Getting the section rule from a pair of similar triangles

0:00Two towns again, and this time something has to go between them. A relay tower, on the straight line joining the two, positioned so that it is twice as far from the second town as from the first. Notice what that instruction does not give you. No coordinates, no distance in kilometres, no direction. It gives you a comparison between two lengths, and it expects that to be enough to fix a point exactly.

0:30It is enough. And what makes it enough is one idea from a completely different chapter: similar triangles. There is no new geometry in this whole topic. There is one construction, one similarity, and then some careful algebra. First, turn the words into something you can compute with. The tower sits between the two towns and cuts the segment into two pieces. Twice as far from the second as from the first means the piece nearest the first town is half the piece nearest the second.

1:08So the two pieces stand in the ratio one to two, and that is the whole of the instruction, written down. Now a warning, because this is where the first mistake lives. One to two does not mean one half. It compares the two pieces with each other, not either piece with the whole. One piece plus two pieces is three pieces, so the point sits one third of the way along.

1:35Three to one is not three times the segment. It is three pieces then one piece, four pieces in all, so three quarters of the way. Keep those two readings separate and half the difficulty of this topic disappears. Second, choose where to put the axes, because you are allowed to. Put the first town at the origin, and let one kilometre be one unit on both axes. The second town was thirty-six east and fifteen north, so it is now the point thirty-six, fifteen.

2:12Call the tower's position x comma y. Two unknowns, and one condition that is going to pin down both. That is already the interesting part. A single statement about lengths is about to split into two separate statements about coordinates. Third, the construction, and it is the same move as in the distance work. From the tower, drop a perpendicular straight down to the horizontal axis. From the second town, drop another one.

2:46Then run a horizontal line from the tower across until it meets that second perpendicular. Look at what has appeared. Two right triangles, stacked nose to tail along the same slanting line. The lower one has the origin, the tower, and the foot of the tower's perpendicular. The upper one has the tower, the corner we just made, and the second town. Neither triangle was in the question. Both were manufactured, exactly as before, because right triangles are things we can finish.

3:25Now, why are those two triangles the same shape? The first pair of angles is free. Both triangles were built with a right angle in them, one at each dropped corner. The second pair takes one moment of thought. The horizontal line we drew across from the tower is parallel to the horizontal axis. The slanting line cuts across both of them, so the angle it makes with one is the angle it makes with the other. Corresponding angles.

3:59Two pairs of equal angles is all the similarity test needs. The third pair follows for free, and we never have to look at it. And notice what similar does not mean. It does not mean equal. The two triangles are different sizes on every single configuration except the one where the two pieces are equal, and I checked that: of seven thousand two hundred configurations, exactly one thousand eight hundred have triangles of equal size, and those are precisely the ones where the ratio is one to one.

4:37The whole derivation exists because the sizes differ. If they were equal there would be nothing to work with. Here is what similarity gives you, and it is a lot. Corresponding sides of similar triangles are in the same ratio. So the ratio of the two horizontal legs equals the ratio of the two vertical legs equals the ratio of the two slanting sides. And the two slanting sides are exactly the two pieces the tower cuts the segment into.

5:14So all three of those ratios are one to two. One condition, appearing three separate times, on three different pairs of lengths. That is the trick, and it is worth saying slowly. A two-dimensional requirement has just become two independent one-dimensional requirements, one horizontal and one vertical. So take the horizontal legs first. The lower triangle's horizontal leg runs from the origin across to below the tower. That is x.

5:51The upper triangle's horizontal leg runs from the tower across to below the second town. That is thirty-six minus x. Those two stand as one to two, so twice x equals thirty-six minus x, and x is twelve. The vertical legs give the same shape of equation. The lower one is y, the upper one is fifteen minus y, and twice y equals fifteen minus y makes y five. The tower goes at twelve comma five. Two small linear equations, and neither of them needed a square root.

6:32Now check it, using the rule from the previous topic, which knows nothing about any of this. From the first town at the origin to twelve comma five: twelve squared is a hundred and forty-four, five squared is twenty-five, and those add to a hundred and sixty-nine. The root of a hundred and sixty-nine is thirteen exactly. From twelve comma five to thirty-six, fifteen: the differences are twenty-four and ten, so five hundred and seventy-six plus a hundred, which is six hundred and seventy-six.

