What are geometric constructions in the context of Cambridge IGCSE Mathematics?

Geometric constructions in Cambridge IGCSE Mathematics refer to the process of creating precise figures using only a compass and a straightedge. This method emphasizes understanding the properties and relationships of geometric shapes, rather than relying on measurements or calculations.

Why are geometric constructions important in learning geometry?

Geometric constructions are important because they help students understand the fundamental principles of geometry. By constructing shapes, students gain insight into the properties and relationships of geometric figures, which enhances their spatial reasoning and problem-solving skills, crucial for the Cambridge IGCSE curriculum.

What tools are typically used in geometric constructions, and why?

The primary tools used in geometric constructions are a compass and a straightedge. These tools are used because they allow for the creation of precise geometric figures without relying on measurements, which helps students focus on the properties and relationships inherent in the shapes themselves.

Can you give an example of a basic geometric construction task?

A basic geometric construction task might involve constructing the perpendicular bisector of a line segment. This involves using a compass to draw arcs from each endpoint of the segment, ensuring the arcs intersect, and then drawing a straight line through the points of intersection, which bisects the original segment at a right angle.

How do geometric constructions differ from drawing or sketching in geometry?

Geometric constructions differ from drawing or sketching because they require precision and the use of specific tools like a compass and straightedge. Unlike freehand drawing, constructions follow strict rules and procedures to ensure accuracy and are used to explore geometric properties rather than simply creating a visual representation.

Why is "Erasing the construction arcs" a common mistake in Geometric Constructions?

Why it happens: Students think a clean diagram looks better. How to avoid it: Construction arcs are the EVIDENCE of the method — leave them visible. Source: 0580/42 — every series

Why is "Using a protractor instead of compasses" a common mistake in Geometric Constructions?

Why it happens: Quicker. How to avoid it: Cambridge mark schemes specifically require compass constructions. A protractor-drawn perpendicular gets zero marks for the construction.

Why is "Using different compass widths for the two arcs" a common mistake in Geometric Constructions?

Why it happens: Forgetting to keep the compass at the same setting. How to avoid it: Set the compass once and don't change it until both arcs are drawn.

Why is "Drawing only one point for a locus that should be a curve or line" a common mistake in Geometric Constructions?

Why it happens: Treating "locus" as one specific point. How to avoid it: A locus is the COMPLETE set of points that satisfy the condition. Usually a line, circle or arc.

Cambridge IGCSE Mathematics (0580)

Geometric Constructions

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Short Notes - Geometric Constructions

Detailed Study Notes

Detailed notes on Geometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Geometric Constructions and Loci — Cambridge IGCSE 0580 Maths Extended (2026)

Ruler-and-compasses constructions for perpendicular bisectors, angle bisectors, perpendiculars, plus loci — sets of points satisfying given conditions. Cambridge wants visible construction arcs, not erased.

At a glance

Tools: STRAIGHT EDGE (ruler) + COMPASSES only. NO protractor.
Perpendicular bisector: equal arcs from each endpoint, line through intersections.
Angle bisector: equal arcs cutting both rays of the angle, then equal arcs from those crossings.
Perpendicular from a point to a line: drop two arcs from the point to the line, then bisect.
Always leave construction arcs visible — they're part of the marked answer.
Locus = set of points satisfying a condition.
Locus equidistant from two points = perpendicular bisector.
Locus equidistant from two lines = angle bisector.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E4.6 — Use ruler and compasses for the construction of: perpendicular bisector of a line segment, angle bisector, perpendicular from a point to a line, perpendicular at a point on a line. Construct simple loci.

Perpendicular bisector of a line segment

Equal arcs from each endpoint, line through their intersections.

Construction.

Set compasses to MORE than half the segment $AB$ .
From $A$ , draw an arc above and below the segment.
From $B$ with the SAME compass setting, draw arcs that intersect the first ones at two points.
Draw a straight line through those two intersections — that's the perpendicular bisector.

Equal-radius arcs from each endpoint cross at two points; the line through them is the perpendicular bisector.

Properties. The perpendicular bisector:

Passes through the midpoint of $AB$ .
Is perpendicular to $AB$ .
Every point on it is equidistant from $A$ and $B$ .

Mark scheme. Cambridge expects:

Arcs visible from BOTH endpoints (showing equal radius).
The perpendicular line drawn through the intersections.

Tip. Don't erase the construction arcs. They're part of the answer.

Open compass to MORE than half $AB$ .
Draw arcs from BOTH endpoints with same setting.
Connect the two intersection points.
Leave arcs visible.

