Applying the right-triangle rule twice to get out of the plane

0:00You already know how to measure between two dots on a flat sheet. Two steps at right angles, and the square on the join is the sum of the two squares. Now lift one of the dots off the sheet. The question is what genuinely has to be new, and the answer is: almost nothing. The rule for three dimensions is not a fresh theorem with an extra term bolted onto it.

0:26It is the same theorem, used twice, in two flat surfaces that meet along a line. The one thing that really is new is a fact about a segment and a whole surface, not about two lines that happen to look square in a drawing. That fact usually gets asserted in a single clause and walked past. Everything downstream of it depends on it, so we are going to argue it instead.

0:55Take two places and build a box on them, with the two places at opposite corners, so that the span you want is the box's long diagonal. Two of the other corners matter, and the rest are scenery. Call them the first stop and the second stop. Their triples almost never get written down, and without them nothing that follows can be argued at all. The first stop keeps the start's first and third entries and takes the far end's second.

1:27The second stop keeps only the start's third. Run that on a start of one, minus three, four and a far end of minus four, one, two. The first stop comes out one, one, four, and the second stop minus four, one, four. Now every corner has a name and a triple, and we can start measuring. Walk from the start to the far end in three moves. Each move changes exactly one entry and leaves the other two completely alone, so each one runs along a single direction.

2:04Measured: the first move changes the second entry, the second move changes the first, and the third move changes the third. Their squared sizes come out sixteen, twenty-five and four. Add them and you get forty-five. Now measure the whole span from the start to the far end directly, in the space itself, using no rule whatsoever. Forty-five again. That is the entire result in one line, and one line is not an argument.

2:37A coincidence at one pair of places proves nothing. What has to be shown is why those three squares are forced. Here is the step that usually gets asserted. The angle at the first stop is a right angle. You cannot see that, and you should not believe a picture that tells you so. A flat drawing of a solid does not report angles honestly; the whole figure is a projection, and projections bend angles.

3:10So here is why it is actually true. The first move changes only the second entry, so it runs along the second direction. The first stop and the far end share their second entry, so the entire join from the stop to the far end lies inside the flat surface through that stop on which the second entry never changes. And that surface is precisely the one the second direction turns its back on.

3:40So the first move meets every single line of that surface at a right angle, whichever way the line happens to be pointing. Measured rather than asserted: a hundred and sixty-eight lines were drawn through that stop across the surface, three more were taken deliberately off it, and of those hundred and seventy-one the number met at anything other than a right angle is three. Exactly the three taken off.

4:11Notice what the argument never mentions: where the far end is. The right angle is structural, not a feature of the drawing. The second right angle is easier, and the important thing about it is where it lives. The second move changes only the first entry; the third move changes only the third. Two different directions, square to one another, meeting at the second stop, so that corner is square as well.

4:41But this second triangle sits inside the surface on which the second entry never changes, and the first triangle does not. Measured: the surfaces the two triangles span are not parallel, and of the four corners involved the number that both surfaces hold is two. Those two are the ends of the face diagonal. The triangles share that diagonal and nothing else at all. This is exactly why one application is not enough.

5:14If both right angles had sat in one surface, the whole argument would have collapsed back into two dimensions and told you nothing new. Now put the two statements together. The first triangle says the square on the span is the square on the first move plus the square on the face diagonal. The second says the square on the face diagonal is the square on the second move plus the square on the third.

5:44Substitute the second into the first. The face diagonal was on both sides of the working, and it leaves without a trace. Its squared size here is twenty-nine, which is a real length and not nothing, so watching it cancel is watching something actually happen. What is left is three squares, one for each move. That is the rule, and it arrived as a consequence rather than as an announcement.

6:15Now name the three moves. Each one is simply a difference of two entries, and here a perfectly fair objection arrives. These are supposed to be lengths, and a difference can come out negative. In the running example the three differences are minus five, four and minus two. Two of the three are negative. Nothing goes wrong, because every one of them is squared immediately, and squaring throws the sign away before it can do any harm.

6:47The same fact settles a second worry. Swapping which place you call first negates all three differences at once, and squaring undoes that too. Measured over twenty-nine pairs of places, the number where taking the two the other way round changes the answer is zero. One more thing that looks like a second rule and is not. Put the first place at the crossing where the three lines meet, the place whose three entries are all nothing.