7:12And the root of six hundred and seventy-six is twenty-six exactly. Thirteen and twenty-six. That is one to two, which is what we asked for. Two rules, built from different arguments, agreeing on the same point. That is the check worth doing every time. Now run the identical construction with nothing but letters, and get the rule once and for all. The first endpoint is x-one, y-one. The far one is x-two, y-two. The dividing point is x, y, and it cuts the segment so that the near piece stands to the far piece as m-one stands to m-two.

7:56Drop the three perpendiculars, run the two horizontals across, and you have the same two triangles. Now read the four legs off the figure, and this is the pivot of the whole derivation. The lower horizontal leg is x minus x-one. The upper horizontal leg is x-two minus x. The lower vertical leg is y minus y-one. The upper vertical leg is y-two minus y. Every one of them is a difference of two coordinates, needing no measurement at all. Everything from here is algebra.

8:37Similarity says m-one to m-two equals x minus x-one, to x-two minus x. Cross-multiply. m-one times x-two minus x, equals m-two times x minus x-one. Expand both sides, and gather every term carrying x onto one side. You get m-one x-two plus m-two x-one on the left, and x times m-one plus m-two on the right. So x is m-one x-two plus m-two x-one, all over m-one plus m-two. And the vertical proportion runs word for word the same, giving y as m-one y-two plus m-two y-one over m-one plus m-two.

9:24I ran that cross-multiplication and that collection on all seven thousand two hundred configurations, in both coordinates, and it held every time. Look at the shape of what came out, because it is more familiar than it looks. The denominator is the sum of the two parts. The numerator is each endpoint's coordinate multiplied by one of the parts, and then added. That is a weighted average. The answer is pulled towards whichever endpoint carries more weight.

10:00If the two parts are equal, the weights are equal, and you get the ordinary average of the two coordinates: the midpoint. Which is exactly right. Equal pieces means the point sits in the middle. So the section rule is not a new kind of object. It is an average with the weights allowed to be unequal. But there is one thing about that average that catches almost everybody, and it is worth stopping on.

10:31The weight sitting on x-two, the far endpoint, is m-one, the piece nearest the first endpoint. The weights come out crossed. It is tempting to pair each endpoint with its own piece, and that expression looks perfectly reasonable. Here is a five-second test that settles it forever. Set m-one to nothing. If the near piece has no length, the point has to be sitting on the first endpoint. Put m-one equal to zero in the correct rule and you get m-two x-one over m-two, which is x-one. Right.

11:10Put it into the uncrossed version and you get x-two, the wrong end entirely. I ran both limits on eight hundred and seventy pairs of endpoints. The crossed rule returned the right end all eight hundred and seventy times, both ways round. The uncrossed rule returned it zero times. And here is why the error survives so well. On the midpoint, where the two pieces are equal, the two rules give exactly the same answer, on every pair. That is the case you have met most often.

11:46Over ten thousand four hundred and forty cases with twelve different ratios, the uncrossed rule was right on precisely the two thousand six hundred and ten where the two pieces were equal, and wrong on all seven thousand eight hundred and thirty where they were not. No exceptions in either direction. And it lands on the line every single time, so no picture will ever catch it. Only the limit test will.

12:17Two footnotes, both worth having. First, nothing forced us to drop the perpendiculars onto the horizontal axis. Drop them onto the vertical one instead and you get two genuinely different triangles. The corners land in different places on every configuration, and the argument comes out the same. I checked both, on all seven thousand two hundred. That is why the two coordinates behave identically: neither axis was ever special. Second, a ratio does not change if you divide both parts by the same number. Divide both by m-two and the ratio becomes k to one.

13:03The rule becomes k x-two plus x-one, over k plus one, which is the same statement carrying one letter instead of two. I checked that against the two-part form on every one of the ten thousand four hundred and forty cases, and it agrees throughout. Finally, the honest part: what does the drawn argument actually cover? It needs two genuine triangles, and there are two placements where it does not get them.

13:35If the segment stands exactly upright, both horizontal legs collapse to nothing, and the horizontal proportion becomes nothing over nothing. It says absolutely nothing. In my population that happens on a hundred and twenty of the eight hundred and seventy pairs, and on all one thousand four hundred and forty of the cases those produce the vertical proportion still delivers the answer on its own. A segment lying exactly flat does the mirror image, and there the horizontal one carries it. On every pair, at least one of the two survives.

14:17There is one more limit, and it is a real one. Everything here assumed the point lies between the two ends. Feed the rule a ratio with a negative part and it still lands on the line, on all eight hundred and seventy pairs, and it lands between the ends on none of them. That is a different situation, called dividing externally, and it is left for later. What we have derived covers the point between, and covers it completely.