Then, since the angles BAC, EDF are equal to two right angles, the angles EDH, EDF are equal to two right angles.

But so are the angles EDH, DHF
Therefore the angles EDF, DHF are equal.
And the alternate angles EDF, DFH are equal.
Therefore the angles DHF, DFH are equal,

and DF is equal to DH.

[1. 29]

[1.6]

Hence the two sides ED, DH are equal to the two sides BA, AC; and the included angles are equal.

Therefore the triangles ABC, DEH are equal in all respects.

And the triangles DEF, DEH between the same parallels are equal.

Therefore the triangles ABC, DEF are equal.

[1.37]

[Proclus takes the construction of Eucl. 1. 24, i.e., he makes DH equal to DF and then proves that ED, FH are parallel.]

(2) Suppose the angles BAC, EDF together less than two right angles. As before, make the angle EDG equal to the angle BAČ, draw FH parallel to ED, and join EH.

In this case the angles EDH, EDF are together less than two right angles, while the angles EDH, DHF are equal to two right angles. [1. 29] Hence the angle EDF, and therefore the angle DFH, is less than the angle DHF

Therefore DH is less than DF.

[1. 19]

Produce DH to G so that DG is equal to DF or AC, and join EG. Then the triangle DEG, which is equal to the triangle ABC, is greater than the triangle DEH, and therefore greater than the triangle DEF.

(3) Suppose the angles BAC, EDF together greater than two right angles.

A

We make the same construction in this case, and we prove in like manner that the angle DHF is less than the angle DFH,

whence DH is greater than DF or AC.

Make DG equal to AC, and join EG.

It then follows that the triangle DEF is greater than the triangle ABC. [In the second and third cases again Proclus starts from the construction in 1. 24, and proves, in the second case, that the parallel, FH, to ED cuts DG and, in the third case, that it cuts DG produced.]

demonstrated that the square on the hypotenuse is equal to the squares on the sides about the right angle" (Symp. vIII. 2, 4).

The story of the sacrifice must (as noted by Bretschneider and Hankel) be given up as inconsistent with Pythagorean ritual, which forbade such sacrifices; but there is no reason to doubt that the first distinct formulation and introduction into Greek geometry of the method of application of areas was due to the Pythagoreans. The complete exposition of the application of areas, their exceeding and their falling-short, and of the construction of a rectilineal figure equal to one given figure and similar to another, takes us into the sixth Book of Euclid; but it will be convenient to note here the general features of the theory of application, exceeding and falling-short.

The simple application of a parallelogram of given area to a given straight line as one of its sides is what we have in 1. 44 and 45; the general form of the problem with regard to exceeding and falling-short may be stated thus:

"To apply to a given straight line a rectangle (or, more generally, a parallelogram) equal to a given rectilineal figure and (1) exceeding or (2) falling-short by a square (or, in the more general case, a parallelogram similar to a given parallelogram)."

What is meant by saying that the applied parallelogram (1) exceeds or (2) falls short is that, while its base coincides and is coterminous at one end with the straight line, the said base (1) overlaps or (2) falls short of the straight line at the other end, and the portion by which the applied parallelogram exceeds a parallelogram of the same angle and height on the given straight line (exactly) as base is a parallelogram similar to a given parallelogram (or, in particular cases, a square). In the case where the parallelogram is to fall short, a duopioμós is necessary to express the condition of possibility of solution.

We shall have occasion to see, when we come to the relative propositions in the second and sixth Books, that the general problem here stated is equivalent to that of solving geometrically a mixed quadratic equation. We shall see that, even by means of 11. 5 and 6, we can solve geometrically the equations

ax + x2= b2,
x2- ax = b2;

but in vi. 28, 29 Euclid gives the equivalent of the solution of the general equations

We are now in a position to understand the application of the terms parabola (application), hyperbola (exceeding) and ellipse (falling-short) to conic sections. These names were first so applied by Apollonius as expressing in each case the fundamental property of the curves as stated by him. This fundamental property is the geometrical equivalent of the Cartesian equation referred to any diameter of the conic and the tangent at its extremity as (in general, oblique) axes. If the parameter of the ordinates from the several points of the conic drawn to the given diameter be denoted by p (p being d'a accordingly, in the case of the hyperbola and ellipse, equal to where dis the length of the given diameter and d' that of its conjugate), Apollonius gives the properties of the three conics in the following form.

(1) For the parabola, the square on the ordinate at any point is equal to a rectangle applied to as base with altitude equal to the corresponding abscissa. That is to say, with the usual notation,

y2=px.

