Why the constructions work

In earlier layers you learned how to construct 60°, 90°, 30°, 45°, bisectors and perpendiculars. You followed the recipes and checked them with a protractor. This layer asks the question a mathematician asks: why do the recipes work, every single time, for any size of compass opening?

A protractor reading can only ever tell you that an angle is about 60°, within a degree or so. A construction, backed by a reason, tells you it is exactly 60°, not 59.9° and not 60.1°. That exactness is the whole point of the compass-and-straightedge game, and it is why these constructions were studied for more than two thousand years.

You will meet three powerful tools of reasoning along the way: equal radii (every point on an arc is the same distance from its centre), congruent triangles (two triangles with the same three sides fit exactly on top of each other) and equidistant points (points the same distance from two places). With those three ideas, every construction in this topic can be explained.

Ancient Greek geometers, whose work was collected by Euclid in the Elements (about 300 BCE), allowed themselves just two tools:

• a straightedge: it draws a straight line through two points you already have. It has no markings, so you cannot measure with it.
• a compass: it draws a circle with a chosen centre passing through a chosen point.

Everything else must be built from those two moves. Why so strict? Because the Greeks wanted geometry to rest on the fewest possible assumptions. A line and a circle are the simplest perfect shapes. If a result can be reached using only them, then it is true because of logic, not because somebody's ruler was printed accurately. A construction was really a proof in pictures: each step comes with a reason, and the final figure is guaranteed.

In school we use a ruler with centimetre marks, but in a construction we use it only as a straightedge. The marks are allowed for checking afterwards, never for building.

Recall the recipe. Draw a ray OA. With centre O and any radius, draw an arc cutting OA at P. Keeping the same radius, put the compass point on P and draw an arc cutting the first arc at Q. Join OQ. Then ∠QOA = 60°.

Now the reason. Look at the three points O, P and Q and the three distances between them:

• OP is a radius of the first arc (centre O).
• OQ is also a radius of the first arc, because Q lies on that arc.
• PQ is a radius of the second arc (centre P), and that arc had the same radius.

So OP = OQ = PQ. Triangle OPQ has three equal sides: it is equilateral.

Here is a beautiful consequence. Keep the compass at the same width and walk it round a circle: centre on P, mark Q; centre on Q, mark R; and so on. Each step cuts off a chord equal to the radius, so each step makes an equilateral triangle with the centre, and each step turns through 60° at the centre.

After six steps the total turn is 6 × 60° = 360°, a complete turn, so the sixth mark lands exactly back on the starting point. That is why a compass flower (the six-petal rangoli design) closes perfectly, and why joining the six marks gives a regular hexagon whose side equals the radius.

The recipe for bisecting ∠AOB: with centre O draw an arc cutting the arms at P and Q. With centres P and Q and equal radii, draw two arcs that cross at T inside the angle. Join OT. Then OT bisects ∠AOB.

To see why, join P to T and Q to T. Now there are two triangles, OPT and OQT. Compare their sides:

• OP = OQ: both are radii of the first arc, centre O.
• PT = QT: both are radii of the two equal arcs.
• OT = OT: the two triangles share this side.

Three pairs of equal sides. That is the SSS rule: if two triangles have all three sides equal, they are congruent, meaning one fits exactly on top of the other (possibly after flipping it over). Congruent triangles have equal matching angles, so ∠POT = ∠QOT. OT cuts ∠AOB into two equal halves.

Edge case 1: the arcs do not meet. The second pair of arcs, centred at P and Q, only cross if their radius is more than half of PQ. If the radius is exactly half of PQ, the arcs just touch at the midpoint of PQ (that still works, but it is hard to see). If it is less than half, the arcs never meet and there is no T at all.

How long is PQ? If the first arc has radius r and the angle is θ, then triangle OPQ is isosceles and PQ grows with the angle. For a 60° angle, triangle OPQ is equilateral, so PQ = r. The safe rule in textbooks, "take a radius more than half of PQ", comes straight from this.

Edge case 2: which crossing point? Two circles that cross meet in two points, one on each side of the line PQ. Both lie on the bisector line, because each is equally far from P and from Q. So either one works. If you reuse the very first radius, one crossing point is O itself (O is already r from both P and Q) and the useful one is the other: that is fine, just do not join O to O!

To draw the perpendicular bisector of segment AB: open the compass to more than half of AB. With centre A draw arcs above and below AB; with the same radius and centre B draw arcs that cross them at P (above) and Q (below). Join PQ, meeting AB at M. Then PQ is perpendicular to AB and M is the midpoint of AB.

The reason: AP = BP = AQ = BQ, all equal to the one radius you used. A four-sided shape with four equal sides is a rhombus, so APBQ is a rhombus. The two diagonals of a rhombus always cut each other in half at right angles. The diagonals here are AB and PQ, so PQ meets AB at its midpoint, at 90°.

