Math Core

Long before rulers had reliable markings, Greek mathematicians drew exact figures with just two tools: a compass and a straightedge. These constructions are still worth learning, because each one is really a small proof. Once you understand why a construction works, you understand the geometry behind it.

The rules of the game

A construction is a drawing made with only two tools:

a straightedge, for drawing a line or segment through two points (it has no markings, so you can't measure with it);
a compass, for drawing a circle or an arc with a chosen center and radius. You can also use it to "carry" a length: open it to the distance between two points, then keep that setting.

No protractor, and no reading numbers off a ruler. Every length is copied with the compass and every straight path is drawn with the straightedge. That's what makes a construction exact instead of approximate.

Why constructions work

Every point on a circle is the same distance from its center. So whenever you draw two arcs with the same compass setting, every point on them is the same distance from the centers. Constructions are built from this fact.

Copying a segment

To construct a segment congruent to $\overline{AB}$ :

Draw a ray with endpoint $C$ .
Put the compass point on $A$ and open it until the pencil touches $B$ .
Without changing the setting, put the compass point on $C$ and draw an arc that crosses the ray. Label the crossing $D$ .

Then $CD = AB$ , because both equal the compass setting.

The perpendicular bisector (and the midpoint)

The perpendicular bisector of a segment is the line that passes through its midpoint and is perpendicular to it. This construction also gives you the midpoint.

Open the compass to more than half of $AB$ .
With the point on $A$ , draw arcs above and below the segment.
With the same setting and the point on $B$ , draw arcs that cross the first two. Label the crossings $P$ and $Q$ .
Draw $\overleftrightarrow{PQ}$ . It is the perpendicular bisector, and it crosses $\overline{AB}$ at the midpoint $M$ .

Arcs of the same radius, centered at A and at B, cross at P and Q. Line PQ is the perpendicular bisector of AB, and M is the midpoint.

Why does it work? $P$ is on an arc centered at $A$ and on an arc centered at $B$ with the same radius, so $PA = PB$ . The same goes for $Q$ . A point that is the same distance from both endpoints of a segment always lies on the perpendicular bisector, and two points determine a line, so line $PQ$ is exactly that bisector.

Common mistake

If the compass is opened to less than half of $AB$ , the arcs from $A$ and from $B$ never meet, and you get no points $P$ and $Q$ . Open it wide, and don't change the setting between the two sets of arcs.

Bisecting an angle

To construct the bisector of $\angle V$ :

With the compass point on $V$ , draw an arc that crosses both sides. Label the crossings $X$ and $Y$ .
With the point on $X$ , draw an arc in the interior of the angle.
With the same setting and the point on $Y$ , draw an arc that crosses the one from step 2. Label the crossing $Z$ .
Draw $\overrightarrow{VZ}$ . It bisects $\angle V$ .

X and Y are the same distance from V, and Z is the same distance from X and from Y. Ray VZ splits the angle into two congruent angles.

Why does it work? $VX = VY$ (same arc) and $XZ = YZ$ (same setting), and $VZ$ is shared. So triangles $VXZ$ and $VYZ$ have all three sides equal. Triangles with matching sides have matching angles, so $\angle XVZ \cong \angle YVZ$ . You'll prove this kind of triangle fact carefully in the Congruent Triangles unit.

Copying an angle

To construct an angle congruent to $\angle A$ , with vertex at the endpoint $D$ of a ray:

With the point on $A$ , draw an arc crossing both sides of $\angle A$ at $B$ and $C$ .
With the same setting and the point on $D$ , draw an arc crossing the ray at $E$ .
Set the compass to the distance $BC$ . With the point on $E$ , draw an arc crossing the arc from step 2 at $F$ .
Draw $\overrightarrow{DF}$ . Then $\angle EDF \cong \angle A$ .

The new angle is congruent for the same reason as before: the triangles $ABC$ and $DEF$ have three pairs of equal sides.

A perpendicular through a point

To construct the line through a point $P$ perpendicular to line $\ell$ , turn it into a perpendicular bisector problem.

With the point on $P$ , draw an arc that crosses $\ell$ at two points, $A$ and $B$ . (This works whether $P$ is on $\ell$ or not.)
Construct the perpendicular bisector of $\overline{AB}$ .

Since $PA = PB$ , point $P$ is on that perpendicular bisector, so the line passes through $P$ and is perpendicular to $\ell$ .

Tip

Constructions combine. A perpendicular gives you $90^\circ$ ; bisect it for $45^\circ$ . Bisect a segment twice to split it into four equal parts. Later you'll also construct $60^\circ$ angles (from an equilateral triangle) and, by bisecting, $30^\circ$ angles.

Examples

Worked example: Distance between the arc crossings

In the perpendicular bisector construction, $AB = 8$ and the compass is set to $5$ . How far apart are the crossing points $P$ and $Q$ ?

$M$ is the midpoint, so $AM = 4$ , and $\overline{PM}$ is perpendicular to $\overline{AB}$ . Triangle $AMP$ is a right triangle with hypotenuse $AP = 5$ (the compass setting) and leg $AM = 4$ . By the Pythagorean theorem, $PM = \sqrt{25 - 16} = 3$ . By symmetry $QM = 3$ as well, so $PQ = 6$ .

Worked example: Building a 45° angle

Describe how to construct a $45^\circ$ angle.

Draw a line $\ell$ and pick a point $P$ on it. Construct the perpendicular to $\ell$ through $P$ , which makes a $90^\circ$ angle. Then bisect that right angle. Each half measures $\dfrac{90}{2} = 45^\circ$ .

Practice

Practice 1

Which tools are allowed in a classical construction?

Practice 2

In the perpendicular bisector construction for $\overline{AB}$ , why must the compass be opened to more than half of $AB$ ?

Practice 3

You construct the bisector of a $128^\circ$ angle. What is the measure, in degrees, of each of the two angles formed?