Commentary on Euclid I, Proposition 23
On a given right line, and at a given point in it, to constitute an angle equal to a given right lined angle.
This also is a problem, whose invention according to Eudemus is rather the gain of Oenopides than of Euclid: but it requires the construction of an angle, on a given right line; and at a given point in it equal to another right lined angle. This, then, Euclid necessarily adds, that the given angle must be rectilineal; because it is impossible that an angle can be construed on a right line equal to every angle. For it has been shewn that there are only two curve-lined angles equal to right-lined angles, viz. the angle of a lunular figure, which we have proved equal to every right-lined angle; and the angle of that figure similar to an axe, which is equal to two thirds of a right angle. But a lunular figure of this kind, which is called (πελεκοειδὲς) Pelecoides, is formed from two circles cutting each other through their centres. However, the construction of an angle on a certain right line, causes the constituted angle to become determinate, and not indifferent in species, but forms it either right-lined, or mixt. But since no mixt can be equal to a right-lined angle, it is manifest that this must be perfectly rectilineal. The institutor of the Elements, therefore, simply using the present problem, and constructing a triangle from three right lines, equal to three given lines, accomplishes the thing proposed. But you may receive a more exquisite construction of the triangle, by the following method. Let there be a given right line a b, and a given point in it a, and a given right-lined angle c d e. It is required, therefore, to accomplish the problem. Connect c e, and produce a b on both sides to the points f, g. Then place f a, equal to c d, and d e to a b, and b g to e c. And with the centre a, but interval a f, describe the circle k. And again, as in the preceding, with the centre b, but interval b g, describe the circle l. The circles, therefore, will cut each other, as we have shewn in the last proposition. Let them cut each other in the points m, n, and from these points draw right lines to the centres as in the figure. Because, therefore, f a, is equal to a m, and a n, but c d, is equal to f a; a m, and a n, will be each equal to c d. Again, because b g is equal to b m, and b n, but g b is not unequal to c e; b m, and b n, will be also equal to c e.
But a b is equal to d e. The two therefore, a b, a m, are not unequal to the two d e, d c, and the base b m, is equal to the base c e. Hence, the angle m a b, is equal to the angle at the point d. And again, the two n a, a b, are equal to the two c d, d e, and the base n b, is equal to the base c e. The angle, therefore, n a b, is equal to the angle c d e, and the thing proposed is doubly accomplished: for we have not only constituted one, but two angles, equal to the given angle, on each side of the right line a b; so that in whatever part we may desire the construction to be made, it will be indubitable, and without contradiction. And this we have added to the construction of the elementary institutor.
But we cannot praise the method of Apollonius, because it requires the assistance of the third book. For he receives any angle c d e, and a right line a b, and with the centre d, but interval c d, he describes a circumference c e. In like manner with the centre a, but interval a b, he describes a circumference b f; and intercepting a circumference c e, equal to b f, he connects the right line a f, and affirms that the angles a and d, insisting on equal circumferences, are equal. But it is necessary to pre-assume that a b is equal to c d, in order that the circles may be also equal. We therefore think that a demonstration of this kind requiring posterior propositions, is foreign from an elementary institution; and we give the preference to that of the geometrician, as consequent to principles.