## Proposition 6

#### Explanation of the proof

This proposition is remarkably similar to the last one, II.5, except the point D does not lie on the line AB but on that line extended.

Let b denote the line AB, x denote AD, and y denote BD as in II.5. Then x – y = b (as opposed to x + y = b as in II.5). According to this proposition the rectangle AD by DB, which is the product xy, is the difference of two squares, the large one being the square on the line CD, that is the square of x – b/2, and the small one being the square on the line CB, that is, the square of b/2. Algebraically,

x(xb) = (x – b/2)2 – (b/2)2.

This equation is easily verified with modern algebra, but it's also easily verified in geometry, as done here in the proof.

The geometric proof is primarily an exercise in cutting and pasting. The rectangle AB by DB is the rectangle AM, which is the sum of the rectangles AL and CM. But the rectangles AL, CH, and HF are all equal. Therefore, the rectangle AB by DB equals the gnomon formed by the rectangles CM and HF. That gnomon is the square CF minus the square LG, but the latter equals the square on BC. Thus, the rectangle AB by DB equals the square on DB minus the ssquare on CB.

#### Solution to a quadratic problem

As was II.5, this proposition is set up to help in the solution of a quadratic problem:
Find two numbers x and y so that their difference x – y is a known value b and their product is a known value c2.
In terms of x alone, this is equivalent to solving the quadratic equation x(x – b) = c2. Since this proposition says that x(x – b) = (x – b/2)2 – (b/2)2, the problem reduces to solving the equation

c2 = (xb/2)2 – (b/2)2,

that is, finding CD so that CD2 = (b/2)2 + c2. By I.47, if a right triangle is constructed with one side equal to b/2 and another equal to c, then the hypotenuse will equal the required value for CD. Algebraically, the solutions AD for x and BD for y have the values

This analysis yields a construction to solve the quadratic problem stated above.
 To apply a rectangle equal to a given square to a given straight line but exceeding it by a square. Let AB be the given straight line. Bisect it at C. Construct a perpendicular BQ to AB at B equal to the side of the given square. Draw CQ. Extend AB to D so that BD equals CQ. Then, as described above, AB has been extended to D so that AD times BD equals the given square.

This construction is not found in the Elements, but a generalization of it to parallelograms is proposition VI.29.

#### Use of this proposition

This proposition is used in II.11, III.36, and a lemma for X.29.

Next proposition: II.7

Previous: II.5

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