Equal angles on the same side force concyclicity (Theorem 10)

0:00Here is a question with a trap in it. Given four points, do they all lie on one circle? The trap is in how you would check. You cannot draw the circle through all four and point at it, because whether such a circle exists is exactly what you are trying to find out. Draw it and you have assumed your answer. So the question has to be come at from the side.

0:26Three points that are not in a line have exactly one circle through them - one, not many. That is the foothold, and everything below is built on it. Points that lie on one circle are called concyclic. Four is the first number where the word carries any weight. Any two points lie on endlessly many circles. Any three off a line lie on exactly one. A fourth is the first genuine constraint, and that is why the question only becomes interesting there.

1:02So the four-point question turns into a three-plus-one question. Take three of them, build the circle they determine, and put the fourth on trial. Either it is on that circle or it is not, and there is nothing else it can be. Here is the claim to be proved. Take a segment from A to B. Take two more points, C and D, both on the same side of it, and neither of them on the line it lies along.

1:33Suppose the segment looks exactly as wide from one as from the other - the angle at C equals the angle at D. Then all four points are concyclic. Two conditions, and they are doing quite different jobs. Equal angles is the interesting one, and it is the one the argument works on. The same side is the one almost everybody reads as decoration, and by the end you will see that it is not.

2:05Build the circle you are entitled to. A, B and C are three points, and provided they are not in a line they determine exactly one circle. That word exactly is the engine of everything that follows, and it is worth stopping on. If three points came with a whole family of circles, then finding that D missed one of them would prove nothing at all - there would always be another to try.

2:33Because there is only one, missing it is a real failure with nowhere to hide. So: the circle through A, B and C, drawn. And D somewhere, on trial. Now suppose D is not on it. There are exactly two ways for a point to be off a circle. Inside, or outside. That is not a simplification for convenience. A point is inside, or on, or outside, and there is no fourth option, so knocking out both suppositions knocks out everything except being on it.

3:10That is the shape of the argument, and it is worth naming, because it is a different shape from anything so far. Everything until now was built forwards. This one runs by elimination. Before the formal version, do it with numbers, because the logic lands harder that way. Say the circle reads fifty degrees from this side - every point of the arc in question sees the segment at fifty.

3:41We already know what happens off the circle. A point inside reads more than fifty. A point outside reads less. And D is given as reading exactly fifty. More than fifty is ruled out. Less than fifty is ruled out. Fifty is neither more nor less, so there is nowhere left for D to be except on the circle itself. Three hundred and sixteen places were tried on one side of one chord, and judging by the angle alone placed every single one of them correctly.

4:19The formal version replaces we already know with an argument, and it needs one extra point that is not part of the question at all. Join A to D. That line leaves the circle at A, and where it meets the circle a second time, call that E. E is scaffolding. It exists to make a triangle, and it disappears from the conclusion - people often try to carry it into the answer, and it does not belong there.

4:50If D is inside the circle, the join from A passes through D and out the far side, so E is beyond D. That one is safe every time: a line that leaves a circle and goes through a point inside it has to come out somewhere. If D is outside, the drawing everyone makes has E sitting neatly between A and D. Most of the time it does. Not always.

5:19The join from A only reaches the circle a second time if it actually goes into it, and when D lies out past the tangent at A the line meets the circle behind A instead, which is no use at all. Of two hundred and fifty-six outside places tested, the join from A landed between A and D at a hundred and eighty-eight of them, landed behind A at fifty-seven, and at eleven more only grazed the circle at A and never came back.

5:53So the picture is quietly doing work the words never mention. The repair is small. Join from B instead: that reaches at two hundred and twelve of them, and between the two ends there is not one outside point left over. Join from whichever end gets there. Now the triangle, and take the outside case with E between A and D. E, D and the far end of the segment make a small triangle, and the line from A carries straight on through E to reach D.

6:28So the angle at E, looking back at A and across at B, is an angle made outside that triangle by carrying one of its sides onwards. And an angle made that way is the other two corners added together. Added - so strictly bigger than either of them alone. In particular, strictly bigger than the angle at D. Now close the loop, and this is the step that gets skipped.

6:57E is on the circle. C is on the circle. They are on the same side of the segment. So they read the same angle - that is the constancy from before, showing up here as a tool rather than as a result. Put the chain together. The angle at C equals the angle at E. The angle at E is strictly bigger than the angle at D. And the angle at D was given as equal to the angle at C.

7:26Follow that round and the angle at C comes out strictly bigger than the angle at C. Nothing is bigger than itself, so D is not outside. Run the same three steps with D inside and the inequality simply turns over - now it is D's own angle that is the carried-on one - and it arrives at the same impossibility. Now the condition everybody skips. Same side. Put the two points on opposite sides instead and the whole thing falls apart.

8:02Two points on opposite arcs of a circle do not read the same angle at all - they read angles that add to a straight one, fifty on one side and a hundred and thirty on the other. So on opposite sides, equal angles is simply not what being on a circle looks like. And it fails the other way round too. Take two points placed as mirror images across the segment.

8:29By symmetry they read exactly the same angle, and they are almost never on one circle with the segment's two ends. Out of thirty-nine heights tried, exactly one works - the one where both readings are right angles and the segment is the circle's full width. The condition is not tidying-up. Drop it and the statement is false. Two last things. First, notice what kind of proof that was. Nothing was constructed and nothing was drawn into existence.

9:05Both alternatives were shown to be impossible, and the positive claim is what was left standing when they fell. People sometimes think a contradiction only shows that one particular drawing is wrong. It does not, because the supposition was never about a drawing in the first place, so neither is its collapse. Second: now that the four points are known to share a circle, the figure they make has a name.

9:34Join them up in order round the circle and you have a cyclic quadrilateral - a four-gon whose four corners all sit on one circle. And that figure carries a property of its own, which is exactly where this goes next.