PrepShorts · Study sheet · Class 7 Mathematics · Chapter 6, Constructions and Tilings
Chapter 6 · Constructions and Tilings
Bisecting an angle, and halving 90° to get 45°
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Bisecting an angle is not a second construction to memorise. It is the first one, wearing different clothes.
The idea
The chapter does not hand down a method and then look for something to use it on; it starts with a design it wants and lets the design dictate the mathematics. An eight-armed rosette forces its arms 45° apart, 45° is not an angle any compass can conjure, and the only route to it is halving something you already have. So the general question — how do you halve any angle? — is asked because a picture demanded it. And the answer turns out to be the segment construction wearing new clothes: mark off equal arms so the two triangles share three equal sides, let SSS do the work, and the halving falls out as corresponding parts. Same idea, second application.
What you should be able to do
- Work out, from an eight-armed design, that its supporting lines must sit 45° apart
- Explain why 45° has to be reached by halving rather than by drawing
- Describe the pair of triangles whose congruence would bisect a given angle
- Explain how marking OA = OB and cutting equal arcs makes those triangles congruent by SSS
- Carry out the three-step bisection on any drawn angle
- Construct 45° from 90°, and use it to build the eight-armed figure
- Investigate what happens when the arcs are cut on the far side of the vertex
- Say which angles repeated bisection can and cannot reach
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| bisection | cutting something into two identical parts | printed in bold in §6.1, Part II, p.137; applied to angles on pp.143–144 |
| angle bisector | the ray from the vertex that splits an angle into two equal angles | printed in §6.1, Part II, p.144 |
| supporting lines | lines drawn only to place a design, and erased from the result | printed in §6.1, Part II, pp.140, 143 |
| SSS congruence condition | three pairs of equal sides, which force congruence | printed in §6.1, Part II, pp.144, 147 |
| congruent | said of two figures that match part for part | printed in §6.1, Part II, pp.143–144 |
| vertex | the corner point where an angle's two arms meet | the plural vertices is printed on p.152; the singular is added here, and on pp.143–144 the chapter names the corner by its letter O instead |
| arc | part of a circle drawn from a centre at a fixed radius | printed in §6.1, Part II, pp.140, 144 |
| radius | the compass opening an arc is drawn with | printed in §6.1, Part II, pp.140, 144 |
| compass | the instrument that holds one distance while it turns | printed in §6.1, Part II, pp.143–144 |
| petal | one leaf-shaped lobe of the designs built here | printed in §6.1, Part II, pp.144–145 |
| Math Talk | the chapter's flag on a question meant to be argued aloud | printed in §6.1, Part II, p.144 |
Where people slip up
- "Bisecting an angle is a different skill from bisecting a segment." Both are the same move: build equal lengths so that two triangles must be congruent, then read the equality you wanted off the congruence. Say this out loud; it is the chapter's design.
- "The arcs from A and B must have the radius used to mark A and B." They need not. The chapter's step 2 says only that the radius must be long enough and the same for both arcs.
- "C has to lie between the arms." Question 3 exists to unsettle exactly that. The two possible crossing points and the vertex are collinear, so the line is unchanged; what changes is which angle the ray on that side sits inside.
- "A protractor would be quicker, so this is just a school exercise." The chapter has restricted itself to two tools since p.140, and the point is that the result is exact rather than read off a scale.
- "Repeated bisection will reach any angle you like." It reaches halves of halves. Question 4 is placed there to make a student notice the limit, not to be answered by "yes".
- "Eight arms must mean eight separate constructions." Four lines through one point give eight rays. Building 90° once and bisecting twice gets the whole skeleton.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 3 Q1, Figure it Out · 3 Q2, Figure it Out · 3 Q3, Figure it Out · 3 Q4, Figure it Out · 3 Q6
Transcript1,442 words
Here is a figure worth wanting: eight narrow leaves, all the same, radiating from one centre. It is the kind of thing you would find carved into a door, or laid into a floor. Now try to draw it, and the trouble starts immediately. Not with the leaves. The leaves are arcs, and you can draw arcs. The trouble is where to put them. Eight things spaced evenly around a point is a demand about angles, and you have no protractor.
So strip the picture back to what has to be built first. Under a design like this there are always some lines that never appear in the finished thing. You draw them to place everything, and then you rub them out. Here they are four straight lines, all crossing at the centre. Four lines through one point give you eight rays coming out of it, because each line runs both ways.
So eight arms do not mean eight separate constructions. They mean four lines. Which is already a better problem than the one we started with. How far apart do those rays have to be? Go all the way round the centre and you have turned through three hundred and sixty degrees. The design cuts that turn into eight equal pieces. So divide. Three hundred and sixty, into eight. Forty-five degrees between one arm and the next, every time, all the way round.
That number was not chosen. It was forced, by the decision to have eight arms. And now the whole design has come down to one question: how do you build an angle of forty-five degrees? You cannot. Not directly. A compass draws circles. Nothing about a circle hands you forty-five degrees. But look at the number for a moment. Forty-five is half of ninety. And ninety you can already build, anywhere you like, out of nothing but a compass and a straight edge.
