The First Six Books of the Elements of Euclid

by Euclid; John Casey

legendre’s and hamilton’s proofs of euclid, I. xxxii.

The discovery of the Proposition that “the sum of the three angles of a triangle is equal to two right angles” is attributed to Pythagoras. Until modern times no proof of it, independent of the theory of parallels, was known. We shall give here two demonstrations, each independent of that theory. These are due to two of the greatest mathematicians of modern times—one, the founder of the Theory of Elliptic Functions; the other, the discoverer of the Calculus of Quaternions.

Legendre’s Proof.—Let ABC be a triangle, of which the side AC is the greatest. Bisect BC in D. Join AD. Then AD is less than AC [I. xix. Ex. 5]. Now, construct a new triangle AB′C′, having the side AC′ = 2AD, and AB′ = AC. Again, bisect B′C′ in D′, and form another triangle AB′′C′′, having AC′′ = 2AD′, and AB′′ = AC′, &c. (1) The sum of the angles of the triangle ABC = the sum of the angles of AB′C′ [I. xvi. Cor. 1] = the sum of the angles of AB′′C′′ = the sum of the angles of AB′′′C′′′, &c. (2) The angle B′AC′ is less than half BAC; the angle B′′AC′′ is less than half B′AC′, and so on; hence the angle B⁽ⁿ⁾A⁽ⁿ⁾ will ultimately become infinitely small. (3) The sum of the base angles of any triangle of the series is equal to the angle of the preceding triangle (see Dem. I. xvi.). Hence, if the annexed diagram represent the triangle AB⁽ⁿ⁺¹⁾C⁽ⁿ⁺¹⁾, the sum of the base angles A and C⁽ⁿ⁺¹⁾ is

equal to the angle B⁽ⁿ⁾C⁽ⁿ⁾; and when n is indefinitely large, this angle is an infinitesimal; hence the point B⁽ⁿ⁺¹⁾ will ultimately be in the line AC, and the angle AB⁽ⁿ⁺¹⁾C⁽ⁿ⁺¹⁾ will become a straight angle [I. Def. x.], that is, it is equal to two right angles; but the sum of the angles of AB⁽ⁿ⁺¹⁾C⁽ⁿ⁺¹⁾ is equal to the sum of the angles ABC. Hence the sum of the three angles of ABC is equal to two right angles.

Hamilton’s Quaternion Proof.—Let ABC be the triangle. Produce BA to D, and make AD equal to AC. Produce CB to E, and make BE equal to BD; finally, produce AC to F, and make CF equal to CE. Denote the exterior angles thus formed by A′, B′, C′. Now let the leg AC of the angle A′ be turned round the point A through the angle A′; then the point C will coincide with D. Again, let the leg BD of the angle B′ be turned round the point B through the angle B′, until BD coincides with BE; then the point D will coincide with E. Lastly, let CE be turned round C, through the angle C′, until CE coincides with CF, and the point E with F. Now, it is evident that by these rotations the point C has been brought successively into the positions D, E, F; hence, by a motion of mere translation along the line FC, the line CA can be brought into its former position. Therefore it follows, since rotation is independent of translation, that the amount of the three rotations is equal to one complete revolution round the point A; therefore A′ + B′ + C′ = four right angles; but

		A + A′ + B + B′ + C + C′	= six right angles [I. xiii.];
	hence	A + B + C	= two right angles.