A line drawn parallel to a side cuts the other two in matching ratios

0:00A triangle is a polygon, so the test we already have applies to it. Two triangles are the same shape when their matching corners are equal and their matching sides are in one ratio. Count what that asks for. Three angles to compare, and three ratios to compute. Six checks. Six checks, every time, for the simplest figure there is. And there is something suspicious about that. A triangle is rigid in a way a four-sided figure is not. Push on the corner of a square and it becomes a rhombus. Push on the corner of a triangle and nothing gives.

0:43Three sides fix a triangle completely. Three angles fix its shape completely. So six checks is almost certainly more than we need. Getting six down to something smaller is what the next stretch of this subject is for. And it starts, unexpectedly, with a single fact about area. The claim that started this is about two and a half thousand years old. Thales, working in the eastern Mediterranean, is said to have held that if two triangles have their matched angles equal, then any two matched sides stand in the same ratio.

1:24Equal angles force proportional sides. That is one half of our six checks doing the work of the other half. Nobody knows how he argued for it. But the result underneath it still carries his name, and it is not about two triangles at all. It is about one triangle, and one line drawn across it. Draw a line inside a triangle, parallel to one of the sides. It will cut the other two sides.

1:55The claim is that it cuts them in the same ratio as each other. And that is what we are going to prove. Before proving anything, it is worth watching it happen. So, a construction. Start with an angle. Two arms, meeting at a point. Call that point A. Along one arm, step off five equal lengths - stepped with a compass, so the five are equal by construction and not by eye.

2:24The fifth mark is the far one; call it B. The third mark, two steps short of it, call D. So A to D is three steps and D to B is two. A ratio of three to two, known exactly, because we made it. Now from B draw any line at all across to the other arm. Call where it lands C. And through D draw a second line, this one parallel to the first. It crosses the other arm between A and C. Call that E.

2:54Now measure A to E, and E to C. The claim is that they come out three to two as well. And they do. Every time. Not just for the line we happened to draw. 6 different lines from B, landing in 6 different places, and the ratio comes out three to two on all 6. Move the mark as well. Draw the parallel through the first mark, one step against four, and the other arm splits one to four. Through the second mark, two against three, and the other arm gives two to three.

3:324 marks, 6 lines, 24 constructions, landing on 19 different points, and every one agrees. So is that a proof? No. And it is worth being precise about why not. A measurement tells you what happened this time, to within the width of a pencil line. There are infinitely many lines we could have drawn, and we checked 24 of them. What we have is very good evidence, and no account whatever of why.

4:03Evidence points at a theorem. It is not one. So here is the statement, carefully. If a line is drawn parallel to one side of a triangle, and it meets the other two sides at two distinct points, then it divides those two sides in the same ratio. Read the middle clause again, because it is doing real work. Two distinct points. A line through the apex meets both of the other sides - at the apex, the same point twice. There is nothing there to take a ratio of, so the statement rules it out.

4:42Everything else about the line is free. It can sit anywhere between the apex and the third side. It does not have to be halfway. One hypothesis: parallel. One conclusion: matching ratios. That is the whole theorem. Now the proof, and the first thing to say is what it uses. Not lengths, cleverly rearranged. Area. Here is the figure. A triangle, apex A at the top, B at the bottom left, C at the bottom right.

5:16D is a point on the side A B, E is a point on the side A C, and the line D E is drawn across. Two things get added, and neither is obvious in advance. First: join B to E, and join C to D. Two segments straight across the figure, crossing somewhere in the middle. Second: drop two perpendiculars. From D onto the side A C. From E onto the side A B.

5:45Those are not the altitudes of the triangle - neither of them starts at a corner. They are heights for the small triangles, not for the big one. That is the whole construction. Two segments joined, two perpendiculars dropped. Look at the perpendicular from E onto A B. Call its length h. Two triangles sit on that line with their apex at E. One stands on the piece A D: its area is half of A D times h. The other stands on the piece D B: half of D B times h.

6:24The same h. One perpendicular serving both, because both have the same apex and both stand on the same straight line. Now divide one area by the other. The half cancels. The h cancels. What is left is A D over D B - the ratio we care about, arriving as a ratio of areas. And notice what has not been used. Nothing about parallel lines. This is true wherever D and E happen to sit.

6:57Base three and base two with a height of 4: areas of 6 and 4, which is three to two. Change the height to 10: 15 and 10, still three to two. The height was never going to matter. Now the same thing on the other side, with the other perpendicular - the one from D onto A C. Call it k. Two triangles stand on the line A C with their apex at D. One on the piece A E, one on the piece E C.