So the design does not need a way to make forty-five. It needs a way to cut something in half. Which is a much older kind of question, and one you have answered before. State it in general, because doing it only for ninety would be a waste. You are handed an angle. Any angle, any size, drawn by somebody else. Two arms, meeting at a corner. Call the corner O.
You want the ray from O that splits the angle into two equal halves. No measuring. No reading anything off a scale. Just a compass and a straight edge, and an answer that is exact rather than close. Think about what would settle it, before thinking about how to draw anything. Suppose the answer ray is already there, and pick any point on it. Call that point C. Now there are two triangles sitting at the corner, sharing the side from O to C.
One on each side of it. If those two triangles were congruent, matching part for part, then the two angles at O would have to match too. And two matching angles that add up to the whole thing are its two halves. So the drawing problem has turned into a different one. Build two triangles that must be congruent. Three matching sides are enough to force it. That is the strongest of the congruence conditions.
One of the three is free: the side from O to C is in both triangles. It is the same side. So you need two more, and you have to manufacture them. Start at the corner. Open the compass, and cut both arms at the same distance from O. Call those two marks A and B. Now O to A and O to B are equal, because they are the same compass opening.
One pair down. And something has just happened that is easy to walk straight past. Look at what is on the paper now. Two marks, and a point that is equally far from both of them. You have seen that before. The set of points equally far from two marks is a line: the one that cuts the segment between them in half and crosses it square. So the corner O is already on that line. Not near it. On it.
And here is the thing worth stopping for. That line is the answer. The line that halves the angle is the same line that halves the segment from A to B. Which means you are not learning a new construction at all. You are finishing one you have already built. A line needs two points, and you are holding one of them. So the arcs owe you exactly one more.
Open the compass again, to any width you like, as long as it is the same from both marks and long enough to reach. It does not have to be the width you used for A and B. In fact that width is the one to avoid, because one of the two crossings it gives you is the corner itself, which you already had. Strike an arc from A and an arc from B. Where they cross is C.
C is equally far from A and from B, because both arcs had one radius. There is the second pair of sides. Three matching sides, so the triangles are congruent, so the two angles at O are equal. And the halving is not something you did. It fell out. So the whole thing is three steps, and it works on any angle you are handed. One: cut both arms at the same distance from the corner. Two: equal arcs from those two marks, crossing at C.
Three: draw the line from the corner through C. Done. Now run it on a right angle. Two marks, two arcs, one line, and the ninety has become two angles of forty-five degrees each. Exact, not measured. Nobody read a scale at any point. Do it once more on one of those halves and you have twenty-two and a half degrees, which is as deep as the eight-armed design ever needs to go.
Here is a question that unsettles people, and it should. What if you strike the two arcs the other way, so that they cross out past the corner instead of between the arms? Try it. Same two marks, bigger opening, arcs crossing on the far side. You get a point out there beyond O, and if you draw the line from the corner through it, it is the same line as before.
Which should not be a surprise any more, because both crossings are equally far from A and B, and so is the corner. All three are on that one line, so there is only one line to draw. But say the next part carefully, because it is where the easy answer goes wrong. The line is the same. The ray pointing out to the far side is not inside your angle at all. It is halving the angle opposite.
One more question, and this one has a limit hiding in it. Halving is now free. So what can you reach? From a right angle: forty-five, then twenty-two and a half, then eleven and a quarter, and on down. Add a second starting angle, the sixty degrees a triangle with three equal sides hands you, and you can add and subtract as well as halve. That opens up a lot. Fifteen degrees, seven and a half, sixty-seven and a half.
Now try for sixty-five point five degrees, and you will not get there. Not ever, by any route. You can get close. Sixty degrees plus a right angle halved four times is sixty-five point six two five, which misses by an eighth of a degree. Halve again and you halve the miss, and again, and again, and it never once lands. Which leaves you with two things to build, and the tools to build both.
The first is the figure we started with. Two lines square to each other, halve both of the right angles, and the four lines are there. Then eight arcs, two to a leaf, and rub out the skeleton. The second is a square with four leaves inside it, meeting at the centre and pointing at the corners. Each one is bounded by two arcs, and the arcs are centred on the middles of the sides.
Ask for them as large as the square allows and the radius stops being yours to choose: any shorter and they do not reach the centre, any longer and they break out of the square. That is what the halving bought. Not a fact to remember. A design you can actually draw.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- The perpendicular bisector, and the equidistance property that justifies itClass 7 · Ch 6, Constructions and Tilings
- Constructing a 90° angle at a chosen point on a lineClass 7 · Ch 6, Constructions and Tilings
- SSS: three sidelengths fix a triangle completelyClass 7 · Ch 1, Geometric Twins
Comes up again in
- Constructing 60° from an equilateral triangle, and the arches built on itClass 7 · Ch 6, Constructions and Tilings
- Regular hexagons, and why the angles round a point must total 360°Class 7 · Ch 6, Constructions and Tilings
Either side of this one
- A stretched rope as compass and straightedge: the Śulba-Sūtra constructionsClass 7 · Ch 6, Constructions and Tilings
- Copying an angle, and why triangle congruence proves it worksClass 7 · Ch 6, Constructions and Tilings