Length and distance from the centre are the same fact twice (Theorems 6–8)

0:00Cut a circle out of paper and fold it inwards once. Open it up: the crease is a chord. Now fold again, bringing the two ends of that chord together. Open it up. Two creases, and they cross. The second crease runs through the centre and meets the first squarely, at the chord's middle. So you can point at something you could only define before. That is how far the chord stands from the centre, and it can be measured with a ruler.

0:32Be exact about which distance that is. It is the perpendicular one - the segment dropped from the centre at right angles. Not the distance to one of the chord's ends. That is the radius for every chord, so it says nothing about which one you picked. The perpendicular distance is the shortest way from the centre to the chord, and the only one that changes as the chord moves.

1:01So each chord carries two numbers: its length, and how far out it sits. Whether those two numbers are independent is the question. They are not. Here is how the guess arrives. Trace a chord and the centre onto transparent paper, and turn the sheet about the centre. The chord lands somewhere new. Its distance has not changed. Do it again, with a shorter chord. Every time, equal chords come out the same distance from the centre.

1:34And here is the part worth stopping on. That is not a proof. It is a pile of examples, and a pile of examples says nothing about every chord of every circle. A demonstration tells you what to go and prove. It does not do the proving. So prove it. Two chords of one circle, given equal in length. Join all four ends to the centre. Each chord now sits in a triangle with two radii, and all four radii are the same length.

2:09The third sides are the chords themselves, given equal. Three sides against three sides, so the triangles are congruent. The distance we want is the height from the centre onto each chord - and in congruent triangles, corresponding heights are equal. So the distances are equal. That is the theorem. Checked on every pair of equal chords in the sample: three thousand six hundred and seventy nine pairs, congruent every time, heights equal every time.

2:45There is a second argument, worth having as well. Split each chord at its middle - exactly where the perpendicular from the centre lands. Now take one half of each figure. A right angle at the foot. The radius as the longest side. Half a chord as the other side. Equal chords have equal halves, and all radii are equal, so both match across the two half-triangles. Right angle, longest side, one other side - and the third side has nowhere left to go. That third side is the distance.

3:23It really is a different argument: this one needs the midpoint fact first, and the other does not. Now swap the given and the goal. Two chords standing the same distance from the centre. Are they the same length? The picture is identical - same right angles, same equal radii. What has moved is which fact is handed to you and which must be earned. This time the equal distances are given, the equal halves follow, and the chords come out equal.

3:58Same tool, other direction - and it must be run, because swapping a statement's halves can turn a true one false. Equal chords, equal distances. Equal distances, equal chords. One figure, two statements, two arguments. Put numbers on it. A circle of radius five, and a table. A chord ten units long runs through the centre, so its distance is nothing. A chord eight units long sits three out, and a chord six units long sits four out.

4:33A chord four units long sits between four and five out - not a whole number, so leave it bracketed. Read the rows together: ten, eight, six, four going down; nothing, three, four, over four going up. They move in opposite directions, and that is no coincidence about these four chords. The third theorem. Of two chords of different lengths, the longer is nearer the centre. People get this backwards, because a long chord feels further round the circle.

5:07Each chord gives a right triangle whose longest side is the radius - the same radius both times. So the distance squared plus the half-length squared comes to the same total in both. The longer chord has the longer half, so the bigger square - and if one square is bigger the other must be smaller, because the total is fixed. So the longer chord has the smaller distance. That last step deserves a moment. Going from a comparison of squares back to the numbers works only because both are lengths, and lengths are positive.

5:46Minus three squared beats two squared, and minus three does not beat two. Here, positivity is guaranteed. Sorted by length, the distance fell at every one of three hundred and forty six steps, with no exception. All three theorems come out of one equation. The perpendicular halves the chord and makes a right triangle: radius, distance, half the chord. So distance squared plus half-length squared is the radius squared - and on a fixed circle that total never moves.

6:24Equal lengths force equal distances because the total is fixed, and equal distances force equal lengths for the same reason. And if one goes up the other must come down, which is the third theorem. It also settles something students assume: a chord on a fixed circle does not carry two free numbers. Choose the length and the distance is decided, and the other way round. Two free numbers is one too many.

6:55Measured across the sample: three hundred and forty seven lengths, three hundred and forty seven distances, and exactly that many pairings between them. The equation also says where the family stops, at both ends. Push the distance to nothing and the chord runs through the centre, with length twice the radius. That is the diameter, and no chord beats it, because a distance under nothing is not available. In the sample the longest chord was exactly twice the radius, and all nineteen of those stood at no distance at all.

7:36Now the other way. Slide a chord away from the centre and it shortens. When the distance reaches the radius the length has reached nothing, and the chord has collapsed to a single point on the circle. That is the end of the range rather than a chord - and every distance measured came out under the radius, every length above nothing. Which brings the trap. One chord is twice as far out as another. Is the near one twice as long?

8:09Circle of radius five. Put one chord one unit out and another two units out. Their lengths squared come to ninety six and eighty four. Ninety six is not four times eighty four, so the ratio is not two. The ratio of the squares is eight to seven, which makes the nearer chord longer by under a tenth. Not double. Not half. About seven per cent. And there is no hidden factor either: doubling the distance on one circle gives nine different ratios.

8:44Worse: sometimes the farther chord does not exist. Three units out doubles to six, which is past a radius of five altogether. The relation runs through squares, and squares do not do proportions. Trust the equation and you can read anything off it. Radius seven, chord six out: the length squared is fifty two, so the chord is between seven and eight. Radius thirteen, five out, and the chord is twenty four exactly.

9:19Backwards: a chord of sixteen standing six out puts the radius at ten. Here is a better one. Two parallel chords, ten and twenty four, same side of the centre, seven apart. Find the radius. Their distances differ by seven, and their squares differ by a hundred and forty four take twenty five - one hundred and nineteen. A difference of squares is the difference times the sum, so the sum is one nineteen over seven - seventeen. A pair differing by seven and adding to seventeen is twelve and five.

9:58So the radius squared is a hundred and sixty nine, and the radius is thirteen. Forced, not found. One last consequence. Of all the chords through a point inside, the shortest is the one at right angles to the line from the centre - because that chord stands farthest out, and farther out means shorter. Which is the same fact, one more time.