Commentary on Proposition III.31¶
Back to Proposition III.31.
As already stated, this proposition is immediately deducible from III. 20 if that theorem is extended so as to include the case where the segment is equal to or less than a semicircle, and where consequently the “angle at the centre” is equal to two right angles or greater than two right angles respectively.
There are indications in Aristotle that the proof of the first part of the theorem in use before Euclid’s time proceeded on different lines. Two passages of Aristotle refer to the proposition that the angle in a semicircle is a right angle. The first passage is Anal. Post. II. 11, 94 a 28: “Why is the angle in a semicircle a right angle? Or what makes it a right angle? (τίνος ὄντος ὀρθή;) Suppose A to be a right angle, B half of two right angles, C the angle in a semicircle. Then B is the cause of A, the right angle, being an attribute of C, the angle in the semicircle. For B is equal to A, and C to B; for C is half of two right angles. Therefore it is in virtue of B being half of two right angles that A is an attribute of C; and the latter means the fact that the angle in a semicircle is right.” Now this passage by itself would be consistent with a proof like Euclid’s or the alternative interpolated proof next to be mentioned. But the second passage throws a different light on the subject. This is Metaph. 1051 a 26: “Why is the angle in a semicircle a right angle invariably (καθόλου)? Because, if there be three straight lines, two forming the base, and the third set up at right angles at its middle point, the fact is obvious by simple inspection to any one who knows the property referred to” (ἐκεῖνο is the property that the angles of a triangle are together equal to two right angles, mentioned two lines before). That is to say, the angle at the middle point of the circumference of the semicircle was taken and proved, by means of the two isosceles right-angled triangles, to be the sum of two angles each equal to one-fourth of the sum of the angles of the large triangle in the figure, or of two right angles; and the proof must have been completed by means of the theorem of III. 21 (that angles
in the same segment are equal), which Euclid’s more general proof does not need.
In the Greek texts before that of August there is an alternative proof that the angle BAC (in a semicircle) is right. August and Heiberg relegate it to an Appendix.
“Since the angle AEC is double of the angle BAE (for it is equal to the two interior and opposite angles), while the angle AEB is also double of the angle EAC, the angles AEB, AEC are double of the angle BAC. But the angles AEB, AEC are equal to two right angles; therefore the angle BAC is right.”
Lardner gives a slightly different proof of the second part of the theorem. If ABC be a segment greater than a semicircle, draw the diameter AD, and join CD, CA.
Then, in the triangle ACD, the angle ACD is right (being the angle in a semicircle); therefore the angle ADC is acute. But the angle ADC is equal to the angle ABC in the same segment; therefore the angle ABC is acute.
Euclid’s references in this proposition to the angle of a segment greater or less than a semicircle respectively seem, like the part of III. 16 relating to the angle of a semicircle, to be a survival of ancient controversies and not to be put in deliberately as being an essential part of elementary geometry. Cf. the notes on III. Def. 7 and III. 16.
The corollary ordinarily attached to this proposition is omitted by Heiberg as an interpolation of date later than Theon. It is to this effect: “From this it is manifest that, if one angle of a triangle be equal to the other two, the first angle is right because the exterior angle to it is also equal to the same angles, and if the adjacent angles be equal, they are right.” No doubt the corollary is rightly suspected, because there is no necessity for it here, and the words ὅπερ ἔδει δεῖξαι come before it, not after it, as is usual with Euclid. But, on the other hand, as the fact stated does appear in the proof of III. 31, the Porism would be a Porism after the usual type, and I do not quite follow Heiberg’s argument that, “if Euclid had wished to add it, he ought to have placed it after I. 32.”
It has already been mentioned above (p. 44) that this proposition supplies us with an alternative construction for the problem in III. 17 of drawing the two tangents to a circle from an external point.
Two theorems of some historical interest which follow directly from III. 31 may be mentioned.
The first is a lemma of Pappus on “the 24th problem” of the second Book of Apollonius’ lost treatise on νεύσεις (Pappus VII. p. 812) and is to this effect. If a circle, as DEF, pass through D, the centre of a circle ABC, and if through F, the other point in which the line of centres meets the circle DEF, any straight line be drawn (and produced if necessary) meeting the circle DEF in E and the circle ABC in B, G,
then E is the middle point of BG. For, if DE be joined, the angle DEF (in a semicircle) is a right angle [III. 31]; and DE, being at right angles to the chord BG of the circle ABC, also bisects it [III. 3].
The second is a proposition in the Liber Assumptorum, attributed (no doubt erroneously as regards much of it) to Archimedes, which has reached us through the Arabic (Archimedes, ed. Heiberg, II. pp. 520—521).
If two chords AB, CD in a circle intersect at right angles in a point O, then the sum of the squares on AO, BO, CO, DO is equal to the square on the diameter.
For draw the diameter CE, and join AC, CB, AD, BE.
Then the angle CAO is equal to the angle CEB. (This follows, in the first figure, from III. 21 and, in the second, from I. 13 and III. 22.) Also the angle COA, being right, is equal to the angle CBE which, being the angle in a semicircle, is also right [III. 31].
Therefore the triangles AOC, EBC have two angles equal respectively; whence the third angles ACO, ECB are equal. (In the second figure the angle ACO is, by I. 13 and III. 22, equal to the angle ABD, and therefore the angles ABD, ECB are equal.)
Therefore, in both figures, the arcs AD, BE, and consequently the chords AD, BE subtended by them, are equal. [III. 26, 29]
Now the squares on AO, DO are equal to the square on AD [I. 47], that is, to the square on BE.
And the squares on CO, BO are equal to the square on BC.
Therefore, by addition, the squares on AO, BO, CO, DO are equal to the squares on EB, BC, i.e. to the square on CE. [I. 47]