Turning it around: matching ratios force the line to be parallel

0:00Every statement of the form 'if this, then that' has a twin, made by swapping the two halves over. If it is raining, the ground is wet. Swap them: if the ground is wet, it is raining. The first is true. The second is not - somebody may have washed a car. That twin is called the converse, and the example makes one point: it is a new question. Proving a statement tells you nothing about its converse. Sometimes both hold, sometimes only one. You have to go and find out.

0:36We have a theorem about a triangle: a line across it, parallel to one side, cuts the other two in the same ratio. Its converse swaps the halves over. Suppose a line cuts the two sides in the same ratio. Must it be parallel to the third? That is a different question, and this video is the answer to it. First, why the question is not already settled. The proof of the forward theorem went through area. Join two corners, drop two perpendiculars, and one triangle gets measured twice on two different bases.

1:17The pieces on one side come out as a ratio of areas, and so do the pieces on the other. So the two ratios agree. Read that chain from the top and it starts with the line being parallel. Everything after is a consequence. Now try to read it backwards, starting from the two ratios agreeing. The chain does not run that way. The step that used parallelism used it to say two triangles have the same height. Knowing the ratios agree does not hand that back.

1:54The forward proof has no reverse gear. If the converse is true, it will need an argument of its own. Before proving anything, watch it happen. Draw an angle - two arms meeting at a point A. Along the first arm, step off five equal lengths with a compass: five marks, equal by construction and not by eye. Along the second arm, do the same, with a different step length, so nothing that follows can be an accident of the two arms matching.

2:29Join the far mark on one arm to the far mark on the other. That is the long segment across the corner. Then join first mark to first mark, second to second, third to third, fourth to fourth. Four short segments, each cutting the two arms into two pieces. The question is what those pieces do. Take the first join. On each arm it leaves one step behind it and four in front. One to four.

3:00The second: two behind, three in front. Two to three. The third: three to two. The fourth: four to one. Four segments, four ratios, each the same on both arms - because the marks were stepped off, so the counting is exact. And every one of the four comes out parallel to the long segment. Not nearly parallel. Parallel. The four are balanced about one: a quarter pairs with four, two thirds pairs with three halves. What they are not is evenly spread, and there is a wide gap in the middle where nothing was tested.

3:40So this is evidence. Good evidence. It is not a proof. Here is the trouble with four successes. Plenty of statements are true four times and false the fifth. To see how easily that happens, try other conditions on the same figure in place of the ratios. Eleven triangles of very different shapes, twelve places for the point on one side and twelve for the point on the other, chosen independently. One thousand five hundred and eighty-four configurations, of which one hundred and thirty-two happen to be parallel.

4:20Now run six conditions across all of them, each in both directions. 'The cut bisects both sides' implies parallel, every time - no exceptions in the whole census. Turn it around and it collapses: one hundred and twenty-one of the parallel cuts do not bisect anything. 'The two pieces at the apex are equal in length' fails in both directions. Four configurations have equal pieces and are not parallel; one hundred and twenty are parallel with unequal pieces.

4:58So conditions on this figure genuinely can hold one way and fail the other. Which is exactly why the ratio condition needs proving rather than believing. Here is the statement we are going to prove. Take a triangle A B C, a point D on the side A B and a point E on the side A C. Suppose A D over D B equals A E over E C - the two sides cut in the same ratio.

5:29Then the segment D E is parallel to B C. Two things in that wording are load-bearing. D and E lie on the sides, not on the lines extended past them, and they are two distinct points, one on each side. Drop either and the statement stops being true. Later we will see precisely where it breaks. The hypothesis is about lengths. The conclusion is about direction. Nothing in the hypothesis mentions direction at all, and that is the gap the proof has to cross.

6:07The strategy is the one you reach for when a statement gives you almost nothing to work with. Suppose it fails. Suppose the ratios agree, and D E is not parallel to B C. That supposition is thin. But it can be made to collide with something, and the something has to be built. So build it. From D, draw the line that is parallel to B C. Not a line we hope is parallel - the parallel one, drawn deliberately.

6:39It runs across the triangle and meets the side A C somewhere. Call that point E prime. Draw E prime a good distance from E. Draw them on top of each other and the argument has nothing left to do. Two segments now leave D: the one we were given, and the one we drew. The whole proof is the comparison between them. The constructed segment D E prime is parallel to B C by construction. So the forward theorem applies to it.

7:13It says A D over D B equals A E prime over E prime C. The forward theorem is not being run backwards here. It is used exactly as it stands, on a line that is parallel because we made it parallel. Run that on all one thousand five hundred and eighty-four configurations - construct the parallel line from D each time, and check. It holds in every single one.