Angle bisector

Equal arcs from the vertex cut both rays. Equal arcs from those crossings meet on the bisector.

Construction.

From the vertex $V$ of the angle, draw an arc that crosses BOTH rays. Call those intersections $P$ and $Q$ .
From $P$ with the same (or any chosen) compass setting, draw an arc inside the angle.
From $Q$ with the SAME setting, draw another arc inside the angle. Let them meet at $X$ .
Draw the line $VX$ — that's the angle bisector.

An arc from V marks P and Q; equal arcs from P and Q meet at X; line VX bisects the angle.

Properties. The angle bisector:

Cuts the angle into two EQUAL halves.
Every point on it is equidistant from the two rays of the angle.

Mark scheme. Cambridge wants:

Arcs from the vertex showing where the rays were cut.
Arcs from $P$ and $Q$ meeting at $X$ — visible.
The bisector line drawn from $V$ through $X$ .

Arc from vertex cuts both rays at $P$ and $Q$ .
Equal arcs from $P$ and $Q$ meet at $X$ .
Bisector is line through $V$ and $X$ .
Leave arcs visible.

Perpendicular from a point to a line

Drop two arcs from the point onto the line, then perpendicular-bisect them.

Construction (point $P$ off the line $\ell$ ).

From $P$ , draw an arc that crosses $\ell$ at TWO points $A$ and $B$ .
Construct the perpendicular bisector of $AB$ using the standard method.
The perpendicular bisector passes through $P$ — it's the perpendicular from $P$ to $\ell$ .

Arc from P marks A and B on the line; bisecting AB gives the perpendicular from P, meeting ℓ at 90°.

Construction (point $P$ ON the line).

From $P$ , draw arcs that cross $\ell$ on either side at $A$ and $B$ (equal distances).
Construct the perpendicular bisector of $AB$ . It passes through $P$ and is perpendicular to $\ell$ .

Off the line: arc from $P$ cuts line at $A, B$ . Bisect $AB$ .
On the line: arcs from $P$ on both sides cut line at $A, B$ . Bisect $AB$ .
Either way: end with a perpendicular bisector construction.

Loci

A locus is a set of points satisfying a condition. Constructions reveal the standard ones.

A locus (plural: loci) is the set of all points that satisfy a given geometrical condition.

Standard loci.

Condition	Locus
Equidistant from a fixed point $P$	Circle centred at $P$
Equidistant from two points $A, B$	Perpendicular bisector of $AB$
Equidistant from two intersecting lines	Pair of angle bisectors
At a fixed distance $r$ from a line	Two parallel lines at distance $r$
Inside a region (e.g. "within 5 cm of $A$ ")	A disc / region

Method for combined loci.

Draw each locus separately on the same diagram.
Identify the OVERLAP (intersection) — that's the locus satisfying ALL conditions.
Shade the answer region (often Cambridge wants a shaded area).

Worked. Find the locus of points within $4$ cm of $A$ AND closer to $B$ than to $C$ .

Locus 1: disc of radius $4$ cm centred at $A$ .
Locus 2: half-plane containing $B$ , separated by the perpendicular bisector of $BC$ .
Final: the part of the disc lying in that half-plane.

The answer is the overlap: inside the circle AND on the B side of the perpendicular bisector.

Locus = set of points satisfying a condition.
Equidistant from a point: circle.
Equidistant from two points: perpendicular bisector.
Equidistant from two lines: angle bisector.
At fixed distance from a line: parallel lines.
Combined loci: overlap of individual loci.

Quick recap

Tools: ruler + compass only. No protractor.
Perpendicular bisector: equal arcs from each endpoint, line through intersections.
Angle bisector: equal arcs from vertex cut rays, then equal arcs from those.
Perpendicular from a point: drop two arcs to the line, bisect.
Locus: set of points satisfying a condition.
Combined loci: overlap.
ALWAYS leave construction arcs visible.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Perpendicular bisector — line through midpoint at $90\degree$ ; locus equidistant from two endpoints.
Angle bisector — line cutting an angle into two equal halves; locus equidistant from the two rays.
Locus — the set of all points satisfying a given condition.

How it’s examined

Constructions appear most years on Paper 4 as a 4-6 mark item — typically asking for the perpendicular bisector or angle bisector, then identifying a locus on the same diagram. Examiner reports flag missing or erased construction arcs as the recurring failure. The arcs are PART of the answer.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E4.6); 0580/42 Oct/Nov 2024 — Q12 (locus + bisector); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Geometric Constructions

Step-by-step solutions to past-paper-style questions on geometric constructions, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Geometric Constructions

Almost every geometric constructions exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Graph or diagram

Recognise it by

An instruction to construct, bisect, draw or sketch the locus / shade the region — a ruler-and-compasses task, never a numerical "find".