7:30Half of A E times k, and half of E C times k. Divide; the half and the k cancel; what is left is A E over E C. So we have both ratios now, each as a ratio of two areas. But look at which four triangles those are, because one of them appears twice. The triangle A D E is the top area in the first pair, on base A D with height h. And the top area in the second pair, on base A E with height k.

8:04Same triangle, two different bases, two different perpendiculars, one area. That is the pivot of the whole proof, and it means the two fractions share a numerator. Two fractions sharing their top. A D over D B is one area over another. A E over E C is that same area over a third. If the two bottoms were equal, the two fractions would be equal and we would be finished. So: are they?

8:36The two bottoms are the triangle B D E and the triangle D E C. They stand on the same segment, D E. That is the base they share. And their remaining corners are B and C, both of which lie on the line B C - which is parallel to D E. That is the hypothesis, and this is the only place in the entire argument where it is used.

9:03Two triangles on one base, with their far corners on a line parallel to that base, have the same height and therefore the same area. Their shapes can be completely different. One tall and lean, the other squat. Same base, same height, same area. So the two bottoms are equal. And that finishes it. The tops are the same triangle. The bottoms are equal. So both fractions equal the same thing, and therefore they equal each other. A D over D B is A E over E C, which is what we set out to prove.

9:44Step back and look at the shape of that argument, because it is worth having. A ratio of lengths is hard to move between two different triangles. There is nothing to hold on to. Turn it into a ratio of areas and suddenly there is, because area does not care about shape. That is what the perpendiculars were for. They convert a length into an area, and the reason they cancel is that each one gets used twice.

10:14The parallel hypothesis then does exactly one job, right at the end, and it is a job about area rather than about length. Used in exactly one place is a claim worth testing rather than asserting. So here is the whole argument run on 972 configurations. 12 triangles of very different shapes, 9 positions for the first point and 9 for the second, chosen independently. Because the two points are placed with no reference to each other, only 108 of the 972 give a line parallel to the third side. Most are not parallel at all.

11:02Now: the two area identities. Both hold on all 972. Every configuration, parallel or not. Those steps never needed the hypothesis. And the pivot - the same triangle measured on two bases with two perpendiculars - holds on all 972 as well. Lift that second point off the side it is meant to sit on and it holds on none of them, so it is a real identity and not a tautology.

11:36The one step that does need the hypothesis is the equal bottoms, and that one is exactly parallel: equal wherever the line is parallel, unequal wherever it is not, with no exceptions in either direction. So the hypothesis really is spent once. Everything else was always going to be true. What the theorem says and what people hear are not always the same, so let us be careful. It does not say the parallel line cuts the sides in half. Three to two is a perfectly ordinary answer.

12:15In the same 972 configurations, 108 cut the first side into two equal pieces, and only 12 of those were parallel lines. Bisecting is one case among infinitely many, not the content. And there is a second, more dangerous misreading. You may write the ratio in either of two forms. Piece over piece, on both sides. Or piece over whole, on both sides. Both are correct, and they are genuinely different numbers: three parts against two is three to two, three parts against the whole five is three to five.

12:58What you may not do is mix them. A D over A B equals A E over E C is not a form of this theorem. Across those 972 configurations that mixed statement is true of 48 - and not one of the 48 is a line parallel to the third side. It is not a weaker theorem. It is a different claim, false where this one is true. One more thing, about the step everything hinged on.

13:30The two bottom triangles have equal areas. It is tempting to hear that as: they are the same triangle. They are not, and it matters, because the proof would be much weaker if it needed them to be. Same base, same height, same area. That is all we claimed and all we used. Their side lengths can be completely different. Across all 972 configurations, the two come out actually congruent in 9 - and all 9 are in the one triangle that had two equal sides to start with.

14:07So congruence is a coincidence of symmetry. Equal area is the rule. Equality of area is a much weaker condition than sameness of shape, and weaker conditions are more useful, because more things satisfy them. The proof worked precisely because it asked for the weakest thing that would do the job. So where does that leave the six checks? We now have a way of turning a statement about a parallel line into a statement about ratios, and back again. That is the lever.

14:44A pair of triangles with equal angles can be slid one inside the other so that one is cut off by a line parallel to a side - and this theorem then hands you the proportional sides for free. Which is Thales's claim, no longer a claim. That is how six checks becomes two, and then one. Not by finding a shortcut, but by proving that the checks were never independent.

15:12All of it rests on a line drawn parallel to a side, and on the single observation that two triangles between the same parallels have the same area. One fact about area. Everything else was arithmetic.