Sound, gave them freely,* regardless of the inconvenience of the saliva that flowed through the badly cicatrized orifice over the chin, but rather laughed when one revolted at the sight, and delighted in thrusting the tongue through the opening, at the same time, that they winked the eyes. Nor are they particular what they substitute for the labrets. One man, we are informed by Captain Cook, appeared before him with two iron nails projecting from them, like prongs; and another endeavoured to make a large brass button answer the purpose of a labret. Through the septum of the nose, awls, and large cod-fish hooks are thrust, and the women appropriate ear-rings and thimbles as decorations to their dress.‡ The perforations are made at the age of manhood, by an incision sufficiently large to admit a quill, about half an inch below the corners of the mouth, which has the effect of depressing the under lip, and keeping the mouth open.§ The orifice is enlarged from year to year, until it reaches half an inch in diameter; and in more advanced years, is not unfrequently of a much larger size. Captain Beechy obtained from a native of Schismareff Inlet a finely polished jade that was three inches in length by an inch and a half in width. For some time after the operation has been performed, it is necessary to turn the cylindrical pieces of ivory frequently, that they may not adhere to the festering flesh. In time, this action becomes as habitual to them, as that of turning the mustaches is to a Mussulman.|| On the Theory of Parallel Lines. By HENRY MEIKLE, During the long succession of ages which have elapsed since the origin of geometry, many attempts have been made and treatises written, though with little success, to demonstrate the important theorem which Euclid, having failed to prove, has styled his 12th Axiom, and which is nearly equivalent to assuming that the three angles of every triangle amount to two right angles. Among the more distinguished and ex *Cook, Beechy. § Franklin. Cook. + Franklin, || Beechy. tensive writers on the subject in our own times, may be reckoned M. Legendre in his Elements, and in the Memoires de l'Academie, as after noticed; and Colonel T. Perronet Thompson, in his "Geometry without Axioms," and in his more recent tract proposing a proof by help of the logarithmic spiral. To the former of his treatises, Colonel Thompson has appended critical notices of thirty of the more plausible methods which have, at very different times, been given by other authors as demonstrations. His criticisms, though neither always the first which have been made on these methods,* nor yet all new, are generally just, and even fatal to them. He strongly objects, though rather metaphysically, to employing infinite quantities, as is done by M. Bertrand, who reasons upon a numerous set of areas, each of which is of infinite magnitude, and such that, while on one side it has no limit of any kind, it is, on other sides, bounded by infinite lines stretching immeasurably beyond the fixed stars. But, from other and more elementary considerations, I shall afterwards briefly shew Bertrand's method to be a complete failure. Colonel Thompson himself, however, does not seem to have been aware that, in his own attempt at a sort of mechanical demonstration, in which he supposes a straight line to preserve its parallelism whilst "travelling" laterally, no finite line could suffice for such “travelling line." Again, in calling in the aid of the logarithmic spiral, he has assumed that triangles formed partly of unequally curved arcs of that spiral, and partly of arcs of unequal circles, are equal to one another, and identical with rectilinear triangles. This, like most of the attempts on Euclid's axiom, evidently assumes far more than the whole thing to be proved. Nay, supposing the curved sides of those triangles were of the same lengths as if formed of straight lines, it would still be necessary to prove, what is not even true, that their angles are the same. But these are by no means the only objections to which his demonstrations are liable. Much of the form of the present essay has been adopted with a view to brevity; various minute and commonplace details are omitted which every tyro can readily supply, and occasionally some abbreviations and modes of reasoning are adopted, which, though not very common at that stage of geometry, are well known, and have no dependence on what is to be proved. But independently of this, the question is here treated in a very different manner from anything I had previously met with on For instance, in the Philosophical Magazine for December 1822, I had greatly anticipated him in refuting the 30th, or Mr Ivory's, method; for I have there shewn that it depends on the very liberal assumption that four straight lines may always be made to meet round any given polygon, so as completely to inclose it, let it have ever so many sides: which is far more complicated and less evident than Euclid's axiom. this subject. For, so far as I am aware, it had not till now been shewn that, if, in so much as one triangle, the sum of the angles differed from 180°, a definite relation behoved to subsist between the areas and angles of all triangles. It is to be hoped that such a relation, which I have here deduced from that supposition, and endeavoured to follow it up to an absurdity, may yet lead to other and preferable modes of demonstration. PROP. I. Triangles, whose areas are equal, have the sums of their angles equal. Fig. 2. Case 1. To prove this of any two triangles, A BC, DEF, which have their bases equal, as well as their areas; bisect A B, BC in G and H, and to G H or its extension draw the perpendiculars AI, BK, CL. Then the triangles AIG, BK G, having two angles and a corresponding side respectively equal, are (Eucl. i, 26) every way equal; and for the like reason, the triangles BKH, CLH, are equal. Hence, the triangle ABC is equal to the quadrilateral AILC, and its three angles are equal to the two angles IAC, ACL. The three perpendiculars drawn to IL