA Higher Maths

1. Why Constructions Matter

A ruler and a pair of compasses are surprisingly powerful tools. With nothing else, you can split a line exactly in half, drop a perfect right angle from any point, halve any angle, and produce an exact 60° — all without a protractor.

examiners test these skills directly: they ask you to construct, which means draw accurately with instruments and leave every arc visible. The arcs are your working — they prove you used the correct method rather than guessing with a protractor.

There are four constructions you need to master: the perpendicular bisector of a line segment, a perpendicular from a point to a line, a perpendicular at a point on a line, and the bisector of an angle (which also unlocks the 60° construction). Each one follows a clear sequence of steps — learn the sequence and the construction takes care of itself.

2. The Perpendicular Bisector of a Line Segment

The perpendicular bisector does two things at once: it finds the midpoint of a line segment and crosses it at exactly 90°. The key insight is that any point equidistant from both endpoints of a segment lies on its perpendicular bisector — and that is exactly what the compass arcs locate.

Worked example: Constructing the Perpendicular Bisector of AB (6 cm)

Line segment AB has length 6 cm. Construct its perpendicular bisector and mark the midpoint M.

1. Draw AB = 6 cm with a ruler. — This is the segment you are bisecting. Label both endpoints clearly.
2. Open compasses to approximately 4 cm (more than half of 6 cm). — The radius must be greater than half of AB, otherwise the arcs will not cross. [M1 — correct method setup]
3. Compass point on A. Draw one arc above AB and one arc below AB. — Keep the radius fixed throughout. These arcs mark all points 4 cm from A.
4. Compass point on B, same radius. Draw arcs above and below AB, crossing the arcs from step 3. — The crossing points are equidistant from A and B — that is the geometric property that makes this work. [M1 — arcs from both endpoints]
5. Rule a straight line through the two intersection points. — This line is perpendicular to AB and passes through its midpoint. [A1 — accurate bisector with arcs visible]
6. Label M where the bisector crosses AB. Check: AM = MB = 3 cm. — A quick measurement confirms the bisector is accurate. The midpoint should be exactly 3 cm from each end.

Answer: The perpendicular bisector passes through M (the midpoint of AB at 3 cm) and crosses AB at 90°. All arcs must remain visible.

The perpendicular bisector is the foundation for the next two constructions. Once you understand that crossing arcs locate a point equidistant from two fixed points, the logic behind perpendiculars from and at a point becomes straightforward.

3. Constructing a Perpendicular to a Line

There are two versions of this construction, and it is worth being clear about which is which. The first is a perpendicular from a point that sits off the line — you are dropping a right angle down to the line. The second is a perpendicular at a point that already sits on the line — you are raising a right angle up from it.

Both versions use the same underlying idea as the perpendicular bisector: find two points on the line that are equidistant from your given point, then bisect the segment between them. The difference is only in where you place the initial compass arcs.

3b. Perpendicular at a Point on a Line

Both perpendicular constructions are really the perpendicular bisector in disguise. Recognising that pattern means you only need to learn one core technique and apply it in two slightly different setups.

4. Bisecting a Given Angle

Bisecting an angle means splitting it into two equal halves. The method uses the same crossing-arcs logic: find two points equidistant from the vertex (one on each arm of the angle), then find a point equidistant from both of those — and that point lies on the bisector.

Worked example: Bisecting Angle ABC

Angle ABC is drawn with vertex at B. Construct the bisector of angle ABC.

1. Place compass point on B. Draw an arc that crosses both arms of the angle. Label the crossing points P (on arm BA) and Q (on arm BC). — This arc ensures P and Q are the same distance from B — that equal distance is what makes the construction fair. [M1 — arc from vertex crossing both arms]
2. Place compass point on P. Open to a radius greater than half of PQ. Draw an arc in the interior of the angle. — You are now looking for a point equidistant from P and Q — exactly like finding the perpendicular bisector of PQ.
3. Place compass point on Q, same radius. Draw an arc in the interior of the angle, crossing the arc from step 2. Label the intersection R. — R is equidistant from P and Q, which means it is equidistant from both arms of the angle. [M1 — crossing arcs from P and Q]
4. Rule a line from B through R and extend it as a ray. — This ray bisects angle ABC. Both halves are equal. [A1 — accurate bisector with all arcs visible]

Answer: Ray BR bisects angle ABC. Each half-angle equals half of the original angle ABC. All arcs must remain visible.

5. Constructing an Angle of 60°

Here is where the constructions connect to something elegant. An equilateral triangle has all sides equal and all angles equal to 60°. If you construct an equilateral triangle using compasses, you automatically produce a 60° angle — no protractor needed.

The method works because when two circles of the same radius intersect, the triangle formed by their two centres and either intersection point is equilateral. That is the geometric guarantee behind the construction.

Worked example: Constructing a 60° Angle at Point A

Construct an angle of exactly 60° at point A on a line.

1. Draw a straight line and mark point A on it. Mark a second point B on the line to the right of A. — AB will become one side of the equilateral triangle. The length of AB sets the compass radius for the whole construction.
2. Open compasses so the radius equals the length AB. — This radius stays fixed for both arcs. Keeping it constant is what guarantees all three sides will be equal. [M1 — compass set to AB]
3. Compass point on A. Draw an arc above the line, roughly where the third vertex of the triangle will be. — Every point on this arc is exactly AB away from A.
4. Compass point on B, same radius. Draw an arc above the line, crossing the arc from step 3. Label the intersection C. — Point C is exactly AB away from both A and B, so triangle ABC is equilateral. [M1 — crossing arcs from A and B]
5. Rule a line from A through C. — Angle CAB = 60° because triangle ABC is equilateral. [A1 — accurate 60° angle with arcs visible]

Answer: Angle CAB = 60°. Triangle ABC is equilateral with all sides equal to AB. All construction arcs must remain visible.

Notice the link back to the angle bisector: if you bisect the 60° angle you have just constructed, you produce a 30° angle. Combining constructions like this is a common exam extension — and it flows naturally from the techniques you have already practised.

6. Practice: Apply the Methods

The best way to consolidate these constructions is to work through them in sequence, checking each one against the key accuracy markers. The problems below cover all four required constructions.

7. Exam Technique and Common Mistakes

Construction questions are among the most mark-efficient on the paper: the method is fixed, the steps are short, and the marks are predictable. The mistakes that cost students marks are almost always the same ones — and all of them are avoidable.