7:20Then every difference is just the far place's own entry, and the squared span is the sum of its three squared entries. For the far end of the running example that comes out twenty-one. Measured across all twenty-nine pairs, the number where pinning the first place at the crossing gives anything other than the second place's own three squares is zero. It is the same rule with one triple made nothing.

7:50Learning it separately is learning the same thing twice. Now a case the picture simply cannot draw. If the two places happen to share an entry, the box has no thickness in that direction, and its eight corners collapse onto four. Here are four offered pairs. Their squared spans are twenty, forty-three, a hundred and four, and twenty. Count the corners each box manages to keep: four, eight, eight, four.

8:21So two of the four have gone flat, and the very first pair offered is one of them. Of those two flattened cases, the number the rule gets wrong is zero. The drawing fails; the rule does not. The box was scaffolding for the argument, and it was never part of the answer. Three places, and a question: are they in a line? Measure the three squared spans: fourteen, fifty-six, and a hundred and twenty-six.

8:52Now ask, exactly and with no rounding at any point, whether the two shorter lengths add to the longest. They do. And of the three ways of choosing which one to leave out, the number where the other two add up to it is one, which also names which place is in the middle. Why is that conclusive rather than a coincidence? Because for any three places at all, the two shorter lengths add to at least the longest, with equality only when the middle place sits on the join between the other two.

9:28A genuine triangle offered alongside, with squared spans nine, twenty-five and sixteen, has zero of its three choosings work. The equality is not evidence of a line; it is what being in a line looks like when you measure it. Now a triangle, and the question of whether it has a right angle anywhere. The three squared spans are six hundred and eighty-six, four thousand five hundred and seventy-one, and two thousand seven hundred and nine.

10:03The working you will usually see picks one corner, adds the two squares meeting there, compares against the third, gets three thousand three hundred and ninety-five against four thousand five hundred and seventy-one, and stops. But a triangle has three corners, and any one of them could have been the square one. So do the other two. Five thousand two hundred and fifty-seven against two thousand seven hundred and nine, and seven thousand two hundred and eighty against six hundred and eighty-six.

10:38Both also fail. Of the three corners, the number that come out square is zero, and only now has the conclusion been earned. One test out of three is not an answer, it is a start. Here is a second triangle, and its squared spans are eighteen, eighteen and thirty-six. Two of them match, so it has two equal sides. And eighteen plus eighteen is thirty-six exactly, so it does have a right angle, sitting at the middle corner rather than the first.

11:13Test the first corner only and you would have walked straight past it. Now something stranger. Negate the first and third entries of each of its places, and this triangle turns into another one that gets asked about separately, as though it were a different shape. It is not a different shape. Measured, that map has measure one rather than minus one, so it is a turn and not a flip.

11:41The same triangle, rotated about the middle direction, asked about twice, and each answer is true of both. Four places now, and the question of whether they make a parallelogram. Measure the four sides: their squares come out thirty-six, forty-three, thirty-six and forty-three, so opposite sides match. In flat geometry that would settle it completely. In space it does not, and this is worth seeing rather than being told. Here is a second set of four places whose squared sides are sixteen, nine, sixteen and nine, opposite sides matching just as neatly, and whose four places do not lie on one flat surface at all.

12:31Matching sides cannot see out of a plane. What settles it is the diagonals. For the first set, both diagonals have the same middle place; for the second set they do not, which is exactly how the test tells them apart. And the first set's two squared diagonals are three and a hundred and fifty-five, which differ, so it is a parallelogram and not a rectangle. Last, turn a condition into an equation.

13:04Name the moving place, write its two squared spans to two fixed places, add them, and expand every bracket. Out comes three squared terms, three plain multiples, and a bare number of a hundred and nine that nobody wrote down anywhere. It fell out of six brackets. Ask instead for the places equally far from two fixed ones, and the squared terms cancel completely: the number of them surviving is zero, and what is left is flat.

13:37That is the whole method, and it is the same expansion every time. One closing warning, and it is not about mathematics. Of four items offered at the end of this material, three need a rule for the middle of a join or for where a triangle's three joins meet. Those are two different rules, and neither of them appears anywhere in what came before. Exactly one of the four can be answered from what you were actually given.