SAS: one angle plus the two sides that enclose it

0:00You already know four rules for deciding that two triangles are identical copies of each other. Three sides. Two sides and the angle between them. Two angles and a side. And, for right triangles, the hypotenuse and one other side. Similarity asks a weaker question - same shape, any size - so each has a weaker partner: ratios in place of equal lengths. One thing crosses the other way. Equal angles are enough for similarity and were never enough for congruence.

0:37Run a hundred and eight pairs of triangles through both questions. Thirty-six come back equiangular. Only nine have three matching side lengths. Today is the second rule on that list, and it turns on a single word that is very easy to read straight past. Start by drawing. Take an angle - any opening you like - and mark two lengths along its arms. Two units on one, four on the other.

1:08Now draw a second angle with the same opening, and mark three units and six along its arms. Join each pair of endpoints and you have two triangles. Two over three is two thirds. Four over six is two thirds. The two ratios agree, and the marked opening is the same in both. Measure the four remaining angles and they come out equal in pairs. Squaring the lengths to keep the arithmetic exact, both ratios come out four over nine - two thirds, squared - and of the six ways to match the corners, exactly one works.

1:48The drawing is not a proof, but it is a strong hint. Here is the rule the drawing is hinting at. If one angle of a triangle equals one angle of another, and the two sides enclosing that angle are in one ratio with the two sides enclosing the other, the triangles are similar. Two side ratios and one angle - three conditions, the same as the three-sides rule. It is the rule you reach for when a question hands you lengths and one angle together.

2:24And the whole of it hangs on that one word. Enclosing. The angle has to sit at the vertex where those two sides meet. Not somewhere else in the triangle. There. Think of the two sides as two rigid rods, joined at one end by a pivot. A hinge. The rods cannot stretch. The only thing that can change is how far the hinge is open, and the third side is the gap between the two far ends.

2:55Open it wider and the gap grows. Close it and the gap shrinks. One opening, one gap. Take two rods of length five and seven and step the opening through twenty-three positions. You get twenty-three third sides, no two the same. Hold the opening fixed instead and take twenty-three readings, and you get one answer twenty-three times. That is the whole content of the rule. Fix the opening, fix both arms, and the third side has nowhere to go.

3:29Which tells you why the angle has to be the enclosed one. An angle somewhere else is not the hinge between those two rods. It is not holding anything. Now prove it, using the same construction that proves every criterion in this family. Call the small one A B C and the large one D E F, with the angle at A equal to the angle at D, and A B over D E equal to A C over D F.

4:01Do not try to shrink the big one. Plant a copy of the small one inside it. Mark P on the side D E so that D P is exactly as long as A B. Mark Q on D F so that D Q is exactly as long as A C. Then join P to Q. Two things now have to be checked, and they lean on different parts of the hypothesis.

4:26Run the construction twenty-seven times, on twenty-seven different pairs, and only one pattern of answers ever comes back. Inside the big triangle sits D P Q. Compare it with A B C. D P equals A B, by construction. D Q equals A C, by construction. And the angle at D equals the angle at A - that was given. Two sides and the angle between them: that is congruence, and it says D P Q is an exact copy of A B C.

5:04So their third sides match, and so do their remaining corners: P equals B, and Q equals C. That is the only place the given angle is spent, and it is spent buying that congruence. Watch what happens without it. Keep both arms the right length and turn one, so the opening is wrong and nothing else is. Run that eighteen times. The marks still land on the sides. The pieces still come out in the right proportion. And the congruence dies, and every corner it was going to hand over dies with it.

5:43So far the argument has never mentioned E or F. It has to: the claim is about D E F, not D P Q. That is what the second half buys. D P over P E equals D Q over Q F - the given proportion, rearranged. A segment cutting two sides of a triangle in the same proportion runs parallel to the third side. So P Q is parallel to E F.

6:13Parallel lines cutting across D E and D F give equal corresponding angles: P equals E, and Q equals F. Now chain the halves. B equals P, and P equals E. C equals Q, and Q equals F. With A equal to D already, all three corners agree. The parallel connects the copy to the big triangle. The congruence connects it to the small one. Remove either and the chain has a hole in it.

6:46And here is the telling part of that control: with the opening wrong, the segment was still parallel. The parallel never needed the angle. Only the congruence did. Now the failure. Go slowly, because every number in it looks cooperative. Triangle one: sides of two and a half and three, and a marked angle. Triangle two: sides of five and six, and a marked angle of the same size. Two and a half over five is a half. Three over six is a half. Both ratios agree. Both marked angles agree. It looks finished.

