Commentary on Proposition II.6¶
Back to Proposition II.6.
In this case the rectangle AD, DB is “a rectangle applied to a given straight line (AB) but exceeding by a square (the side of which is equal to BD)”; and the problem suggested by II. 6 is to find a rectangle of this description equal to a given area, which we will, for convenience, suppose to be a square; i.e., in the language of geometry, to apply to a given straight line a rectangle which shall be equal to a given square and shall exceed by a square figure.
We suppose that in Euclid’s figure $AB = a$, $BD = x$; then, if the given square be $b^2$, the problem is to solve geometrically the equation [ ax + x^2 = b^2. ]
The solution of a problem theoretically equivalent to the solution of a quadratic equation of this kind is presupposed in the fragment of Hippocrates’ Quadrature of lunes preserved in a quotation by Simplicius (Comment. in Aristot. Phys. pp. 61—68, ed. Diels) from Eudemus’ History of Geometry. In this fragment Hippocrates (5th cent. B.C.) assumes the following construction.
AB being the diameter and O the centre of a semicircle, and C being the middle point of OB and CD at right angles to AB, a straight line of length such that its square is $1\frac{1}{2}$ times the square on the radius (i.e. of length $a\sqrt{\frac{3}{2}}$, where $a$ is the radius) is to be so placed, as EF, between CD and the circumference AD that it “verges towards B,” that is, EF when produced passes through B.
Now the right-angled triangles BFC, BAE are similar, so that
$$BF : BC = BA : BE,$$
and therefore the rectangle $BE, BF = \text{rect. } BA, BC$
$$= \text{sq. on } BO.$$
In other words, EF ($= a\sqrt{\frac{3}{2}}$) being given in length, BF ($= x$, say) has to be found such that
$$\left(\sqrt{\frac{3}{2}}a + x\right)x = a^2;$$
or the quadratic equation
$$\sqrt{\frac{3}{2}}ax + x^2 = a^2$$
has to be solved.
A straight line of length $a\sqrt{\frac{3}{2}}$ would easily be constructed, for, in the figure, $CD^2 = AC \cdot CB = \frac{3}{4}a^2$, or $CD = \frac{1}{2}a\sqrt{3}$, and $a\sqrt{\frac{3}{2}}$ is the diagonal of a square of which CD is the side.
There is no doubt that Hippocrates could have solved the equation by the geometrical construction given below, but he may have contemplated, on this occasion, the merely mechanical process of placing the straight line of the length required between CD and the circumference AD and moving it until E, F, B were in a straight line. Zeuthen (Die Lehre von den Kegelschnitten im Altertum, pp. 270, 271) thinks this probable because, curiously enough, the fragment speaks immediately afterwards of “joining B to F.”
To solve the equation
$$ax + x^2 = b^2$$
we have to find the rectangle AH, or the gnomon NOP, which is equal in area to $b^2$ and has one of the sides containing the inner right angle equal to CB or $\frac{1}{2}a$. Thus we know $\left(\frac{1}{2}a\right)^2$ and $b^2$, and we have to find, by I. 47, a square equal to the sum of two given squares.
To do this Simson draws BQ at right angles to AB and equal to $b$, joins CQ and, with centre C and radius CQ, describes a circle cutting AB produced in D. Thus BD, or $x$, is found.
Now the rectangle AD, DB together with the square on CB is equal to the square on CD, i.e. to the square on CQ, i.e. to the squares on CB, BQ.
Therefore the rectangle AD, DB is equal to the square on BQ, that is, [ ax + x^2 = b^2. ] From Euclid’s point of view there would only be one solution in this case.
This proposition enables us also to solve the equation [ x^2 - ax = b^2 ] in a similar manner.
We have only to suppose that $AB = a$, and AD (instead of BD) $= x$; then [ x^2 - ax = text{the gnomon.} ] To find the gnomon we have its area ($b^2$) and the area, $CB^2$ or $\left(\frac{1}{2}a\right)^2$, by which the gnomon differs from $CD^2$. Thus we can find D (and therefore AD or $x$) by the same construction as that just given.
Converse propositions to II. 5, 6 are given by Pappus (VII. pp. 948—950) among his lemmas to the Conics of Apollonius to the effect that,
if D be a point dividing AB unequally, and C another point on AB such that the rectangle AD, DB together with the square on CD is equal to the square on AC, then
[ AC text{ is equal to } CB; ]
if D be a point on AB produced, and C a point on AB such that the rectangle AD, DB together with the square on CB is equal to the square on CD, then
[ AC text{ is equal to } CB. ]