(2) For the hyperbola and ellipse, the square on the ordinate is equal to the rectangle applied top having as its width the abscissa and exceeding (for the hyperbola) or falling-short (for the ellipse) by a figure similar and similarly situated to the rectangle contained by the given diameter and p.

The form of these equations will be seen to be exactly the same as that of the general equations above given, and thus Apollonius' nomenclature followed exactly the traditional theory of application, exceeding, and falling-short.

PROPOSITION 45.

To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

Let ABCD be the given rectilineal figure and E the given rectilineal angle;

5 thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E;

[1. 42] 10 let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E.

15

[1. 44]

Then, since the angle E is equal to each of the angles HKF, GHM,

the angle HKF is also equal to the angle GHM. [C. N. 1]

Let the angle KHG be added to each;

therefore the angles FKH, KHG are equal to the angles KHG, GHM.

But the angles FKH, KHG are equal to two right angles;

[1. 29]

20 therefore the angles KHG, GHM are also equal to two right angles.

25

Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles;

therefore KH is in a straight line with HM. [1, 14] And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another.

[1. 29] 30 therefore the angles MHG, HGL are equal to the angles HGF, HGL.

Let the angle HGL be added to each;

35

[C. N. 2] [1. 29]

But the angles MHG, HGL are equal to two right angles; therefore the angles HGF, HGL are also equal to two right angles.

Therefore FG is in a straight line with GL.
And, since FK is equal and parallel to HG,

and HG to ML also,

KF is also equal and parallel to ML;

[C. N. 1]

[1. 14]

[1. 34]

[C. N. 1; 1. 30]

[1. 33]

and the straight lines KM, FL join them (at their extremities); 40 therefore KM, FL are also equal and parallel.

Therefore KFLM is a parallelogram.

And, since the triangle ABD is equal to the parallelogram FH,

and DBC to GM,

45 the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.

Q. E. F.

2, 3, 6, 45, 48. rectilineal figure, in the Greek “rectilineal” simply, without "figure," εὐθύγραμμον being here used as a substantive, like the similarly formed παραλληλόγραμμον.

Transformation of areas.

We can now take stock of how far the propositions 1. 43-45 bring us in the matter of transformation of areas, which constitutes so important a part of

5

what has been fitly called the geometrical algebra of the Greeks. We have now learnt how to represent any rectilineal area, which can of course be resolved into triangles, by a single parallelogram having one side equal to any given straight line and one angle equal to any given rectilineal angle. Most important of all such parallelograms is the rectangle, which is one of the simplest forms in which an area can be shown. Since a rectangle corresponds to the product of two magnitudes in algebra, we see that application to a given straight line of a rectangle equal to a given area is the geometrical equivalent of algebraical division of the product of two quantities by a third. Further than this, it enables us to add or subtract any rectilineal areas and to represent the sum or difference by one rectangle with one side of any given length, the process being the equivalent of obtaining a common factor. But one step still remains, the finding of a square equal to a given rectangle, i.e. to a given rectilineal figure; and this step is not taken till 11. 14. In general, the transformation of combinations of rectangles and squares into other combinations of rectangles and squares is the subject-matter of Book II., with the exception of the expression of the sum of two squares as a single square which appears earlier in the other Pythagorean theorem I. 47. Thus the transformation of rectilineal areas is made complete in one direction, i.e. in the direction of their simplest expression in terms of rectangles and squares, by the end of Book 11. The reverse process of transforming the simpler rectangular area into an equal area which shall be similar to any rectilineal figure requires, of course, the use of proportions, and therefore does not appear till vi. 25.

Proclus adds to his note on this proposition the remark (pp. 422, 24423, 6): "I conceive that it was in consequence of this problem that the ancient geometers were led to investigate the squaring of the circle as well. For, if a parallelogram can be found equal to any rectilineal figure, it is worth inquiring whether it be not also possible to prove rectilineal figures equal to circular. And Archimedes actually proved that any circle is equal to the right-angled triangle which has one of its sides about the right angle [the perpendicular] equal to the radius of the circle and its base equal to the perimeter of the circle. But of this elsewhere."

PROPOSITION 46.

On a given straight line to describe a square.

Let AB be the given straight line;

thus it is required to describe a square on the straight line AB.

Let AC be drawn at right angles to the straight line AB from the point A on it [1. 11], and let AD be made equal to AB;

through the point D let DE be drawn 10 parallel to AB,

A

B

E

and through the point B let BE be drawn parallel to AD.

[1. 31]