SSS: three sidelengths fix a triangle completely

0:00There is a wooden triangular frame standing on the grass outside a school. It comes up to about chest height, and somebody wants to make another one exactly like it. The obvious thing to do is trace it. Lay paper over it and follow the edges round. But it is wider than any sheet of paper they have. So tracing is out, and the shape has to be carried across as numbers instead.

0:27Two people are standing beside it. One is holding a measuring tape. The other is holding a protractor. And they disagree about what needs writing down. That disagreement is worth listening to, because both of them are being reasonable. The one with the tape measures the three sides. Forty centimetres, sixty, and eighty. Then comes the claim. That is enough. Put the protractor away. The other one is not convinced. Surely you want the angles as well, just to be safe.

1:04It is a fair worry. Three numbers feels thin for a whole shape. But look at how strong the claim actually is. Anyone, anywhere, who builds a triangle from those three lengths will end up with an exact duplicate of this frame. Not roughly the same. The same. So let us test it, by trying to build one and seeing whether there is any room at all to get it wrong.

1:32Forty, sixty and eighty will not fit on a page, so shrink all three by ten. Four centimetres, six, and eight. That is the same shape at a tenth of the size. Its angles have not moved: about twenty-nine degrees, about forty-seven, and about a hundred and four. The small one is not congruent to the frame, obviously. Same shape, different size. But the question we are asking is a question about shape, so the small version will answer it.

2:05And it has to be small, because the answer is going to be a drawing. Right. Three lengths. Four, six and eight. Build the triangle. Start with any one of the three. Take the six. Draw it as a straight segment and name its two ends A and B. That side is finished. It is never going to move again. Now the four. It has to run from A out to wherever the third corner turns out to be.

2:35So the third corner is four centimetres from A, and we do not yet know in which direction. Every point that is four centimetres from A: that is a circle around A with radius four. Same story for the eight. Every point eight centimetres from B is a circle around B with radius eight. The third corner has to be on both circles at once. So the question is where those two circles cross.

3:06They cross here. And they also cross here. Two points, not one. One of them sits above the line and the other sits below it, and both of them are a centimetre out past A. That is the objection, and it is a good objection. Join each one up and you get two triangles, pointing in opposite directions. Both were built from exactly the same instruction. Both have sides of four, six and eight.

3:37So the three lengths did not hand us one answer. They handed us two. And if those two are genuinely different triangles, the whole claim falls over. There are three ways to settle it, and it is worth having all three. The first: trace one of them and lay the tracing down on the other. The second: cut one of them out and put it on top. Both of those work, and both of them you can do at a desk in about a minute.

4:07But neither one tells you why, and neither one says anything about the next pair of triangles you meet. The third way is the reason. Look again at how the two were made. The same circle drawn from A. The same circle drawn from B. One crossing above the line, one below. Neither side was ever treated differently from the other. Which means the segment A B is a fold line.

4:37Fold the paper along it, and the bottom half lands exactly on the top half. Watch what happens to the corners. A stays where it is. B stays where it is. And the lower crossing point swings up and lands precisely on the upper one. The two triangles coincide. So they are congruent. And a fold is allowed. Turning and flipping are both permitted before you compare two figures. So the construction did not really give two answers. It gave one answer and its flip.

5:12The only freedom left in the whole thing is a freedom that was never going to count. Now go back through that argument and look at what it actually used. It used that the same circle was drawn from each end of the base. It used that the two crossings sit on opposite sides of that base. It never once used the number four, or six, or eight. Call the three lengths a, b and c, and every word of it still stands.

5:43I ran it anyway, over every whole-number triple up to twelve that makes a triangle at all. Eight hundred and seventy of them. In every single one the circles crossed, both triangles came out carrying exactly the lengths that were asked for, and the two crossings were a mirrored pair. Not one exception. But the sweep is only the evidence. The fold is the reason. So here is the result, in one line.

6:16If two triangles have their three sidelengths equal, then the two triangles are congruent. Not might be. Are. Three numbers, and the triangle is pinned. It has a name. The side-side-side condition. S S S. And notice that word, condition. This is not a description of triangles that happen to match. It is a guarantee. Hand me three matching lengths and I will hand you back the congruence. Which is exactly what lets a shape travel across a room, or across a country, as three numbers on a scrap of paper.

6:57Back to the protractor, because the person holding it deserves a proper answer. Would measuring the angles have helped? No. And here is the precise reason why not. Once the three sides are fixed, the three angles are already decided. You can work them out. Across the two hundred and three different triangles in that sweep, every one determined its own three angles, and no two with the same sides ever disagreed.

7:28So writing the angles down as well rules nothing out. It repeats what you were already holding. But be careful with the other direction, because it does not run the same way. Angles on their own are not enough. In that same sweep, twenty-four sets of angles belonged to more than one size, and one set was shared by twelve different triangles. One more idea, and it is the one that does the most work from here on.

8:00Here is a rectangle, with corners A, B, C and D. Draw the diagonal from B across to D. That cuts it into two triangles. A B D, and C D B. Are those two congruent? Count up the equal sides. Opposite sides of a rectangle are equal, so A B equals C D. That is one. And A D equals C B. That is two. The third one is the diagonal itself. B D belongs to both triangles, so of course it equals itself.

8:36A length nobody measured, and it completes the set. That is S S S, and it held for all one hundred and forty-four rectangles I tried. Once you have seen that move you start noticing it everywhere. Here is a kite. Two equal short sides meeting at the top, two equal long sides meeting at the bottom. Draw the line from the top corner down to the bottom corner, and it belongs to both halves at once.

9:08Two pairs given, plus the shared one. Three. S S S again. And there is something extra. Because the two halves are congruent, that line cuts the angle at the top into two equal pieces. I checked that across three hundred and forty-three kites, and it held in every one of them. Here is a stranger figure. A dart, with a notch pushed up between the two lower corners. Same three equalities, same conclusion, and the same shared side doing the work. Three hundred and forty-three of those as well, every one congruent.