AAA and AA: equal angles are enough on their own

0:00Two figures are the same shape when two things hold. Matching corners equal, and matching sides in one ratio. For a triangle that is six separate checks: three angles to compare, three ratios to compute. And the two halves are genuinely independent - for four-sided figures, neither one implies the other. A rectangle two by five and a square three by three. Every corner of both is a right angle, so all four angles match.

0:34But the sides are two against three, then five against three, then two against three again. Not one ratio. Not the same shape. Over sixty-four pairs of rectangles, every pair is equiangular and only twenty-two are similar. Equal angles buy you nothing there. So for four-sided figures you need both halves. The question is whether a triangle is different - and it is. Before anything is proved, a word about how the claim is written, because the notation carries part of the meaning.

1:13Writing that triangle A B C is similar to triangle D E F is not a claim about two triangles. It is a claim about a pairing. A with D. B with E. C with F. The angle at A equals the angle at D, and A B over D E is the ratio. Shuffle the letters on one side and you have asserted something else - usually something false.

1:40Take a scalene triangle and any enlargement of it. Of the six possible orderings of the letters, exactly one is true. Give the triangle two equal sides and two orderings work, because the shape now has a symmetry to exploit. So the order is not labelling. It is the claim, and getting it wrong is getting the answer wrong. Now the experiment that suggests the theorem. Draw a segment three centimetres long, and a second one five centimetres long. Different lengths, deliberately.

2:18At each end of the first, construct an angle - say sixty degrees at one end and forty at the other. The two rays meet at a point above the segment. At each end of the second, construct the same two angles: sixty and forty, in the same order. Both triangles now have a third angle, and neither of them was chosen. Sixty and forty leave eighty, at both apexes.

2:46So the two triangles agree in all three angles. Nothing at all has been said about their sides. One is plainly bigger than the other. The bases are three and five. The question is what the other two sides do. Measure them. The bases are three against five, so that ratio is three fifths. Measure the second pair of sides. Three fifths. Measure the third pair. Three fifths again. Three ratios, computed from three separate measurements, and they land on one number.

3:24And a check beside it: the areas stand in the ratio nine to twenty-five - which is three fifths, squared. A third measurement, made a completely different way, and it agrees. And it forces the drawing. If a side of the small triangle comes out at two point four, the matching side of the large one has to be four. Two point four divided by three fifths. There is no freedom left.

3:54That is a striking result and it is not a proof. One pair of triangles, measured with a ruler, at one choice of angles. A ruler agrees with anything to within a ruler's accuracy. So here is the same question asked properly. Eight triangles of very different shapes, each moved five ways - turned, reflected, and scaled by five different amounts. That gives forty pairs that ought to be similar, and every triangle against every other gives fifty-six that ought not to be.

4:32Ninety-six pairs, and three referees that cannot see each other's answers. The angle referee reads a squared cosine and a sign, and no lengths. The side referee reads squared distances, and no angles. The area referee is a shoelace sum, and reads neither. The report: forty equiangular and in one ratio, fifty-six neither, and the two cells that would break the theorem are empty. Nothing equiangular with the sides out of ratio. Nothing in one ratio without the angles agreeing.

5:10The area referee agrees on all forty and disagrees on all fifty-six. So it is worth asking why. Here is the statement. If two triangles have their matched angles equal - all three pairs - then their matched sides are in one ratio, and the triangles are similar. Angles in. Sides out. One clause doing the work of two. Notice how strange that is. The hypothesis says nothing about size at all. You could satisfy it with a triangle the size of a coin and one the size of a field.

5:51And the conclusion is not that the sides are equal. It is that they are proportional, with the ratio left completely open. That is the trade. Equal angles fix the shape exactly and fix the size not at all. The proof has to get a statement about lengths out of a hypothesis that contains no lengths, and it does it with one construction. Call the small triangle A B C and the large one D E F, with the angles matched in that order.

6:27The construction is this. On the side D E, measure off from D a length exactly equal to A B, and call that point P. On the side D F, measure off from D a length exactly equal to A C, and call that point Q. Both marks land on their sides, because D E and D F are longer. Now look at triangle D P Q against triangle A B C. D P equals A B by construction. D Q equals A C by construction. And the angle between them at D equals the angle at A, because that was one of our three.

7:09Two sides and the angle between them. That is congruence. So triangle D P Q is a congruent copy of A B C, planted inside the larger triangle. The third sides match too: P Q equals B C. Now the step that makes the whole thing work. Because the two triangles are congruent, every angle of one equals the matching angle of the other. In particular the angle at P equals the angle at B.

7:43But the angle at B was equal to the angle at E - that was another of our three. So the angle at P equals the angle at E. And those two are corresponding angles, made by the line D E crossing P Q and crossing E F. Corresponding angles equal means the two lines are parallel. P Q is parallel to E F. That is worth stopping on. We began with three angle equalities and no lengths, and we have manufactured a parallel line inside the larger triangle.