7:28It is not. In the second triangle the marked angle sits between the five and the six. In the first, it does not. The first triangle's two given sides meet at one vertex and the mark is at a different one. It is not the hinge between them, so it is holding nothing. Work out what that third side actually is and it comes out exactly two. Squared, four. For the triangles to be similar, that squared length would have to be a quarter of the second triangle's, which is fifty-three and a half. A quarter of that is thirteen point three seven five.

8:10Four is not thirteen point three seven five. They are not similar, under any matching of the letters at all. And now move nothing except the mark. Same lengths - two and a half and three, against five and six. Same opening. The only change is that in the first triangle the mark now sits between its two given sides. That single move takes its squared third side from four to thirteen point three seven five - exactly a quarter of fifty-three and a half.

8:46And the triangles are similar, under exactly one matching of the letters. Do that for ten pairs of lengths. With the mark off the enclosed corner, seven come out as clean counter-examples: matching angles, both ratios a half, and none similar. Put the mark back and all ten are similar. Same lengths, same angles, opposite answers. So when a question hands you two sides and an angle, ask first not what the numbers are, but where the angle is.

9:21The rule earns its keep when the equal angle arrives for free. Here is the classic way. Two segments crossing at O. The first runs from A through O to B, the second from C through O to D. You are told one thing: O A times O B equals O C times O D. A product, not a ratio - and the rule needs a ratio. Divide both sides by O C times O B. On the left, O A over O C. On the right, O D over O B.

9:56Watch which letters land together: the tempting rearrangement, O A over O B equals O C over O D, is a different statement and it is false here. Build thirty-six of these crossings. The correct rearrangement holds on all of them. The tempting one on none. Now the free angle. Angle A O D and angle C O B are vertically opposite, so they are equal - and they sit exactly between the four sides in that proportion.

10:30So triangle A O D is similar to triangle C O B, all thirty-six times. Write the last two letters the other way round and it is similar zero times. The proportion chose the pairing, not the alphabet. A second place the rule pays. Two similar triangles, and in each draw a median - the segment from a corner to the midpoint of the opposite side. Take A B in the first and P Q in the second. Their ratio is the similarity ratio. Now take A M and P N, the halves of those sides.

11:11Halving the top of a fraction and the bottom changes nothing, so A M over P N is the same number as A B over P Q. That gives two sides in one ratio; the angle at A equals the angle at P; and that angle is enclosed by exactly those two sides. So the half-triangles are similar. And once they are, the two medians themselves are in that same ratio.

11:41Check it on thirty-six similar pairs and all four statements hold every time. Check the last one on seventy-two pairs that are not similar and it holds on none. So a median can stand in for a side wherever a criterion asks for a ratio. Two more things. Right triangles get an extra rule. If the hypotenuse and one other side of one are in the same ratio as the hypotenuse and one other side of another, they are similar - and those two sides need enclose nothing.

12:20Three hundred and twenty-four ordered pairs of right triangles. The test says yes on fifty-four; all fifty-four are similar; and thirty-six of them are different sizes. Run that same two-sides-in-one-ratio test on triangles with no right angle and it says yes seven times - and not one of those seven is similar. The right angle is what removes the second answer. Which brings us to the rule that is not a rule.

12:53Two sides and an angle that is not between them. Take a side of five, a side of four, and a fixed angle at the far end of the five. Solve for the third side and you get two answers - four and a half, and two. Both are real triangles, with the same lengths and the same marked angle, and they are not the same shape. Two answers is one too many, which is why no such rule exists.

13:22One last check, on the whole thing. A hundred and eight pairs of triangles. For each, ask two questions: does one angle match with the two enclosing sides in one ratio, and are they similar? Thirty-six say yes to both. Seventy-two say no to both. And the two mixed boxes - rule satisfied but not similar, similar but failing the rule - are both empty. That emptiness is the theorem. Now run the identical test on four-sided figures.

13:58Three hundred and sixty-one pairs. Rule satisfied and similar: fifty-one. Similar without the rule: twenty-four. And the box that was empty for triangles: twenty-eight. Twenty-eight pairs where every condition the rule asks for is met and the figures are different shapes. A quadrilateral has a fourth vertex, and the rule says nothing about where it goes. A triangle has no fourth vertex. That is the whole reason the rule is a rule.

14:33So: one angle, and the two sides that meet at it. Not any angle. The one the two sides are holding.