Compass and Straightedge Constructions

The two tools

Tool	What it may do
compass	draw a circle or arc from a center; carry a length from one place to another
straightedge	draw a line through two points; it has no markings

No measuring is allowed. That restriction is the point: a figure made by construction is exact, because a compass carries a length itself rather than a reading of it.

Each construction below is followed by the reason it works. The steps are what you do; the reason is why you can trust the result, and it is always a theorem you have already proved.

Copying a segment

Bisecting a segment

Construct the perpendicular bisector of $\overline{AB}$.

1. Open the compass to more than half of $AB$.
2. Draw an arc centered at $A$, above and below the segment.
3. Keep the same opening. Draw an arc centered at $B$ crossing the first two.
4. Label the crossings $P$ and $Q$, and draw line $PQ$.

Line $PQ$ is perpendicular to $\overline{AB}$ and passes through its midpoint $M$.

Why it works: $P$ lies on both arcs, so $PA$ and $PB$ each equal the compass opening, and $PA = PB$. The same holds for $Q$. Points equidistant from $A$ and $B$ lie on the perpendicular bisector — the converse proved earlier — so both $P$ and $Q$ are on it. Two points determine a line, so line $PQ$ is the perpendicular bisector.

The opening must be more than half of $AB$. Less than that, and the arcs from $A$ and $B$ never meet.

Bisecting an angle

Bisect $\angle A$.

1. Draw an arc centered at $A$ crossing both sides. Label the crossings $B$ and $C$.
2. Draw an arc centered at $B$ inside the angle.
3. With the same opening, draw an arc centered at $C$ crossing it. Label the crossing $D$.
4. Draw ray $\overrightarrow{AD}$.

Why it works: consider $\triangle ABD$ and $\triangle ACD$.

Statement	Reason
$AB = AC$	same compass opening from $A$
$BD = CD$	same compass opening from $B$ and $C$
$AD = AD$	reflexive property
$\triangle ABD \cong \triangle ACD$	SSS
$\angle BAD \cong \angle CAD$	CPCTC

The ray splits $\angle A$ into two congruent angles, which is what bisecting means.

Copying an angle

Copy $\angle A$ onto a ray starting at $P$.

1. Draw an arc centered at $A$ crossing both sides at $B$ and $C$.
2. With the same opening, draw an arc centered at $P$ crossing the ray at $Q$.
3. Open the compass from $B$ to $C$.
4. With that opening, draw an arc centered at $Q$ crossing the second arc at $R$.
5. Draw ray $\overrightarrow{PR}$. Then $\angle QPR \cong \angle BAC$.

Why it works: $AB = PQ$ and $AC = PR$ (the first opening), and $BC = QR$ (the second). So $\triangle ABC \cong \triangle PQR$ by SSS, and the angles at $A$ and $P$ are corresponding parts.

Perpendiculars and parallels

Perpendicular through a point $P$ on a line. Draw an arc centered at $P$ crossing the line at two points, then construct the perpendicular bisector of the segment between them. It passes through $P$, because $P$ is equidistant from the two points.

Perpendicular through a point $P$ not on the line. Draw an arc centered at $P$ crossing the line twice, then bisect that segment the same way. $P$ is again equidistant from the two crossings, so the bisector passes through it.

Parallel through a point $P$ not on the line.

1. Draw any line through $P$ crossing the given line, forming an angle there.
2. Copy that angle at $P$, in the corresponding position.
3. The new side through $P$ is parallel to the given line.

Why it works: the construction makes corresponding angles congruent, and congruent corresponding angles force the lines to be parallel — the converse of the corresponding angles postulate.

Why a hexagon’s side equals the radius

Draw a circle centered at $O$. Join $O$ to two neighboring vertices of a regular hexagon inscribed in it. The angle at $O$ is $\tfrac{360°}{6} = 60°$, and the two segments from $O$ are radii, so they are equal.

That triangle is isosceles with a $60°$ top angle, so its base angles are $\tfrac{180° - 60°}{2} = 60°$ as well. All three angles are $60°$, so the triangle is equilateral, and the hexagon’s side equals the radius.

That fact is the whole construction.

1. Draw a circle centered at $O$.
2. Keep the compass at the radius. Put its point anywhere on the circle and mark an arc on the circle.
3. Move the point to that mark and repeat, until you have six marks.
4. Join consecutive marks. That is the regular hexagon.

Equilateral triangle. Construct the hexagon’s six marks, then join every other one. Each side skips a mark and spans $120°$ at the center, and the three are equal.

Square. Draw a diameter. Construct its perpendicular bisector through $O$, which is a second diameter at right angles to the first. The four ends are the square’s vertices: the diameters are equal, perpendicular and bisect each other, which makes the figure a square.

Worked examples

Common mistakes

Practice problems