7:43Now put the two facts side by side. We assumed A D over D B equals A E over E C. The theorem just gave A D over D B equals A E prime over E prime C. The left-hand sides are identical. So the right-hand sides are equal to each other. A E over E C equals A E prime over E prime C. Two points on the side A C, and the same ratio.

8:14E and E prime both lie on the segment A C. Each cuts it into two pieces, and the two cuts make the same ratio. If a ratio can only be made by one point of a segment, then E and E prime are the same point. And if they are one point, the segment D E is the segment D E prime - which was parallel to B C by construction.

8:41That contradicts the supposition. So the supposition was wrong, and D E was parallel all along. Assume failure, construct the line that cannot fail, force the two to collide. That is the whole shape of it. But it rests entirely on one sentence that has not been justified yet: a ratio can only be made by one point. Start from what we have. A E over E C equals A E prime over E prime C.

9:14Add one to both sides. That is legitimate - adding the same thing to equal quantities leaves them equal. On the left, one is E C over E C, so the left becomes A E plus E C, all over E C. And A E plus E C is the whole side A C, because E sits between A and C. So the left is A C over E C.

9:40The right does the same, and becomes A C over E prime C. So A C over E C equals A C over E prime C. Same numerator, so the denominators are the same length. E C equals E prime C. Two points of A C, at the same distance from C, on the same side of it. They are one point. That is the collision, made exact. Adding one turned a ratio of two pieces into a ratio of a piece to the whole - and the whole was shared.

10:16That step deserves a look on its own, because it is the real content of the converse. Slide a point along the segment A C and watch the ratio it makes. Near A it is almost nothing. Near C it grows without limit. Take one hundred and forty-four positions along the segment. They give one hundred and forty-four different ratios, no repeats, and the ratios climb steadily as the point moves.

10:46So the map from point to ratio never doubles back. Name a ratio, and exactly one point of the segment makes it. Now break it. Let the point off the segment, out past A or out past C, and the pinning stops working. Sample thirty positions running well past both ends: only twenty-seven of the ratios are different, and they no longer climb. That is why the statement insisted both points lie on the sides. It is not decoration. It is what makes the last step legal.

11:25The add-one move is worth keeping, because it converts between two ways of writing the same proportion. A D over D B equals A E over E C - piece against piece. And A D over A B equals A E over A C - piece against the whole side. Students often carry these as two separate results. They are one result in two outfits, and here is the changing room.

11:53Start from piece against piece. Invert both sides: D B over A D equals E C over A E. Add one to both sides. The left becomes A B over A D, the right A C over A E. Invert once more, and there it is: A D over A B equals A E over A C. Three moves, each reversible. Watch a number go through them: three halves inverts to two thirds, add one for five thirds, invert again for three fifths.

12:28Now the two theorems start working as a pair, on a shape that is not a triangle at all. A trapezium: four sides, one pair of them parallel. Call it A B C D, with A B parallel to D C. From a point E on one slanting side to a point F on the other, draw a segment parallel to A B. Since A B is parallel to D C, it is parallel to both.

12:56The claim is that A E over E D equals B F over F C. But there is no triangle here to apply anything to. So make one. Draw the diagonal A C. It cuts the trapezium into two triangles, and it crosses E F at a point - call it G. In the first triangle E G is parallel to a side, so A E over E D equals A G over G C. In the second, G F is parallel to a side, so B F over F C equals A G over G C.

13:34Both equal the same ratio on the diagonal, so they equal each other. The diagonal was the bridge. On five trapezia of very different shapes the three ratios agree every time. Move the segment off the matching fraction and they disagree - so the agreement is a result, not an accident of the drawing. One more use, and it is the cleverest. The converse can manufacture a parallel line you were never given.

14:07In a triangle P Q R, take S on P Q and T on P R with P S over S Q equal to P T over T R. And suppose the angle at S in the small triangle equals the angle at R. The ratios give you, by the converse, that S T is parallel to Q R. Nobody said it was parallel. The converse said it. And now that it is parallel, corresponding angles apply: the angle at S equals the angle at Q.

14:43Combine that with what we were given, and the angle at R equals the angle at Q. Equal angles, so the sides opposite them are equal, and the triangle is isosceles. The parallel line existed only to license one angle equality, and then its work was over. Three last segments, and the only question is whether each is parallel. Thirteen tenths against three halves - unequal, so no. Eight ninths on both sides - parallel.

15:17The third gives the whole sides rather than the pieces, so subtract first: both come to nine fifty-fifths. Parallel. The forward theorem hands you ratios. The converse hands you parallelism. Almost every problem on this figure is a choice of which to reach for.