What we are allowed to do, recapped
The previous guide fixed the rules of the game, so we will not re-argue them — only put them to work. A straightedge draws the straight line through two points you already have; it has no marks, so it can never measure. A compass, set to the distance between two points, draws the circle of every point that far from a chosen centre. That is the whole toolkit. Every construction in this rung is just a sequence of those two moves, and the points where the resulting lines and circles cross become your new material.
One subtlety from guide 1 is worth keeping in view. The classical compass is a collapsing compass: lift it off the page and it snaps shut, so you cannot use it as a pair of dividers to carry a fixed length across the paper. It feels like a crippling restriction, yet guide 1 showed that a collapsing compass can do everything a rigid one can. That is exactly why our very first real task below is to copy a segment — once we can transport a length honestly, the collapse stops mattering.
Copying a segment, copying an angle
Start with the most basic transport. You are given a segment AB and a fresh point C on a line, and you want a point D on that line with |CD| equal to |AB| — that is copying a segment. Open the compass to span from A to B, plant the needle at C, and swing an arc cutting the line; the crossing is D. Why is |CD| equal to |AB|? Because a single compass opening, by definition, is one fixed radius — A to B and C to D are radii of the same-sized circle, so they are equal. With a rigid compass that is the whole story; with a collapsing one you first build the length using the trick from guide 1, then copy as here.
Copying an angle is the same spirit, one step richer. To reproduce angle BAC at a new vertex P on a given ray, first draw an arc centred at A; it meets the two sides of the angle at two points, say X and Y. Without changing the opening, draw the same-radius arc centred at P, crossing your new ray at X'. Now set the compass to the chord |XY| — the straight gap between where the first arc hit the two sides — and swing that length from X' to mark Y'. The ray from P through Y' completes the copy. This is copying an angle, and it is the move that lets you transplant a known angle anywhere on the page.
The reason it works is pure SSS again. Look at triangle AXY and triangle PX'Y'. The two short sides are equal because both arcs used the same opening (|AX| = |PX'| and |AY| = |PY'|), and the third sides are equal because we deliberately copied the chord (|XY| = |X'Y'|). Three equal pairs of sides, so the triangles are congruent, so their angles at A and at P match — meaning m(angle BAC) equals the angle you just built at P. No measuring, no protractor, and yet the angle is reproduced exactly.
Bisecting a segment — and the perpendicular bisector
Now we halve. To carry out bisecting a segment AB, open the compass to more than half of |AB|, plant the needle at A and draw a full arc above and below the segment, then do the same from B with the same opening. The two arcs cross at two points; call them P and Q. The straight line through P and Q crosses AB at its exact midpoint M, and as a bonus it crosses at a right angle — so line PQ is the perpendicular bisector of AB, the single construction that both halves the segment and stands square to it.
Why P and Q land where they do is the loveliest idea in the rung. We chose the same compass opening from both A and B, so each of P and Q is equally far from A as from B: |PA| = |PB| and |QA| = |QB|. The perpendicular bisector is precisely the set of all points equidistant from A and B, so any point with that equal-distance property must lie on it — which forces both P and Q onto the line, and the line drops straight through the midpoint at a right angle. We are not checking a picture; we are using the defining property of the perpendicular bisector.
construction you supply you produce key fact used copy a segment AB, point C D with |CD| = |AB| one radius, one length copy an angle angle BAC, ray same angle at P SSS on the two arc-triangles bisect a segment AB midpoint M + perp bisector equidistant points lie on it bisect an angle angle BAC ray splitting it in two SSS on a mirrored pair erect a perpendicular line, point on/off the right-angle line symmetry across the foot
Bisecting an angle
Halving an angle uses the same equidistance instinct, turned sideways. For bisecting an angle BAC, put the compass needle at the vertex A and draw an arc that crosses both sides, at points X and Y. Then, with any convenient opening, draw two arcs of equal radius — one centred at X, one centred at Y — and let them meet at a point Z inside the angle. The ray from A through Z splits angle BAC into two equal halves. One construction, and the angle is cut perfectly in two.
The proof is SSS one more time. Compare triangle AXZ with triangle AYZ. Side AX equals side AY (both struck with the first, vertex-centred arc); side XZ equals side YZ (both struck with the second, equal-radius arc); and side AZ is shared by the two triangles. Three equal pairs of sides, so the triangles are congruent, so the angle at A in the first equals the angle at A in the second — which is exactly to say ray AZ divides angle BAC into two equal angles. Notice how the whole figure is symmetric across line AZ; the construction simply builds that mirror in.
Dropping and erecting a perpendicular
A perpendicular is a line meeting another at a right angle, and there are two everyday versions. Erecting a perpendicular means starting from a point P that lies on a line and raising a line through P at right angles to it. Dropping a perpendicular means starting from a point P off the line and finding the line through P that meets it square — the shortest path from the point to the line, hitting it at the foot of the perpendicular. Both are the same idea, erecting a perpendicular, wearing two hats.
Here is dropping a perpendicular from a point P to a line, step by step. The secret is that we never need a fresh idea: we manufacture two points on the line that are equally far from P, and then the perpendicular bisector we already know how to build does the rest.
- Plant the compass at P and open it wide enough to cross the line in two places; swing the arc to mark those crossings A and B. By construction |PA| = |PB|, so P is equidistant from A and B.
- Now bisect segment AB exactly as before: equal arcs from A and from B meet at a point Q on the far side of the line from P.
- Draw line PQ. Since both P and Q are equidistant from A and B, both lie on the perpendicular bisector of AB — so line PQ IS that perpendicular bisector.
- Therefore PQ meets the original line at a right angle, and the crossing point F is the foot. The drop is complete, and m(angle PFA) = 90 degrees by the perpendicular-bisector property.
Erecting a perpendicular at a point P that is already on the line is even quicker, and it reuses the very same parts. Mark two points A and B on the line equally far from P on either side, by swinging one arc from P that cuts the line twice; now |PA| = |PB|, so P is the midpoint of AB. Then build the perpendicular bisector of AB through P with the two-arc method, and the line you raise stands at a perfect right angle to the original. Once you see it, every perpendicular in geometry is the perpendicular bisector wearing a disguise — which is why mastering one construction quietly hands you three.
Why these five matter, and what is next
Step back and notice how few genuinely new ideas you used. Copying rests on a single radius; copying an angle, bisecting an angle, and the whole perpendicular business all collapse to SSS congruence and one defining property — equidistance — of the perpendicular bisector. These five are the alphabet of construction. Inscribing a circle in a triangle, building a regular hexagon, dividing a segment into equal parts, the 17-gon ahead: every one of them is spelled out of bisecting, copying, and dropping perpendiculars.
An honest word on precision. A construction is exact in principle: the proofs above use no approximation, so the midpoint really is the midpoint and the right angle really is 90 degrees. What is approximate is your hand — a slipped needle or a fat pencil line blurs the ideal point. Geometry separates these cleanly: the theorem is about the perfect figure, and the smudges are a fact about pencils, not about the mathematics. That distinction is what lets the impossibility results later be airtight.
Next in the rung we put the alphabet to work on shapes. Guide 3 inscribes regular polygons — the easy hexagon, then the astonishing case of Gauss's 17-gon — and asks which regular polygons are even constructible at all. From there the rung turns philosophical but rigorous: guide 4 pins down exactly which lengths a compass and straightedge can reach, and guide 5 uses that to prove three ancient problems, trisecting the angle among them, genuinely impossible under these rules. Everything starts from the five moves you now own.