Mathematics · Plane geometry · Grade 7

Compass and straightedge

With an unmarked straightedge and a compass, how do you halve an angle or bisect a segment?

The situation

A carpenter has only a compass and an unmarked straightedge: he lays off equal segments, bisects an angle, drops a perpendicular to a board — without measuring, yet exactly. The ancient Greeks built all of geometry with these two tools. Why is the point where two arcs cross exactly the point we need? From the four basic constructions, how do we build a triangle, a parallelogram, a square and even a “mean proportional” segment?

Topic

Compass-and-straightedge construction: a segment, an angle, a bisector, a perpendicular, a triangle and a quadrilateral. A carpenter’s compasses and string: marking an equal segment, bisecting an angle, dropping a perpendicular, constructing a triangle from its three sides

Key ideas

Construction
Angle bisector
Perpendicular bisector

Formulas

P Q = A B

Equal segment: we set the compass to $A B$ and draw an arc from $P$ on the ray — because the setting is kept, $P Q = A B$ .

△ O M K = △ O N K \Rightarrow ∠ M O K = ∠ K O N

Angle bisector: $O M = O N$ , $M K = N K$ (equal arcs), $O K$ is common: by the third criterion (SSS) the triangles are congruent, so the angles are equal.

P A = P B, Q A = Q B \Rightarrow P Q ⟂ A B, A O = O B

Perpendicular bisector: the points $P$ , $Q$ where equal arcs from $A$ and $B$ cross are equally distant from $A$ and $B$ — the line $P Q$ is perpendicular to $A B$ and bisects it.

C H = \sqrt{a \cdot b}

Mean proportional: in a semicircle on the diameter $a + b$ , the perpendicular raised at the point where the two segments meet gives $C H^{2} = A H \cdot H B$ .

Theory

Tools and the rule

In a construction the compass does only two things: it draws a circle (an arc) with a given centre and radius, and it carries the length of a segment in its “setting”. The straightedge has no scale: it only draws a straight line through two points. There is no measuring — that is why the result is exact and provable. In Grade 3 you drew a circle and a six-petal pattern with a compass: if, without changing the setting, you step the point round the circle, the arcs fit exactly six times (the radius fits round the circle six times).

Equal segment, copying an angle, bisector

Equal segment: set the compass to $A B$ and draw an arc from $P$ on the ray — $P Q = A B$ . Copying an angle: an arc from $O$ cuts the sides at $M$ and $N$ ; with the same radius, an arc from $O_{1}$ gives $M_{1}$ ; set the compass to $M N$ and draw an arc from $M_{1}$ — this gives $N_{1}$ . $△ O M N$ and $△ O_{1} M_{1} N_{1}$ are congruent by three sides, so the angles are equal. Bisector: equal arcs from $M$ and $N$ cross at $K$ ; $△ O M K = △ O N K$ (third criterion) — $O K$ bisects the angle. The “why it works” of every construction is congruence of triangles.

Perpendicular and perpendicular bisector

Perpendicular from a point to a line: an arc from $P$ cuts the line at $M$ and $N$ ; equal arcs from $M$ and $N$ meet at $Q$ ; $P Q ⟂ l$ (both $P$ and $Q$ are equally distant from $M$ and $N$ ). Perpendicular at a point on a line: lay off equal segments on both sides of the point, then use the same method. Perpendicular bisector: equal arcs from $A$ and $B$ with radius greater than $\frac{A B}{2}$ cross at $P$ and $Q$ ; the line $P Q$ meets $A B$ at its midpoint $O$ at a right angle. This solves both bisecting a segment and finding its midpoint.

Triangles and quadrilaterals

A triangle is built from three elements — it is exactly the congruence criteria that tell us how much information is enough: three sides (crossing arcs), two sides and the included angle (copy the angle and lay off the side on the ray), a side and two angles (two rays crossing), a leg and the hypotenuse (a perpendicular, the leg, then an arc with the hypotenuse as radius — Grade 8). A parallel line — by copying a corresponding angle; a parallelogram — an angle, two sides and the equality of opposite sides; a square — a perpendicular and four equal sides.

Advanced constructions

Mean proportional segment (Grade 9): lay $A H = a$ and $H B = b$ on one line, draw a semicircle on $A B$ as diameter, and the perpendicular at $H$ cuts the semicircle at $C$ . The angle in a semicircle is a right angle, and in a right-angled triangle the altitude to the hypotenuse satisfies $C H^{2} = A H \cdot H B$ — so $C H = \sqrt{a b}$ : for $a = 2$ , $b = 8$ we get $C H = 4$ . Homothety (a similar polygon): draw rays from the centre $O$ through the vertices and lay off the distance $k$ times on each — the new polygon is similar to the given one with ratio $k$ .

About the model

On the screen the arcs are drawn with the compass, and you place the points: when a point comes near an intersection it “snaps” to it (on a real drawing this is marking exactly with the pencil tip). The straightedge has no scale: lengths are carried only with the compass; the numbers in the measurement fields are for checking the drawing. On a real drawing each construction gives an error of ± 0.5 mm; on the screen the calculation is exact. Not every problem can be solved with compass and straightedge: trisecting an angle or squaring the circle is impossible (this was proved in the 19th century).

Did you know?

The first three postulates of Euclid's “Elements” describe exactly the work of the straightedge and compass: drawing a straight line through two points, extending it, and drawing a circle with a given centre and radius.
The French mathematician Pierre Wantzel proved in 1837 that trisecting an angle and doubling the cube are impossible with compass and straightedge; that squaring the circle is impossible became certain in 1882 (Lindemann, the transcendence of π).

Inside the experiment

The topic is explained step by step with animation, from simple to advanced: from a real-life picture to the formula and the proof. Then you drag the points yourself to check. Inside are missions, a quiz and a guided tour.

More experiments in this section

Point, segment, angle — Grade 7
Congruent triangles — Grade 7
Parallel lines — Grade 7
Angles of a triangle — Grade 7
Pythagoras’ theorem — Grade 8