8:21And a line parallel to one side of a triangle is exactly what the previous theorem knows how to handle. P Q is parallel to E F inside triangle D E F, so the proportionality theorem applies to it directly. It says D P over P E equals D Q over Q F. Piece against piece, on both of the cut sides. That is not yet what we want. We want the whole sides in it, not the pieces.

8:55So use the conversion. Invert both sides, add one to both sides, invert again, and piece against piece becomes piece against the whole. D P over D E equals D Q over D F. And now translate back: D P was A B and D Q was A C. A B over D E equals A C over D F. Two of the three ratios, out of a hypothesis that mentioned no lengths.

9:26Repeat the construction on a different corner and the third ratio follows. The theorem is proved. That argument has several joints, and each one is a place it could be wrong. So it was run. On all forty similar pairs the construction was carried out and every step checked separately: the two marked lengths, the angle between them, the third sides, the planted segment parallel, the proportionality theorem, the conversion, and the ratio at the end. All forty gave the same seven answers.

10:04But a proof that only runs where it succeeds says nothing about its own steps, so it was run where it must fail. Stretch the two sides out of the corner D by different amounts. The corner angle is untouched, so exactly one angle pair still agrees, and the congruence still holds - two sides and the included angle are still right. And the planted segment is no longer parallel. Twenty-four times out of twenty-four.

10:35Turn the second arm before stretching it, so the angle at D is wrong too, and the congruence itself fails - the third sides come out unequal, on all twenty-four. Every joint fails when it should. That is what makes the forty successes mean something. Now a simplification that costs nothing. The criterion asks for three angle equalities. Suppose you check only two. The three angles of a triangle add to a straight angle. So if two of yours agree with two of mine, the third has no freedom left. Mine is a hundred and eighty minus the other two, and so is yours, and those are the same subtraction.

11:23Sixty and forty leave eighty. There is no other option. Over the whole population of ninety-six pairs, the number of matched corners that agree was counted. Fifty-two pairs agree in none. Four agree in exactly one. Forty agree in all three. Exactly two: zero. Never once, in ninety-six chances. And that cell is one the same test can fill - on six general quadrilaterals, where four angles are matched, exactly two agree ten times. In a triangle it simply cannot happen. So the three-angle criterion is a two-angle criterion, and nobody checks the third.

12:11Here is what it looks like in use. Two parallel segments, one above the other. Join the left end of the top one to the right end of the bottom one, and the right end of the top to the left end of the bottom. The two joins cross at a point in the middle. That makes two triangles, one pointing up and one pointing down, meeting at the crossing point.

12:37Each join crosses the two parallels, so the angles it makes at the top and the bottom are alternate angles, and alternate angles are equal. Two equalities, one from each join. The angles at the crossing point are vertically opposite, so equal as well. Two angles was all we needed, so the triangles are similar - and the third equality was free. Tilt one segment so the two are no longer parallel and watch what survives. Exactly one equality: the vertically opposite pair, which never depended on the parallels. Both alternate pairs are gone. That is precisely what parallelism was buying.

13:26And here is why anybody cares. A girl ninety centimetres tall walks away from a lamp-post at one point two metres a second. The lamp is three point six metres up. How long is her shadow after four seconds? She has walked one point two times four: four point eight metres. The post stands vertical and so does she, so the angle at the foot of the post and the angle at her feet are both right angles.

13:57And the angle at the tip of the shadow belongs to both triangles - the same angle, seen twice. Two equal angles, so the large triangle is similar to the small one. No measuring required. So the whole ground distance over her shadow equals the post's height over hers. Three point six over nought point nine is four. Call the shadow x. Four point eight plus x, all over x, equals four. So four point eight equals three x, and the shadow is one point six metres.

14:34One trap, and it catches people every time. Ninety centimetres is nought point nine metres. Put ninety into that ratio against three point six and you get a shadow of minus five metres - behind her, and longer than she has walked. The units have to agree before the numbers meet. One more, and then what all of this does and does not tell you. A pole six metres tall casts a shadow four metres long. At the same moment, a tower casts a shadow twenty-eight metres long. How tall is the tower?

15:13The sun is far enough away that its rays arrive parallel, so they make the same angle with the ground for both. Both stand vertical. Two equal angles again. So the tower's height over its shadow equals the pole's over its shadow. Six times twenty-eight, over four. Forty-two metres. Take that ratio the other way up and you get six sevenths of a metre for a tower, which is the sort of answer that tells you to check.

15:45Now the caution. Equal angles force the sides into one ratio. They say nothing whatever about what that ratio is. The pole and the tower are equiangular and one is seven times the other. Equal angles fix shape and leave size completely open. That is exactly why this is useful. It lets you measure a tower with a stick, because the one thing you can copy from a tower without touching it is an angle.