Commentary on Proposition I.9¶
Back to Proposition I.9.
It will be observed from the translation of this proposition that Euclid does not say, in his description of the construction, that the equilateral triangle should be constructed on the side of DE opposite to A; he leaves this to be inferred from his figure. There is no particular value in Proclus’ explanation as to how we should proceed in case any one should assert that he could not recognise the existence of any space below DE. He supposes, then, the equilateral triangle described on the side of DE towards A, and hence has to consider three cases according as the vertex of the equilateral triangle falls on A, above A or below it. The second and third cases do not differ substantially from Euclid’s. In the first case, where ADE is the equilateral triangle constructed on DE, take any point F on AD, and from AE cut off AG equal to AF. Join DG, EF meeting in H; and join AH. Then AH is the bisector required.
Proclus also answers the possible objection that might be raised to Euclid’s proof on the ground that it assumes that, if the equilateral triangle be described on the side of DE opposite to A, its vertex F will lie within the angle BAC. The objector is supposed to argue that this is not necessary, but that F might fall either on one of the lines forming the angle or outside it altogether. The two cases are disposed of thus.
Suppose F to fall as shown in the two figures below respectively.
Then, since FD is equal to FE, the angle FDE is equal to the angle FED. Therefore the angle CED is greater than the angle FDE; and, in the second figure, a fortiori, the angle CED is greater than the angle BDE. But, since ADE is an isosceles triangle, and the equal sides are produced, the angles under the base are equal, i.e., the angle CED is equal to the angle BDE. But the angle CED was proved greater: which is impossible.
Here then is the second case in which, in Proclus’ view, the second part of I. 5 is useful for refuting objections.
On this proposition Proclus takes occasion (p. 271, 15—19) to emphasize the fact that the given angle must be rectilineal, since the bisection of any sort of angle (including angles made by curves with one another or with straight lines) is not matter for an elementary treatise, besides which it is questionable whether such bisection is always possible. “Thus it is difficult to say whether it is possible to bisect the so-called horn-like angle” (formed by the circumference of a circle and a tangent to it).
Trisection of an angle.¶
Further it is here that Proclus gives us his valuable historical note about the trisection of any acute angle, which (as well as the division of an angle in any given ratio) requires resort to other curves than circles, i.e. curves of the species which, after Geminus, he calls “mixed.” “This,” he says (p. 272, 1—12), “is shown by those who have set themselves the task of trisecting such a given rectilineal angle. For Nicomedes trisected any rectilineal angle by means of the conchoidal lines, the origin, order, and properties of which he has handed down to us, being himself the discoverer of their peculiarity. Others have done the same thing by means of the quadratrices of Hippias and Nicomedes, thereby again using ‘mixed’ curves. But others, starting from the Archimedean spirals, cut a given rectilineal angle in a given ratio.”
(a) Trisection by means of the conchoid.¶
I have already spoken of the conchoid of Nicomedes (note on Def. 2, pp. 160—1); it remains to show how it could be used for trisecting an angle. Pappus explains this (iv. pp. 274—5) as follows.
Let ABC be the given acute angle, and from any point A in AB draw AC perpendicular to BC.
Complete the parallelogram FBCA and produce FA to a point E such that, if BE be joined, BE intercepts between AC and AE a length DE equal to twice AB.
I say that the angle EBC is one-third of the angle ABC.
For, joining A to G, the middle point of DE, we have the three straight lines AG, DG, EG equal, and the angle AGD is double of the angle AED or EBC.
But DE is double of AB; therefore AG, which is equal to DG, is equal to AB. Hence the angle AGD is equal to the angle ABG. Therefore the angle ABD is also double of the angle EBC; so that the angle EBC is one-third of the angle ABC.
So far Pappus, who reduces the construction to the drawing of BE so that DE shall be equal to twice AB.
This is what the conchoid constructed with B as pole, AC as directrix, and distance equal to twice AB enables us to do; for that conchoid cuts AE in the required point E.
(b) Use of the quadratrix.¶
The plural quadratrices in the above passage is a Hellenism for the singular quadratrix, which was a curve discovered by Hippias of Elis about 420 B.C. According to Proclus (p. 356, 11) Hippias proved its properties; and we are told (1) in the passage quoted above that Nicomedes also investigated it and that it was used for trisecting an angle, and (2) by Pappus (iv. pp. 250, 33—252, 4) that it was used by Dinostratus and Nicomedes and some more recent writers for squaring the circle, whence its name. It is described thus (Pappus iv. p. 252).
Suppose that ABCD is a square and BED a quadrant of a circle with centre A.
Suppose (1) that a radius of the circle moves uniformly about A from the position AB to the position AD, and (2) that in the same time the line BC moves uniformly, always parallel to itself, and with its extremity B moving along BA, from the position BC to the position AD.
Then the radius AE and the moving line BC determine at any instant by their intersection a point F.
The locus of F is the quadratrix.
The property of the curve is that, if F is any point, the arc BED is to the arc ED as AB is to FH.
In other words, if φ is the angle FAD, ρ the radius vector AF and a the side of the square,
$$(\rho \sin \phi)/a = \phi/\frac{1}{2}\pi.$$
Now the angle EAD can not only be trisected but divided in any given ratio by means of the quadratrix (Pappus iv. p. 286).
For let FH be divided at K in the given ratio.
Draw KL parallel to AD, meeting the curve in L; join AL and produce it to meet the circle in N.
Then the angles EAN, NAD are in the ratio of FK to KH, as is easily proved.
(c) Use of the spiral of Archimedes.¶
The trisection of an angle, or the division of an angle in any ratio, by means of the spiral of Archimedes is of course an equally simple matter. Suppose any angle included between the two radii vectores $OA$ and $OB$ of the spiral, and let it be required to cut the angle $AOB$ in a given ratio. Since the radius vector increases proportionally with the angle described by the vector which generates the curve (reckoned from the original position of the vector coinciding with the initial line to the particular position assumed), we have only to take the radius vector $OB$ (the greater of the two $OA$, $OB$), mark off $OC$ along it equal to $OA$, cut $CB$ in the given ratio (at $D$ say), and then draw the circle with centre $O$ and radius $OD$ cutting the spiral in $E$. Then $OE$ will divide the angle $AOB$ in the required manner.