Chapter 13.1 — Exercise 13.1 — Basic Constructions

Why an Arc of Equal Radius Always Gives 60°

Draw ray OA. With O as centre, draw an arc of any radius crossing OA at P. With the same radius and P as centre, draw a second arc crossing the first at Q. Since OP, OQ, and PQ are all the same radius, △OPQ is equilateral — and every angle of an equilateral triangle is 60°, so ∠POQ = 60° automatically, with no protractor ever touching the page. This is really the same isosceles/equilateral-triangle reasoning used constantly since Chapter 7, now run in the opposite direction from usual: instead of starting with a triangle and deducing its angles, the construction starts by forcing three sides equal on purpose, specifically so that the 60° angle comes out guaranteed rather than needing to be measured or assumed.

Bisecting a Compass Angle Splits It Exactly in Half

Once ∠POQ = 60° is drawn, its angle bisector OM splits it into two 30° angles — and this is where every other angle in this exercise comes from, since 90° is reached by carefully adding two separate 60° arcs plus one further bisection (∠PON = 60° + 30° = 90°), while every remaining angle smaller than 60° is reached simply by bisecting downward repeatedly.

Every single row in this table reduces to the same two moves: bisect an already-constructed angle to halve it, or place two already-constructed angles side by side to add them. No new construction technique is ever introduced past the very first 60° arc — the entire table above is really just repeated halving and adding, chained in different combinations to land on each specific target. Notice, too, which angles are conspicuously missing from the table: 20°, 40°, and 50°, for instance, can never be reached this way no matter how cleverly the halving and adding are combined, since neither is obtainable from 60° or 90° through any finite sequence of bisections and additions alone — a limitation of compass-and-straightedge construction that goes well beyond what this one exercise happens to ask for, but is worth knowing exists.

Proving 45° Really Is 45°, Not Just Close

The construction steps alone don't fully prove the resulting angle is exactly 45° — that needs the same triangle argument used for 60°, chained twice. Since △OPQ is equilateral, ∠POQ = 60°. Bisecting it carefully with ray OR gives ∠POR = ∠ROQ = 30° exactly. Bisecting ∠ROQ once again with ray ON gives ∠RON = ∠NOQ = 15° precisely. Adding the two now-known pieces together: ∠PON = ∠POR + ∠RON = 30° + 15° = 45° exactly.

This chain — construct, bisect, bisect again, then add — is the exact template behind every angle in the table above; only the specific combination of additions and bisections changes from one target angle to the next, never the underlying justification. Writing out this chain explicitly for every single target angle in the table above would be repetitive rather than instructive — once the 45° case is fully understood, the 22½° case is simply one further bisection nested on top of it, the 75° and 105° cases are simply one further addition, and every remaining entry follows the identical logic without needing its own separate proof written out in full.

An Equilateral Triangle, Straight From Two Arcs

Draw segment AB of length 4.5 cm. With A and B each taken as centre and radius 4.5 cm, draw two separate arcs meeting at point C. Join AC and BC together. Since AB = AC (both equal the exact same compass radius), ∠C = ∠B; since AB = BC, ∠C = ∠A. Combining both, ∠A = ∠B = ∠C, which forces BC = AC = AB — △ABC is equilateral.

The proof leans on "sides opposite equal angles are equal" twice, in each direction — first using equal sides to get equal angles, then immediately using those equal angles to force the third side equal too, closing the loop so that all three sides and all three angles end up equal simultaneously. Nothing about this proof depends on the specific 4.5 cm measurement chosen — the identical argument works for any radius at all, which is exactly why the construction method itself, arcs of equal radius from both endpoints, guarantees an equilateral triangle every single time it's carried out, regardless of the particular length picked for the base segment.

An Isosceles Triangle From a Base and Two Equal Base Angles

Draw segment AB of length 6 cm exactly, using a ruler. Draw rays AX and BY so that ∠BAX = ∠ABY = 45° (any equal angle works). Their intersection point is C. In △ABC, ∠A = ∠B by construction, so the sides opposite them are equal: BC = AC. Two sides being equal is already fully enough on its own to make △ABC a genuine isosceles triangle.

This construction is really the mirror image of the equilateral one above: there, two equal sides were fixed first and equal angles followed; here, two equal angles are fixed first and an equal pair of sides follows instead — the same "sides opposite equal angles" fact, just entered from the opposite direction. Choosing 45° for both base angles here was arbitrary — any two equal angles would work identically, producing a differently-shaped isosceles triangle each time, tall and narrow for a small shared angle, short and wide for a large one, while the underlying proof that BC = AC stays exactly the same regardless of which specific angle value happened to be chosen for the construction.

From Angles to Full Triangles

Every construction here built a single angle or a simple triangle from a small set of independent measurements. Exercise 13.2 tackles a harder category — triangles defined by a sum or a difference of two sides rather than the sides themselves, needing a genuinely new construction idea — an auxiliary point built specifically to turn a sum or difference condition into an ordinary length that a compass can actually draw — before the triangle can even be started.