How to approach it

Use the matching standard construction: equal arcs from each end for a perpendicular bisector; an arc cutting both arms then equal arcs for an angle bisector; an equilateral triangle for $60\degree$ . A locus is a complete line, circle or arc — translate "equidistant from two points" to a perpendicular bisector and "equidistant from two lines" to an angle bisector, then take the intersection of loci for a region.

Common trap

Erasing the construction arcs, using a protractor instead of compasses, or changing the compass width between paired arcs. Examiner reports flag all three — the visible equal arcs are the evidence the method scores on.

1Construct the perpendicular bisector of a segment
CoreGraph or diagram• perpendicular bisector
Question
Construct the perpendicular bisector of segment $AB$ .
Step-by-step solution
1. Step 1
  Open compass to more than half $AB$ .
2. Step 2
  From $A$ , draw an arc above and below the line.
3. Step 3
  From $B$ with the SAME radius, draw two more arcs — they intersect the first ones at two points.
4. Step 4
  Join those two intersection points. That line is the perpendicular bisector.
Answer
Two pairs of equal arcs from $A$ and $B$ ; the line through their intersections.
Examiner tip
Mark schemes require visible construction arcs — never erase them. The arcs are the EVIDENCE that you used compasses.
2Construct the bisector of an angle
CoreGraph or diagram• Adapted from 0580/42 May/Jun 2024 Q5• angle bisector
Question
Bisect $\angle BAC$ .
Step-by-step solution
1. Step 1
  From $A$ , draw an arc cutting both $AB$ and $AC$ — call those points $P$ and $Q$ .
2. Step 2
  From $P$ and $Q$ with the SAME radius (or any radius), draw two arcs that intersect inside the angle at $X$ .
3. Step 3
  Draw line $AX$ — it bisects $\angle BAC$ .
Answer
Line $AX$ where $X$ is the intersection of equal arcs from $P$ and $Q$ .
3Drop a perpendicular from a point to a line
ExtendedGraph or diagram• perpendicular from point
Question
Construct the perpendicular from point $P$ (off the line) to line $\ell$ .
Step-by-step solution
1. Step 1
  From $P$ , draw an arc cutting $\ell$ at two points $A$ and $B$ .
2. Step 2
  Construct the perpendicular bisector of segment $AB$ — it passes through $P$ .
Answer
Perpendicular bisector of $AB$ where $A, B$ are arc-intersections from $P$ .
4Find a locus
ExtendedGraph or diagram• locus
Question
Sketch the locus of points equidistant from two parallel lines $4$ cm apart.
Step-by-step solution
1. Step 1
  Equidistant from two parallel lines = a line parallel to both, exactly midway between them.
Answer
A line parallel to both, $2$ cm from each.
5Construct an angle of $60\degree$
ExtendedGraph or diagram• Adapted from 0580/22 Oct/Nov 2023 Q11• 60 degrees, equilateral
Question
Construct an angle of exactly $60\degree$ at point $A$ on a given line.
Step-by-step solution
1. Step 1
  From $A$ , draw an arc of any convenient radius $r$ that cuts the line at a point $B$ .
2. Step 2
  Without changing the compass, from $B$ draw another arc of the same radius $r$ . It cuts the first arc at a point $C$ .
3. Step 3
  Triangle $ABC$ has all three sides equal to $r$ , so it is equilateral — therefore $\angle BAC = 60\degree$ .
4. Step 4
  Draw the ray $AC$ — this gives the $60\degree$ angle at $A$ .
Answer
Equilateral construction: arcs of equal radius from $A$ and from a point $B$ on the line intersect at $C$ ; $\angle BAC = 60\degree$ .
Examiner tip
The examiner report flags candidates who measure $60\degree$ with a protractor — full marks require the compass-and-equilateral method, with all construction arcs left visible.
6Construct a triangle given three sides (SSS)
ExtendedGraph or diagram• SSS, triangle construction
Question
Construct triangle $ABC$ with $AB = 7\,\text{cm}$ , $BC = 5\,\text{cm}$ and $AC = 6\,\text{cm}$ .
Step-by-step solution
1. Step 1
  Draw segment $AB = 7\,\text{cm}$ with a ruler.
2. Step 2
  From $A$ , draw an arc of radius $6\,\text{cm}$ above $AB$ .
3. Step 3
  From $B$ , draw an arc of radius $5\,\text{cm}$ . The two arcs intersect at $C$ .
4. Step 4
  Join $AC$ and $BC$ . Triangle $ABC$ is complete.
Answer
Draw $AB = 7\,\text{cm}$ , then arcs of radius $6\,\text{cm}$ from $A$ and $5\,\text{cm}$ from $B$ intersect at $C$ .
