ASA and AAS: two angles are enough to fix the third

0:00Here is a different set of three measurements to try. One side, and the angle at each of its two ends. The side B C is five centimetres. At B there are fifty degrees, and at C there are thirty. Notice where that side sits. It runs from B to C, joining the two marked corners. It is the included side, and that word is going to matter again. So is that enough to pin the triangle down? Draw it and see.

0:31Put the five centimetres down first, exactly as before. Then swing a ray up from each end, at whichever angle is marked there. From B, a ray at fifty degrees. From C, a ray at thirty. Both go up on the same side of the base, and they lean towards each other. So they meet. And they meet at exactly one point. There is no arc here, and nothing swinging, so there is nothing that can cross twice.

1:02Two straight rays leaning together cross once, or never, and never twice. They only fail to meet if the two angles have already used up a hundred and eighty between them. Fifty and thirty come to eighty, so there is plenty of room left. One meeting point means one triangle. Everybody drawing this gets the same one. That is the third condition, and it gets a name as well. Angle, side, angle. A S A.

1:35And again the letters are telling you where things have to sit. The side is the middle letter, so the side goes in the middle. Between the two angles. Once those rays meet, the rest of the triangle is settled without anybody choosing anything. The third angle is a hundred degrees, because all three have to reach a hundred and eighty. And the other two sides come out at about two point five four and three point eight nine centimetres.

2:06Nobody measured those. They were decided the moment the side and the two angles were fixed. Now here is a figure where all of this earns its keep. Four points. A up on the left, C up on the right, B down on the left, D down on the right. Draw the segment from A across to D, and the segment from B across to C. They cross somewhere in the middle. Call that crossing point O.

2:35And you are told exactly one thing about O. It is the midpoint of both of those segments. Midpoint of A D, so A O equals O D. Midpoint of B C, so B O equals O C. The question is how the length A B compares with the length D C. And nobody is going to measure either of them. Count what you have. Two equal pairs of lengths, and that is everything you were given.

3:05There are two triangles sitting in that figure. A O B on one side, and D O C on the other. For each of them you now know two sides. You need a third thing. And there is a third thing, right at the crossing, that nobody had to measure. The angle A O B and the angle D O C face each other across the point where the two lines cross.

3:31Angles facing each other like that are always equal. Every crossing, every time. It costs nothing. It is not a given and it is not a measurement. It is just true of crossings. So now each of those two triangles has three known parts. So which condition is this? It is tempting to answer A S A, because that is what we just met. But look at what you actually hold, and where each piece of it sits.

4:02A O is a side. O B is a side. And the angle at O is the corner where those two meet. Side, angle, side. This one is S A S. Only one angle in each triangle is known here, and A S A needs two of them. The two conditions are easy to confuse when they turn up next to each other. So do not go by the heading. Read the figure, and count the parts you were handed.

4:31S A S fires, and the two triangles are congruent. Now write that congruence down carefully, because the order of the letters decides everything. A pairs with D, because A O and O D are the sides that were equal. O pairs with itself. And B pairs with C. Triangle A O B is congruent to triangle D O C. That is the only ordering of it that is true.

5:01Now read off the third side of each. In the first name, A and B are the first and last letters. In the second name, those two positions hold D and C. So A B and D C are corresponding sides. And corresponding sides are equal. That was the question. And the answer arrived without anybody measuring a thing. Now the awkward case. Two triangles, tall and narrow, drawn side by side.

5:33In the first one, thirty-five degrees up at the apex A, and seventy-five down at the corner C. The side B C, along the bottom, is four centimetres. The second has the same three. Thirty-five at X, seventy-five at Z, and four along Y Z. Now check where that side sits, the way we have been checking all along. C is one of the marked corners, so B C does touch one of them.

6:03But A is the other marked corner, and B C does not go anywhere near A. The side is not between the two marked angles. So this is not A S A. And this is where the reflex from the last case wants to take over. The side was in the wrong place before and the whole thing fell apart, so surely it does again. It does not. And the reason is that angles behave differently from sides.

6:33You were handed two angles. And a triangle's three angles always come to a hundred and eighty. So the third one is not missing. It is only not written down yet. Thirty-five plus seventy-five is a hundred and ten. Take that from a hundred and eighty. Seventy degrees. And that is arithmetic, not a protractor, so it is exact. Write seventy in at B in the first triangle, and at Y in the second.

7:04Now look at the two figures again. Nothing has moved. The only difference is one number written into each of them. But read off what you are holding now. Seventy degrees at B equals seventy at Y. The side B C equals the side Y Z. And seventy-five at C equals seventy-five at Z. Angle, side, angle. And this time the side really is between the two angles. B and C are the marked corners now, and B C is exactly the side that joins them.

7:41This is the problem from the start of this video, wearing different numbers. It was never a new problem. It was that one, with a piece left unwritten. So how safe is that? I swept it, over exactly the range I swept last time. Every whole-number side up to twelve, and every pair of angles in five degree steps. Fourteen thousand seven hundred sets of measurements. Seven thousand one hundred and forty of them gave exactly one triangle.

8:16Seven thousand five hundred and sixty gave none, because the two angles had already spent the hundred and eighty. And the number that gave two different triangles was zero. Not one case in fourteen thousand. Set that beside two sides and an angle off to the side, where three hundred and eighty cases came apart. With two angles in hand the third comes free, so nothing is ever left open.

8:46So there are five sets of measurements that settle a triangle completely. Three sides. Two sides with the angle between them. Two angles with the side between them. Two angles with a side somewhere else. And a right angle, with the longest side and one more. The fourth of those is the one we just did, and it usually gets called A A S. But notice what it is not. It is not a fifth independent fact to be remembered.

9:17It is A S A with one subtraction done first. Take the subtraction away and nothing at all is lost. And that is worth more than the name. Some conditions are genuinely new, and some are old ones in disguise. Telling those two apart is the real skill. The arithmetic was never the hard part.