SSS: sides in proportion drag the angles into agreement

0:00Two figures are the same shape when two things hold: matching corners equal, and matching sides in one ratio. For triangles we have seen one clause swallow the other: equal angles alone force the sides into one ratio. Now turn it round. You are handed no angles at all - just six lengths, three sides of one triangle and three of another. You divide, matched side by matched side, and the three fractions come out the same. No angle has been mentioned.

0:36Does that force the corners to agree? That is a separate question, and at this point you do not know. The answer is yes, and the proof is the same construction as before, walked through the door in the other direction. And when it lands, the two clauses of similarity have collapsed into either one - for triangles, and nothing else. Start with the reason you should not simply assume it.

1:05A square with sides two point one, and beside it a rhombus with sides four point two - a pushed-over square: four equal sides, corners that are not right angles. Every side of the square over its partner in the rhombus gives one half. Four sides, four ratios, one number. The side clause is perfectly satisfied. And not one corner agrees - the square has four right angles and the rhombus has none. So proportional sides do not force equal angles.

1:41Run it as a census. Thirteen four-sided figures - nine rhombi and four rectangles - every one against every one, a hundred and sixty-nine pairs. Sixty-six have all four sides in one ratio with the corners disagreeing. Twenty-four have equal corners with the sides out of ratio. Both failures happen constantly. So whatever makes triangles different is not something four-sided figures share. Here is the experiment. Draw a triangle with sides three, six and eight centimetres. Three plus six is nine, which just clears eight, so it exists - but it is flat.

2:27You can check that without measuring. A formula turns three squared side lengths into sixteen times the squared area, and for three, six and eight it returns nine hundred and thirty-five. Positive, so the triangle is real. Hand it three, six and nine and it returns exactly zero - those lengths lie flat in a line. Now a second triangle with sides four point five, nine and twelve. No angle was constructed. No angle was copied. Only lengths were used.

3:05Three over four point five. Six over nine. Eight over twelve. Every one of them is two thirds. The sides are in one ratio by construction. The question is what the corners did while you were not looking. Measure them, and all three pairs agree. There is a check that needs no protractor. Three side lengths already fix a triangle's angles - that is what rigidity means - so the corners follow from the six numbers you drew with.

3:39Do that for both and the three pairs come out equal exactly, not nearly. And a third opinion, from something that is neither an angle nor a ratio of two sides: the area. The two areas stand in the ratio four to nine - two thirds squared, which is what you would expect if every length in the second triangle really is three halves of its partner. Three separate measurements of three separate things, and they agree.

4:10But this is one pair of triangles, drawn once, and a drawing agrees with almost anything. It suggests the theorem. It does not establish it. So here is the same question asked properly. Nine triangles of very different shapes - right, obtuse, tall and thin, wide and flat, one nearly straight - each moved four ways: turned, reflected, scaled. That gives thirty-six pairs that ought to be similar, and every triangle against every other gives seventy-two that ought not. A hundred and eight.

4:50Two referees, neither able to see the other's answer. The side referee reads squared distances and no angles. The angle referee reads a squared cosine and the sign of a dot product, and no lengths. The report: thirty-six in one ratio and equiangular, seventy-two neither. The cell that would break the theorem - sides in one ratio with the corners disagreeing - is empty. And it is empty because of the shape, not because the test cannot fill it. Hand the same routine those thirteen four-sided figures and it fills that cell sixty-six times.

5:34One routine, two answers. That is the difference between a triangle and everything else, measured rather than asserted. Here is the statement. If two triangles have their matched sides in one ratio, then their matched angles are equal, and the triangles are similar. Sides in. Angles out. The mirror image of the criterion that ran the other way. One thing to keep straight. An older rule uses the same three letters: three equal sides make two triangles congruent.

6:12Congruence wants the sides equal. This wants them proportional. Congruence is the special case where the ratio happens to be one. Among the thirty-six similar pairs, nine are congruent - the ones never scaled - and the other twenty-seven are similar without being congruent. The two rules are not competitors, because the proof of this one calls on the congruence rule at its very last step. Call the smaller triangle A B C and the larger one D E F, with the sides in one ratio in that order.

6:51Same construction as before. On the side D E, measure off from D a length exactly equal to A B, and call it P. On D F, measure off a length equal to A C, and call it Q. Then join P to Q. Both marks land, because the ratio is less than one, so D E and D F are longer than A B and A C. Last time this construction started from three equal angles. This time it starts from three equal ratios.