have in effect been proved to be equal; and in the same way as it has been shewn of ABC, every other triangle which could have A C for a base, and whose sides would be bisected by GH produced, if necessary, must have its perpendiculars which are drawn to this bisecting line, as well as its area, and sum of angles respectively, equal to those of ABC. Bisect, therefore, IL in M, and on it erect the perpendicular MN=AI; join AN, NC, cutting IL in O and P. Then, nearly in the same way as above, the triangles AIO, NMO, NMP, CLP, being found equal, first in pairs, and then all four; the triangle ANC is isosceles, and has its area equal to AILC= A BC, and its three angles equal to the same two angles as before, or to the three angles of ABC. In like manner, construct an isosceles triangle DQF DEF. Then the areas and bases of the two isosceles triangles ANC, DQ F, being respectively equal, there is no alternative but their angles must agree; for otherwise, a copy of the one triangle being constructed on the base of the other, their areas behoved to differ. Consequently, the triangles ABC, DEF, have the sums of their angles equal. Case 2. When ABC, DEF (figs. 1 & 3) are any two equal triangles which have no equal bases; bisect DE, L Fig. 3. E K H M N above, DL FM, and therefore, the triangles DIL, KIN, having two angles and a corresponding side respectively equal, we have also KN DL FM; and since, therefore, the triangles KNO, FMO, have likewise two angles and a corresponding side equal, L M bisects FK in O. Hence, as noticed under the first case, the triangle DKF has its area and the sum of its angles equal to those of DEF, since their sides are bisected by the same line. Again, by the first case, the triangles DK F, ABC, have the sums of their angles equal; for DK is equal to one side of A B C, and their areas are equal. Consequently, any two triangles ABC, DEF, whose areas are equal, have the sums of their angles equal. Fig. 4. PROP. II. If, in one triangle, the sum of the angles differed from 180°, so it would in every triangle; the difference would always have the same sign, and be proportional to the area. D B F Case 1. Let A B C be a triangle whose angles, if possible, fall short of 180°, and let its area be bisected by BD. Then each of these halves will (Prop. 1.) have the same amount of angles; but whether they were halves or unequal A E parts, their six angles would evidently be equal to the three angles of ABC, together with 180° at the point D: so that, the six angles of the two parts will always be less than 360° by the same quantity that the three angles of ABC are less than 180°. Consequently, the defect of half the sum of the six from 180° will just be half the defect of the three angles of ABC from 180°. In the same way, if B E bisect the area A BD, the angular defect of each part from 180° will be half of the defect for ABD, or one-fourth of that for A B C; and so on, for the continual bisection of the whole till each part be less than any given area. But however unequal the two parts may be into which any one of such bisecting lines divides the whole or a part of the area A B C, the same relation will still subsist. For when the several parts of the areas and of the angular defects are proportional, so must any corresponding sums of such parts. Thus area E B C will be proportional to its angular defect. VOL. XXXVI. NO. LXXII.—APRIL 1844. X The like evidently follows, when, on the extension of any side of A B C, as, for instance, on that of A C produced to F, an addition BCF equal to one or more of the foresaid parts of ABC (which have their areas proportional to their angular defects) is made to the area ABC; są that, area A B F will likewise be proportional to its angular defect. Hence, in every triangle, the area and the angular defect follow the same pro portion.* Case 2. In the same way, it may be shown that, if one triangle had the sum of its angles greater than 180°, so would every other, and that the excess (as in spherics) would be proportional to the area. SCHOLIUM. Since it could still more readily be shewn, either as above or in several other ways, that if one triangle had the sum equal to 180°, so would every other; it is obvious that, in respect of 180°, every triangle has the sum of its angles of the same kind, whatever that may be, But there are several known methods of proving that the sum can never exceed 180°, The grand difficulty has always been equivalent to prov= ing that it can never be less. Thus, since, however small might be the amount of some two angles of a triangle, we may always increase those two to be ever so little short of 180°, and yet, according to Euclid's 12th axiom, the sides which have been thereby altered, if produced, will con= tinue to meet, forming a still greater triangle; it is evident that his axiom is equivalent to assuming that the three angles of even the greatest of triangles cannot be sensibly less than 180°. From this second proposition, too, would follow the converse of the first one, if ever the sum differed from 180°. PROP. III. The three angles of every triangle are equal to two right angles. M Fig. 5. H Let ABC be any equilateral triangle ; bisect its angles, which obviously will divide it into three equal triangles, each having an angle of 120° at the point D. Produce DA, DO to E and F, making AE CF ¦ AC; join A F. Then each of the angles ACF, EAF being exterior in respect of ADC, exceeds this obtuse angle. Hence A F exceeds AC, whilst EF being greater than AF, is still greater than ▲ Ɑ or twice A E. Produce BD to bisect A C in G, and EF in H. Let the lines KL, KM, each of which is equal A E, form an angle equal to half the least angle, which any equilateral triangle can ever have: for as long. as no two sides of an equilateral triangle can coincide in one straight line, its angle must have some magnitude, and therefore so must the area * Although, strictly speaking, such reasoning is only applicable to commensurable areas, it is more than sufficiently general and exact for the present purpose. |