Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12)

0:00Four points on a circle. Join them up and every corner sits on the circle. But you had to choose an order to join them in, and the choice is not innocent. Take them in the order they come round the circle and the sides close up into a proper shape. Take two of them out of turn and the sides cross over each other. Same four points, two different figures.

0:26Everything that follows depends on which of those you drew. And that is the same question as which side of a line a corner stands on. So build it properly, in order. Here is the claim. Take the two corners that are opposite each other - not next to each other, opposite, the two that no side joins. Those two angles add to a straight angle. A hundred and eighty degrees.

0:55And then the other pair does exactly the same. One warning first, because this is where the mistakes live. The four angles of any four-cornered shape add to a full turn. Stretched ones, lopsided ones, ones nowhere near a circle. Every single shape tested came to three hundred and sixty degrees. So that total tells you nothing whatsoever. It is the pairs that carry the information. Now the reason, and it starts with some bookkeeping worth slowing down for.

1:31Leave two opposite corners alone for a moment and look at the other two. Those other two cut the circle into two arcs. One of your corners sits on one arc, and the other sits on the other. That is not luck. It is exactly what taking the points in order bought you, and it held in every case tested. So each corner gets paired with the arc it is not standing on.

1:58One corner, one arc, and the two arcs between them are the whole circle. Each of those two angles is half of something. Put the centre in and draw the two radii out to the corners you left alone. The angle they make at the centre is the arc's own angle - how much of the turn that arc takes up. And a corner standing anywhere off that arc reads exactly half of it.

2:26Half, no matter where on its own arc you slide the corner to. That is the fact everything here rests on. So one angle is half its arc. The opposite angle is half the other arc. Two halves, of two arcs, that together make one circle. There is a catch, and it is the best thing in the proof. Sweep a radius from one corner round to the other, going the way that passes through your first corner.

2:56Sometimes that sweep goes more than halfway round. The angle at the centre comes out bigger than a hundred and eighty degrees - a reflex angle. You may have been taught that an angle stops at a straight line; here that instinct would sink the proof. The doubling still holds - it is just being read the long way round. And it is not random when this happens. A hundred and twenty six of the four hundred and ninety five shapes needed the long reading, and they were exactly the ones whose corner was obtuse.

3:32Obtuse corner, reflex arc, every time. Now put it together. The two arcs make up the whole circle. So their two angles at the centre make one complete turn. Three hundred and sixty degrees, between them, always. Each corner reads half of its arc. Half of one, plus half of the other, is half of the two together. Half of a complete turn is a straight angle. A hundred and eighty degrees, and there is the claim.

4:04The whole theorem is one turn, halved. Run it on the other pair and you get the same answer. All four hundred and ninety five shapes, both pairs, no exceptions. Go back to the order you joined them in. In a proper shape, two opposite corners sit on opposite sides of the line joining the other two. Opposite sides, opposite arcs, and the angles are supplementary. Join the same four points in the wrong order and both corners end up on the same side of that line.

4:41Same side, same arc. And two points on the same arc do not give supplementary angles - they give equal ones. True in all four hundred and ninety five crossed versions; supplementary in none. So these are twin results, and the side of the line decides which one you get. Same side, equal. Opposite sides, adding to a straight angle. With that, a whole family of questions becomes arithmetic. One corner is a hundred degrees, so the corner across from it is eighty.

5:19Two opposite corners given as two x plus ten and three x minus twenty: set their total to a straight angle and x is thirty eight, so the angles are eighty six and ninety four. And now the test. Could eighty, a hundred and ten, a hundred, and seventy be the corners of a shape on a circle? Check the pairs, not the total. Eighty and a hundred; a hundred and ten and seventy.

5:49Both make a straight angle, so yes. Now try ninety, eighty, seventy and a hundred and twenty. Still three hundred and sixty altogether - and both pairs fail. No circle will take them. That last step took something you have not been given. Knowing that a shape on a circle has supplementary opposite angles does not tell you that supplementary opposite angles put a shape on a circle. That is the statement turned around, and turning a statement around is not free.

6:25So it has to be earned, and it is earned by elimination. Take three of the corners. They are not in a line, so exactly one circle passes through them. Suppose the fourth corner is off it. Then it is outside that circle or inside it, and there is nothing else available. Knock out both and it has nowhere left to be. Take the outside case, and here the usual drawing quietly cheats.

6:55The side running from a corner on the circle out to the stray corner is drawn dipping back through, crossing it at a helpful point in between. It does not have to. Of seventy five outside positions tested, that side dipped back through for fifty eight of them - and for the other seventeen it left the circle at once and never returned. For those the point is not there at all, and the argument has nothing to stand on.

7:26The repair is cheap. Come at it from the other end of the shape instead, and between the two ends every one of the seventy five is covered. The inside case, usually left as an exercise, turns out to be the tidy one: fourteen positions, and the line always leaves the circle beyond the corner. No hole at all. Now the squeeze. It needs one fact about triangles. Carry a side of a triangle out past a corner and the angle you open up outside is the other two angles added together.

8:06So it is strictly bigger than either of them on its own. Strictly - that word does the work. Outside the circle, the crossing point reads the segment strictly wider than the stray corner does. That was true for all fifty eight. Inside, it tips the other way and the stray corner reads strictly wider. True for all fourteen. But the crossing point is on the circle, so its angle is already pinned by the theorem, and the supposition says the stray corner matches it.

8:40One angle ends up strictly bigger than itself. Nothing is. Both cases fall, and being on the circle is what is left. What does all that buy you? Carry a side out past a corner of a shape on a circle and the angle you open up equals the angle at the corner opposite - not the one next door. A parallelogram drawn on a circle has to be a rectangle: its opposite angles are equal and supplementary at the same time, so all four are right angles.

9:17One caution to end on. There is a prettier route - cut the shape into four triangles from the centre, use the equal radii, and the answer falls out with no arcs at all. It is lovely, and it is not general. It works only when the centre lies inside the shape, which was three hundred and six of the four hundred and ninety five. The arc argument covered all of them.

9:44That is usually the trade: the elegant proof, or the one that holds everywhere.