7:26A B over D E is the ratio, so D P over D E is the ratio. A C over D F is that same ratio, so D Q over D F is too. Subtract each from one and invert, and D P over P E equals D Q over Q F. Piece against piece, on both cut sides. And now a theorem we already have does the work. A line cutting two sides of a triangle in the same ratio must be parallel to the third side - the converse of the proportionality theorem, proved on its own terms.

8:08P cuts D E and Q cuts D F in the same ratio. So P Q is parallel to E F. Notice what just happened. We began with three lengths against three, and we have manufactured a parallel line. This is the exact mirror of the other proof. There, a congruence came first and produced the parallel. Here the ratios produce the parallel, and a congruence is still to come.

8:37Once P Q is parallel to E F, the angle at P equals the angle at E and the angle at Q equals the angle at F, as corresponding angles. Two equal angles is enough, so triangle D P Q is similar to triangle D E F - by the criterion proved just before this one. And that hands back something the proportionality theorem could not. The proportionality theorem speaks only about the two sides that were cut. It says nothing about P Q, the segment we drew.

9:17But similarity does. Because D P Q is similar to D E F, all three ratios agree: D P over D E, D Q over D F, and P Q over E F. So P Q over E F is the ratio. And the hypothesis we started from says B C over E F is the ratio too. Same denominator, same value. So P Q equals B C. Not proportional to it - equal to it.

9:49That is the step the whole proof exists for. Everything before it converts a proportion into the equality of one length, because equality is what congruence needs and a proportion does not give. The middle of this argument is spent buying one equals sign. Now finish. Triangles A B C and D P Q have D P equal to A B, D Q equal to A C, and P Q equal to B C.

10:21Three sides equal. Congruent, by the older rule. And congruent triangles have all their angles equal. So A equals D, B equals E, and C equals F. Which is what we wanted. That argument has seven joints, so it was run. On all twenty-seven cases every step was checked separately, and all twenty-seven agreed. Then it was run where it must fail. Keep the two arms out of D exactly right - A B and A C still in the ratio - and put only the third side wrong.

10:58The marks land. The pieces come out in one ratio. The planted segment is still parallel. The inner triangle is still equiangular with the outer one. And the extraction fails. P Q and B C are two different lengths, so there is no congruence and no angle arrives. Six steps pass and the seventh does not - which is exactly the claim that the seventh is where the work happens.

11:30Step back, because something has changed that is easy to read past. For four-sided figures neither clause implies the other. The rhombus is still standing there, four sides in one ratio and every corner wrong. For triangles, the angle criterion says equal angles give you the ratios. This one says the ratios give you the angles. Put those together and the two clauses have become interchangeable. Verify either one and the other is automatic.

12:05So the six checks you started with are never six. Three angles are enough - two, in fact. Three ratios are enough. You take whichever three you can measure. And that is about triangles and nothing else. The same two sizes, two point one against four point two, drawn as triangles instead, do force the angles. The rigidity of the triangle is the whole of the difference, and this theorem is what that rigidity looks like written down.

12:40Now two places where you use it. A triangle with sides three point eight, six, and three root three. Another with sides twelve, seven point six, and six root three. Similar? Compare them in the order written and you get nonsense - three point eight over twelve is nowhere near six over seven point six. But three point eight and seven point six are a half. Six and twelve are a half. And three root three over six root three is a half, because the root three cancels.

13:16So they are similar - and the pairing is A with R, B with Q, C with P. The order is backwards, and the order is the claim. Say the first triangle has angles eighty and sixty marked, and you want the angle at P. Its third angle is a hundred and eighty minus eighty minus sixty - forty - and it sits at C. C pairs with P, so the angle at P is forty. Take the alphabetical pairing instead and you would have answered eighty.

13:50One more pair, and one more habit worth having. Sides two, two point five and three against four, five and six. Two over four, two point five over five, three over six - all one half. Similar, and again not alphabetically. Now a pair that fails. Sides two point seven, two and three, against four, five and six. In the order written: nought point six seven five, nought point four, nought point five. Three different numbers, so that ordering is wrong.

14:25But that is not yet an answer. One ordering fails; there are six, and the pair might still be similar under another. Here is the move that settles all six at once: sort each triangle's sides. A similarity carries the smallest to the smallest and the largest to the largest - nowhere else for them to go. Sorted: two, two point seven, three against four, five, six. The ratios are a half, nought point five four, and a half. Two of the three agree, and it is still not similar.

15:02Which is the whole lesson in a line. Two ratios agreeing settles nothing. Three ratios agreeing settles everything - and that is a theorem, not a definition.