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Proposition 14

To construct a square equal to a given rectilinear figure.
Let A be the given rectilinear figure. It is required to construct a square equal to the rectilinear figure A.
Construct the rectangular parallelogram BD equal to the rectilinear figure A. | I.45

Then, if BE equals ED, then that which was proposed is done, for a square BD has been constructed equal to the rectilinear figure A. But, if not, one of the straight lines BE or ED is greater.
Let BE be greater, and produce it to F. Make EF equal to ED, and bisect BF at G. | I.3 I.10
Describe the semicircle BHF with center G and radius one of the straight lines GB or GF. Produce DE to H, and join GH. | I.Def.18
Then, since the straight line BF has been cut into equal segments at G and into unequal segments at E, the rectangle BE by EF together with the square on EG equals the square on GF. | II.5
But GF equals GH, therefore the rectangle BE by EF together with the square on GE equals the square on GH.
But the sum of the squares on HE and EG equals the square on GH, therefore the rectangle BE by EF together with the square on GE equals the sum of the squares on HE and EG. | I.47
Subtract the square on GE from each. Therefore the remaining rectangle BE by EF equals the square on EH. But the rectangle BE by EF is BD, for EF equals ED, therefore the parallelogram BD equals the square on HE. And BD equals the rectilinear figure A. Therefore the rectilinear figure A also equals the square which can be described on EH. Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilinear figure A.
Q.E.F.

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This proposition puts the finishing touches on the theory of areas of rectilinear figures. Proposition I.45 on application of areas of rectilinear figures allows us to assume that the figure under question is a rectangle. Then the semicircle construction finds what is called the "mean proportional" between the sides of the rectangle. If the sides of the rectangle are denoted a and b, then the mean proportional x between them satisfies the proportion a:x = x:b, and that's equivalent to an equality of areas ab = x2, that is to say, the square on this mean proportional has the same area as the rectangle. Thus, any rectilinear figure can be squared.

This result is an end in itself. It is not used in the rest of the Elements.

The ideas of application of areas, quadrature, and proportion go back to the Pythagoreans, but Euclid does not present Eudoxus' theory of proportion until Book V, and the geometry depending on it is not presented until Book VI. There, proposition VI.13 uses the semicircle to construct the mean proportional of two straight lines, and proposition VI.17 shows the square on the mean proportional equals the rectangle on the two straight lines.

#### Squaring the circle

What about circles and other shapes? The general theory of circles is treated in Book III, but there are no propositions about the areas of circles until book XII. Proposition XII.2 says the areas of circles are proportional to the squares on their diameters. That allows the area of two circles to be compared, but it doesn't answer the question "what's the area of this circle?" in the same way that this proposition does for rectilinear figures. That would require finding a square equal to a given circle.

This problem, "quadrature of the circle," was one of three famous problems that goes back at least to the time of Anaxagoras, about 150 years before Euclid. It is equivalent to constructing a line segment of length pi (relative to a unit length). This problem was solved by ancient Greek geometers but not by means of the Euclidean tools of straightedge and compass; higher curves were required. In fact, by the time of Pappus it was believed that the circle could not be squared using only straightedge, compass, and even the conic sections (parabola, hyperbola, and ellipse). But the ancient Greeks had no mathematical proof that it could not be squared.

That the circle could not be squared with Euclidean tools was not shown until 1882 when Lindemann proved that pi is a transcendental number.

Book II Introduction - Proposition II.13 - Book III Introduction.

© 1996
D.E.Joyce
Clark University