Examiner tip
The examiner report flags candidates whose arcs do not actually intersect — that is a sign of measurement error, since $5 + 6 > 7$ guarantees an intersection by the triangle inequality.
7Construct a perpendicular at a given point on a line
ExtendedGraph or diagram• perpendicular, construction
Question
Construct a line perpendicular to line $\ell$ passing through a point $P$ that lies ON $\ell$ .
Step-by-step solution
1. Step 1
  With centre $P$ , draw arcs cutting $\ell$ at $A$ and $B$ on either side of $P$ (so $PA = PB$ ).
2. Step 2
  Open the compass wider than $PA$ . From $A$ and $B$ with this new radius, draw two arcs that meet at $Q$ above (or below) the line.
3. Step 3
  Join $PQ$ — this line is perpendicular to $\ell$ at $P$ .
Answer
Line $PQ$ , where $Q$ is the intersection of arcs from $A$ and $B$ with $PA = PB$ .
8Locus of points a fixed distance from a point
ExtendedGraph or diagram• locus, circle
Question
Sketch the locus of points exactly $3\,\text{cm}$ from a fixed point $A$ .
Step-by-step solution
1. Step 1
  Points exactly $r$ units from a fixed point form a circle of radius $r$ centred at that point.
2. Step 2
  Set the compass to $3\,\text{cm}$ , place the point on $A$ , and draw the full circle.
Answer
A circle of radius $3\,\text{cm}$ with centre $A$ .
9Locus equidistant from two intersecting lines
ChallengeGraph or diagram• Adapted from 0580/42 May/Jun 2023 Q11• locus, angle bisector
Question
Two straight lines intersect at $X$ . Describe and construct the locus of points that are equidistant from both lines.
Step-by-step solution
1. Step 1
  The locus of points equidistant from two intersecting lines is the pair of angle bisectors of the angles formed at $X$ .
2. Step 2
  Construct the angle bisector of each of the two angle pairs at $X$ using the standard angle-bisector construction (arc from $X$ cutting both lines, then equal arcs from those intersection points meeting inside the angle).
3. Step 3
  The two bisectors are perpendicular to each other and together form the full locus.
Answer
Two perpendicular angle bisectors through $X$ — the pair of angle bisectors of the angles formed by the two lines.
Examiner tip
The examiner report flags candidates who give only one of the two bisectors. "Equidistant from two lines" gives BOTH bisectors — they are perpendicular to each other.
10Region satisfying multiple loci
ChallengeGraph or diagram• Adapted from 0580/42 Feb/Mar 2024 Q9• loci, region, shading
Question
A garden is rectangular $ABCD$ with $AB = 10\,\text{m}$ , $BC = 6\,\text{m}$ . A new flower bed must lie inside the garden, be within $4\,\text{m}$ of vertex $A$ , and be nearer to side $AB$ than to side $AD$ . Describe (with constructions) how to shade the region for the flower bed.
Step-by-step solution
1. Step 1
  Locus 1: "within $4\,\text{m}$ of $A$ " is the interior of a circle of radius $4\,\text{m}$ centred at $A$ . Construct the arc inside the rectangle.
2. Step 2
  Locus 2: "nearer to $AB$ than to $AD$ " is the side of the angle bisector of $\angle DAB$ that contains $AB$ . Since $\angle DAB = 90\degree$ , the bisector makes a $45\degree$ angle with both $AB$ and $AD$ .
3. Step 3
  Construct the angle bisector of $\angle DAB$ using the standard method.
4. Step 4
  The required region is the intersection: inside the quarter circle AND on the $AB$ -side of the angle bisector.
Answer
Shade the region bounded by: the arc of the circle (centre $A$ , radius $4\,\text{m}$ ) inside the rectangle; the angle bisector of $\angle DAB$ ; and the side $AB$ .
Examiner tip
The examiner report flags candidates who shade the wrong side of the angle bisector. "Nearer to $AB$ " means closer to $AB$ — visualise dropping perpendiculars from a test point to each side and choose the side where the perpendicular to $AB$ is shorter.

Key Formulae — Geometric Constructions

The formulae you need to memorise for geometric constructions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Perpendicular bisector property
$\text{Locus of points equidistant from } A \text{ and } B$
When to use
Building point-equidistant-from-two-points loci.
Angle bisector property
$\text{Locus of points equidistant from two lines}$
When to use
Point-equidistant-from-two-lines loci.

