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.]; henceA + B + C = two right angles. Observation.—The foregoing demonstration is the most elementary that was ever given of this celebrated Proposition. I have reduced it to its simplest form, and without making any use of the language of Quaternions. The same method of proof will establish the more general Proposition, that the sum of the external angles of any convex rectilineal figure is equal to four right angles.