Ruler-and-compasses constructions · Construction & loci

A construction is an accurate drawing made with a ruler (only to draw straight lines and measure lengths) and a pair of compasses. “Using a ruler and a pair of compasses only” means no protractor: every angle must be built from arcs. Keep a sharp pencil, and leave every arc showing: the arcs are your working, and they earn marks.

Everything is built from a few basic constructions:

Bisect a lineEqual arcs from A and B; join the crossings

Bisect an angleArc from O, then equal arcs from C and D

Construct 60°Two arcs of the same radius make an equilateral triangle

Drop a perpendicularArc from P cuts the line; equal arcs from Q and R meet at X

Draw a parallelPS = QR and RS = QP make a parallelogram

Try it

Bisect a lineStep through with the buttons

1 of 4stepThe linenow

Start with the line AB.

Pick a construction and step through it. The gold arcs are the step you’ve just taken. The last step says why it works: each one rests on equal radii, so the triangles it makes are equilateral, isosceles or congruent.

Building other angles

Every angle you’ll be asked for comes from $60^\circ$ and $90^\circ$ by adding, and by bisecting:

$60^\circ$ : the basic construction. $120^\circ$ : two $60^\circ$ angles side by side.
$30^\circ$ : bisect $60^\circ$ . $15^\circ$ : bisect $30^\circ$ .
$90^\circ$ : construct $60^\circ$ and $120^\circ$ , then bisect the $60^\circ$ between them; or construct the perpendicular at a point by bisecting a line through it.
$45^\circ$ : bisect $90^\circ$ . $135^\circ$ : $90^\circ + 45^\circ$ .
$75^\circ$ : bisect the angle between $60^\circ$ and $90^\circ$ . $105^\circ$ : bisect the angle between $90^\circ$ and $120^\circ$ .

Dividing a line in a ratio

To divide a line $BC$ in the ratio $m : n$ , draw a second line from $B$ and step off $m + n$ equal lengths along it with the compasses. Join the last mark to $C$ , then draw a line parallel to that join through mark $m$ . It cuts $BC$ in the ratio $m : n$ .

Divide a line in a ratio5 equal steps and a parallel line divide BC in the ratio 3 : 2

Worked example · WAEC 2025

WAEC 2025 · Paper 2 · Q9

Using a ruler and a pair of compasses only, construct $\triangle ABC$ with $|AB| = 7.5\text{ cm}$ , $|BC| = 8.1\text{ cm}$ and $\angle ABC = 105^\circ$ .

Locate $D$ on $BC$ such that $|BD| : |DC| = 3 : 2$ , and through $D$ construct the line perpendicular to $BC$ .

If the perpendicular meets $AC$ at $P$ , measure $|BP|$ (cm).

Sketch, then draw BC
Make a rough sketch with the lengths and the angle marked. Then draw $BC = 8.1$ cm with the ruler: the $105^\circ$ angle is at $B$ , one end of it.
Think first. Which side should you draw first, and why?
(a) The 105° angle at B
Construct $90^\circ$ and $120^\circ$ at $B$ , then bisect the angle between them: $90^\circ + 15^\circ = 105^\circ$ .
Think first. How do you build 105° from 60° and 90°?
Mark A and finish the triangle
Open the compasses to 7.5 cm and, with centre $B$ , cut the $105^\circ$ arm at $A$ . Join $AC$ .
Think first. How do you get AB = 7.5 cm without a protractor?
(b) Divide BC in the ratio 3 : 2
Draw a line from $B$ and step off $3 + 2 = 5$ equal lengths. Join the 5th mark to $C$ , and through the 3rd mark draw a parallel line to cut $BC$ at $D$ . (Check: $BD = \frac35 \times 8.1 = 4.86$ cm.)
Think first. How many equal steps do you need along the extra line?
The perpendicular at D
With centre $D$ , mark two points on $BC$ the same distance either side, then bisect the line between them. The bisector is the perpendicular to $BC$ at $D$ .
Think first. D is on the line. How do you construct a right angle there?
(c) Measure
It meets $AC$ at $P$ . Measure $|BP| \approx 5.4$ cm.
Think first. Where does the perpendicular meet AC?

Your turn

WAEC 2020 · Paper 2 · Q12 (b)

(b)
(i) Using a ruler and a pair of compasses only, (α) construct a triangle $ABC$ in which $|AB| = 8\text{ cm}$ , $|BC| = 9\text{ cm}$ and $\angle ABC = 75^\circ$ ; (β) locate the point $P$ inside $ABC$ such that $|PA| = |PB|$ and $|PA| = 4.5\text{ cm}$ . (ii) Measure: (α) $|CP|$ ; (β) $\angle ACB$ .
Model answer
Draw $AB = 8$ cm, construct $75^\circ$ at $B$ ( $60^\circ$ plus half of the next $30^\circ$ ) and mark $C$ with $|BC| = 9$ cm. $P$ lies on the perpendicular bisector of $AB$ , where an arc of radius $4.5$ cm centred at $A$ cuts it inside the triangle. Measuring gives $|CP| \approx 6.8$ cm and $\angle ACB \approx 48^\circ$ .

Try it on a graph

The accurate construction: A(0, 0), B(8, 0), C(5.67, 8.69), P(4, 2.06).

Worked solution (try it first)

(b)(i)

(α) Draw $|BC| = 9$ cm, construct an angle of $75^\circ$ at $B$ (a $60^\circ$ angle plus half of the $30^\circ$ next to it), and mark $A$ on its arm with $|AB| = 8$ cm.
Join $AC$ .

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