There is an even deeper way to see it. P is the same distance from A as from B; we say P is equidistant from A and B. So is Q. So is M. In fact:

Every point on the perpendicular bisector of AB is equidistant from A and B, and every point equidistant from A and B lies on the perpendicular bisector.

The set of all points that obey a rule is called a locus. The perpendicular bisector is the locus of points equidistant from A and B. The construction simply finds two such points (P and Q) and draws the line through them. Since two points fix a straight line, the whole bisector is found.

That is why you may use a different radius above and below AB: P could come from 4 cm arcs and Q from 6 cm arcs. APBQ is then a kite, not a rhombus, but P and Q are still each equidistant from A and B, so PQ is still the perpendicular bisector.

Notice that both recipes are really the perpendicular bisector construction in disguise. The first step manufactures a segment XY whose perpendicular bisector is the line you want; the second step finds that bisector. Once you see this, you do not need to memorise three separate recipes, only one idea: find two points equidistant from X and Y.

The second recipe also answers a practical question: how far is a point from a road? The distance from a point to a line always means the perpendicular distance, ZN, because it is the shortest.

With the 60° mark Q and the 120° mark R on the same arc, the 90° recipe says: with centres Q and R and equal radii, draw arcs crossing at S. Join OS. Why is ∠AOS exactly 90°?

Look at what you have. O is equidistant from Q and R (both are on the first arc). S is equidistant from Q and R (equal arcs). So O and S both lie on the perpendicular bisector of QR, and OS bisects ∠QOR. Since ∠QOR = 120° − 60° = 60°, the bisector splits it into 30° + 30°. Therefore

∠AOS = 60° + 30° = 90°.

You can also see it as an average: the bisector of the gap between the 60° ray and the 120° ray points to (60° + 120°) ÷ 2 = 90°. This "average rule" is the key to every combination angle in the next chapter.

With exact 60° steps, exact bisection and the straight line (180°), you can reach a large family of angles. Every recipe is a combination of three moves:

1. Step 60° along an arc (to 60°, 120°, 180°).
2. Bisect the gap between two rays you already have (the average rule).
3. Use the straight line: an angle x on one side of a point gives 180° − x on the other.

A good constructor looks for the shortest route, because every extra arc is another chance for a small error.

Sometimes you are given an angle with no idea of its size: a carpenter's corner, a roof pitch, an angle drawn by a friend. Can you make an exact copy of it somewhere else, using no protractor? Yes. The trick is that an angle is completely fixed by an isosceles triangle cut from it: two equal arms and the distance between their ends.

A protractor only goes up to 180°, so a reflex angle (between 180° and 360°) needs a little reasoning. There are two exact methods, and a careful measurer uses one to check the other.

Method 1, the leftover. Measure the ordinary angle on the other side, call it x. The two angles together make a full turn, so the reflex angle is 360° − x.

Method 2, straight line plus extra. Extend one arm backwards through the vertex to make a straight line (180°). Measure the part of the reflex angle beyond that line, call it y. The reflex angle is 180° + y.

If the two answers differ by more than a degree or two, one measurement is wrong.

How accurate can a protractor measurement be? Two things limit it.

1. How far apart the degree marks are. On a protractor of radius 5 cm, the arc for one degree is 2 × π × 5 ÷ 360 ≈ 0.87 mm long. On a 10 cm board protractor it is about 1.75 mm. Your eye can reasonably split a gap of under a millimetre into halves at best, so ±1° is an honest target with a school protractor, and ±0.5° with a big one.

2. How well the arms are drawn and lined up. Suppose the end of an arm is misplaced sideways by just 1 mm. The angle error is the angle whose "opposite over adjacent" is 1 mm over the arm length. For a 4 cm arm that is about 1.43°; for a 10 cm arm only about 0.57°. Longer arms make the same slip matter less, which is exactly why teachers say: extend short arms before you measure.

You do not need trigonometry to believe the numbers below: draw a 10 cm arm, mark a point 1 mm to the side of its tip, join that point to the vertex and measure the new angle with a protractor. The scale drawing gives the same answer as the calculation.

Bisecting any angle is easy. So the Greeks naturally asked: can every angle be split into three equal parts (trisected) with compass and straightedge alone? They also asked two other famous questions:

• Doubling the cube: construct the edge of a cube with exactly twice the volume of a given cube.
• Squaring the circle: construct a square with exactly the same area as a given circle.

For over two thousand years, brilliant mathematicians failed to find constructions. Many found clever methods that cheated slightly: Archimedes trisected any angle using a straightedge with two marks on it, slid into position (a move called neusis), which the strict rules forbid.

In 1837 the French mathematician Pierre Wantzel proved that trisecting a general angle and doubling the cube are impossible with compass and straightedge, and in 1882 Ferdinand von Lindemann's work on π showed that squaring the circle is impossible too. The key idea: every compass-and-straightedge step can only produce lengths built from whole numbers using +, −, ×, ÷ and square roots, and the lengths needed for those problems are not of that kind.