 Our ability to compare two ratios leads to an important result. So you could first of all define that magnitudes in the same ratio are said to be proportional. Now, notice that we have to have two ratios, so A and B, as well as C and D, must have a ratio. However, there's no requirement that B and C have a ratio. They could be entirely different things that we can't find a ratio between. Now, if they did have a ratio, if B and C were the same things, then Euclid noted that a proportion required at least three quantities. And so Euclid then considered, suppose that we have three magnitudes proportional, then the first is said to have the third a duplicate ratio of that which it has to the second. So again, this language of ratio and proportion might not be as familiar as they should be, so let's consider this. If we view our ratios as fractions with positive real numbers, then our proportion A is to B as B is to C. We can read that as A B is equal to B C. And let's see what this duplicate ratio really means. What we want to do is we want to find the ratio of the first to the third and compare it to the ratio of the first to the second. So introducing some horrifically anachronistic algebra. If we multiply both sides by A B and simplify, and over here on the right-hand side that's our ratio of the first to the third, and over on the left-hand side that's our duplicate ratio of the first to the second. And that means the duplicate ratio is what we think of as the square of the ratio. What's worth noting here is that over on the left-hand side we have the ratio between the squares of two quantities, but it's still a ratio between two quantities A and C. What this means is that we can express ratios of squares as ratios of sides. So let's do that. If the two sides of the two squares are A and B, then the ratio of the two squares is the ratio A squared to B squared. Now this is the duplicate ratio of A to B. And so A and B are the first and second terms in the proportion. A is to B as B is to C. And this continued proportion gives us A squared is to B squared as A is to C. And this is important because that euclid generalizes it to the idea of a continued proportion. When four magnitudes are continuously proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually whatever the proportion. Again, let's see what this triplicate ratio is. So I have four magnitudes that are continuously proportional. So again, regarding these ratios as fractions with positive real numbers, then we have. So A is to B as B is to C. And B is to C as C is to D. And again, we'll anachronistically multiply this first ratio by A divided by B. And there's our duplicate ratio. Now if we multiply by A B's again, if we multiply on the left by A B's and on the right by something equal to A B's, let's see, A over B is B over C, which is C over T. So I can multiply on the left by A over B and on the right by C over D. And simplify. And so here I have my triplicate ratio. A cubed is to B cubed as A is to D. That's the first to the fourth. But euclid says we can go on. So suppose I extend our continued proportionality. C is to D as D is to E. And remember these are all equal to A over B. So if I multiply on the left by A over B and on the right by D over E, we get. And now I have the ratio of two fourth powers is equal to the ratio between the first and the fifth. And here's the important thing. If A and B are the lengths of line segments, then C, D, E and so on also has to be line segments. But it's not clear what A to the fourth, B to the fourth are. But the ratio of A to the fourth and B to the fourth is equal to the ratio between two line segments. So we might express a ratio of two cubes. So if our two cubes are A cubed and B cubed, then A and B are the first two magnitudes in a proportionality. A is to B as B is to C. And since these are cubes, we want to continue C is to D. And this continued proportionality gives us the triplicate ratio A cubed is to B cubed as A is to D. And again, remember that A and D are line segments. And so this gives the relationship between two cubes in terms of the relationship between two lines.