Key Definitions and Keywords — Geometric Constructions

Definitions to memorise and the exact keywords mark schemes credit for geometric constructions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Geometric construction
Examiner keyword
A drawing made using ONLY a straight edge (ruler) and compasses — protractors are NOT allowed.
Perpendicular bisector
Examiner keyword
A line that crosses a segment at $90\degree$ through its midpoint.
Angle bisector
Examiner keyword
A line that divides an angle into two equal angles.
Locus
Examiner keyword
The set of all points satisfying a given geometrical condition.

Common Mistakes and Misconceptions — Geometric Constructions

The traps other students keep falling into on geometric constructions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Erasing the construction arcs
0580/42 — every series
Why it happens
Students think a clean diagram looks better.
How to avoid it
Construction arcs are the EVIDENCE of the method — leave them visible.
✕Using a protractor instead of compasses
Why it happens
Quicker.
How to avoid it
Cambridge mark schemes specifically require compass constructions. A protractor-drawn perpendicular gets zero marks for the construction.
✕Using different compass widths for the two arcs
Why it happens
Forgetting to keep the compass at the same setting.
How to avoid it
Set the compass once and don't change it until both arcs are drawn.
✕Drawing only one point for a locus that should be a curve or line
Why it happens
Treating "locus" as one specific point.
How to avoid it
A locus is the COMPLETE set of points that satisfy the condition. Usually a line, circle or arc.

Geometric Constructions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Construction.

Set compasses to MORE than half the segment $AB$ .
From $A$ , draw an arc above and below the segment.
From $B$ with the SAME compass setting, draw arcs that intersect the first ones at two points.
Draw a straight line through those two intersections — that's the perpendicular bisector.

Equal-radius arcs from each endpoint cross at two points; the line through them is the perpendicular bisector.

Properties. The perpendicular bisector:

Passes through the midpoint of $AB$ .
Is perpendicular to $AB$ .
Every point on it is equidistant from $A$ and $B$ .

Mark scheme. Cambridge expects:

Arcs visible from BOTH endpoints (showing equal radius).
The perpendicular line drawn through the intersections.

Tip. Don't erase the construction arcs. They're part of the answer.

Condition

Locus

Equidistant from a fixed point

P

Circle centred at

P

Equidistant from two points

A, B

Perpendicular bisector of

AB

Equidistant from two intersecting lines

Pair of angle bisectors

At a fixed distance

r

from a line

Two parallel lines at distance

r

Inside a region (e.g. "within 5 cm of

A

A disc / region

Geometric Constructions

Short Notes - Geometric Constructions

Detailed Study Notes

At a glance

What you’ll learn

Perpendicular bisector of a line segment

Angle bisector

Perpendicular from a point to a line

Loci

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Geometric Constructions

1Construct the perpendicular bisector of a segment

2Construct the bisector of an angle

3Drop a perpendicular from a point to a line

4Find a locus

5Construct an angle of 60°60\degree60°

6Construct a triangle given three sides (SSS)

7Construct a perpendicular at a given point on a line

8Locus of points a fixed distance from a point

9Locus equidistant from two intersecting lines

10Region satisfying multiple loci

Perpendicular bisector property

Angle bisector property

Geometric construction

Perpendicular bisector

Angle bisector

Locus

✕Erasing the construction arcs

✕Using a protractor instead of compasses

✕Using different compass widths for the two arcs

✕Drawing only one point for a locus that should be a curve or line

Geometric Constructions — frequently asked questions

Geometric Constructions

Short Notes - Geometric Constructions

Detailed Study Notes

At a glance

What you’ll learn

Perpendicular bisector of a line segment

Angle bisector

Perpendicular from a point to a line

Loci

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Geometric Constructions

1Construct the perpendicular bisector of a segment

2Construct the bisector of an angle

3Drop a perpendicular from a point to a line

4Find a locus

5Construct an angle of 60°60\degree60°

6Construct a triangle given three sides (SSS)

7Construct a perpendicular at a given point on a line

8Locus of points a fixed distance from a point

9Locus equidistant from two intersecting lines

10Region satisfying multiple loci

Perpendicular bisector property

Angle bisector property

Geometric construction

Perpendicular bisector

Angle bisector

Locus

✕Erasing the construction arcs

✕Using a protractor instead of compasses

✕Using different compass widths for the two arcs

✕Drawing only one point for a locus that should be a curve or line

Geometric Constructions — frequently asked questions

5Construct an angle of $60\degree$

5Construct an angle of $60\degree$