The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5)

0:00Here is a circle, and a chord across it. Find the middle of the chord, and join it to the centre. First question. Is that join always square to the chord? Second question. Drop a square line from the centre onto the chord - does its foot always land in the middle? Those sound like one question said twice. They are not. One starts from the middle and asks about the angle. The other starts from the angle and asks about the middle.

0:32Each has to be argued on its own, and this video argues both. A chord arrives with two radii already attached, so it always comes with a triangle that has two equal sides. Now add the join from the centre to the middle of the chord. It cuts that triangle into two. Look at what the halves have. Two radii, equal. Two pieces of chord, equal, because we cut at the middle. And the join, in both.

1:04So the join is not just a line drawn into the picture. It is the mirror line of the triangle that was already there. Everything that follows is that one observation, written out. Be careful about what is given and what is wanted. Given: the point on the chord is its middle. That is all. Wanted: the join meets the chord at a right angle. So the right angle must not be drawn into the figure. It is the thing being shown, and a figure that ticks it in has assumed the answer.

1:40Draw the chord, the middle, the two radii and the join, and nothing else. Then earn the tick. Compare the two halves side by side. The radii are equal, because every radius is. The pieces of chord are equal, because the point was the middle. And the join is shared. Three sides against three sides, so the two halves are congruent. You can run it the other way if you prefer, using the equal base angles between each radius and its half of the chord. Side, angle, side. Same conclusion.

2:19Either way everything corresponding now matches, and the pair we want is the two angles at the middle. They correspond, so they are equal. Two equal angles. And here is where the proof actually happens. Equal is not enough. Take the two base angles of that same triangle. They are equal too, in every one of one thousand three hundred and fifty nine chords tested, and not one of them is a right angle.

2:52So being equal cannot be what forces ninety. What forces it is where those two angles sit. The two angles at the middle lie along the chord, one either side of the join, and together they make a straight line. Two angles that make a straight line add to a hundred and eighty. These two are also equal. So each of them is ninety. And nothing in that step was about circles. Equal angles on a straight line would have done it anywhere.

3:25There is a way to see it without any of that. Cut the circle out of paper and fold it so that one end of the chord lands exactly on the other. The crease runs through the centre, crosses the chord at its middle, and crosses it squarely. The same fact, felt rather than argued. The join was a mirror line, and folding is what mirror lines do. But a fold is one example. Fold a different circle and you have two examples.

3:57No number of examples settles a claim about every chord of every circle. The congruence does. Use the paper for the idea and the argument for the proof. Now the second question, and the same picture doing a different job. This time the right angle is given. Drop a perpendicular from the centre onto the chord, and what is wanted is that its foot lands at the middle. The drawing looks identical, which is exactly the trap.

4:30Before, the tick was the conclusion and had to be left off. Now the tick is the given and must be drawn in, and the middle is the thing to prove. That swap is not a formality. A statement can be true while the statement got by swapping its two halves is false. So the converse has to be argued from the beginning. The given has changed, so the tool changes with it.

4:58The two triangles now share the join, they have a radius each, and they have a right angle each. A right angle, a longest side, and one other side. That is enough - but it is worth asking why, because a right angle looks like any other given. Two sides and an angle that is not between them normally settle nothing. In a small grid of points, twenty one bundles of that shape each allow two different triangles.

5:30Pin that angle to ninety, though, and the ambiguity disappears. Of the twenty six right angled bundles in the same grid, not one of them allows two answers. And every single ambiguous bundle turned out to have an acute angle. Not one was obtuse, not one was right. So the right angle is not decoration. It is what makes the bundle decide. Put the two results together and something useful falls out.

6:02The centre is the same distance from both ends of a chord, because both of those distances are radii. The points equally far from two given points make one line - the line that cuts the segment in half at right angles. So a chord's perpendicular bisector is forced to run through the centre. It cannot do anything else. Now take two chords and build both bisectors. Each contains the centre, and two different lines meet in one place, so that place is the centre.

6:38Thirty seven pairs of chords were tried this way, and every single crossing landed exactly on the centre. That is the whole of the paper trick. Fold once, fold again, and the creases cross at the centre - because each crease is a perpendicular bisector, and neither had a choice. The right angle also turns half a chord into a right triangle, and that gives numbers. The radius is the longest side. The distance from the centre to the chord is one shorter side, and half the chord is the other.

7:17So the distance squared plus the half length squared equals the radius squared, for every chord there is. So the distance settles the length completely. Three hundred and forty six different distances were measured, and not one turned up with two lengths. And it runs one way. As the chord moves further out, its length falls at every single step. On a circle of radius five, a chord four out from the centre is six long, and three out it is eight long.

7:50Straight through the centre, distance nothing, and it is ten - the longest chord there is. Now a question those numbers answer, where the marks are lost on one word. A circle of radius five, with two parallel chords. One is six long, the other is eight. Their halves are three and four, so one chord sits four out from the centre and the other sits three out. How far apart are they? It depends entirely on whether they are on opposite sides of the centre or the same side.

8:27On opposite sides, one is four up and the other three down, so they are seven apart. On the same side, one is four up and the other three up, so they are one apart. Seven and one. Same circle, same chords, same distances - and the answer differs by a factor of seven, because of a word. One last thing the axis explains. Draw a triangle with all three corners on a circle, and make two of its sides equal.

8:59Now drop the height from the corner where those equal sides meet, onto the third side. It passes through the centre. Every time. And you already know why. Two equal sides put that corner on the perpendicular bisector of the third side. The centre is on that bisector too. And the height from that corner is that same line. Three descriptions, one line. Of every triangle fifty three points on a circle can make, a hundred and six came out with two equal sides, and in every one of them the height found the centre.

9:38Seventy thousand one hundred and seventy two heights were dropped from some other corner, and not one of them did. The join from the centre to a chord's middle was never just a join. It was the axis the picture already had.