AP; Posium. tio, the point F is in the circumference of a given circle, problem, suppose that there is, and that the following Porissa, we bave a locus. But when conversely it is said, if a proposition is true. Two points A and B, and two jacent to the given points H and M, having to one an- ing as B to d, so this distance to a fourth proportional, it may be demonstrated just as above, that AE" does To confirm the truth of the preceding theory, it may not pass through H, contrary to the supposition. The be added, that Professor Dugald Stewart, in a paper read point to be found is therefore in the line E"B, which is a considerable time ago before the Philosophical Society given in position. Now if from E there be drawn EP of Edinburgh, defines a porism to be “A proposition parallel to AE', and ES parallel to BE', BS : SE :: BL affirming the possibility of finding one or more condi SE X BL PE=AF tions of an indeterminate theorem ;" where, by an in :LN= and AP: PE :: AF:FG=BS * determinate theorem, he means one which expresses a PEX AF SEX BL relation between certain quantities that are determinate therefore FG:LN:: :: PEX AF AP BS LN is compounded of the ratios of AF to BL, PE to LM is given in magnitude. Now LM is parallel to The following porism is the first of Euclid's, and the BE, a line given in position ; therefore M is in a line first also which was restored. It is given here to ex QM, parallel to AB, and given in position; therefore emplify the advantage which, in investigations of this the point M, and also the line KLM, drawn through it kind, may be derived from employing the law of conti- parallel to BE', are given in position, which were to be nuity in its utmost extent, and pursuing porisms to found. Hence this construction : From A draw AE' those extreme cases where the indeterminate magnitudes parallel to FK, so as to meet DE in E'; join BE', and increase ad infinitum. take in it BQ, so that a : B :: HF : BQ, and through This porism may be considered as having occurred in Q draw QM parallel to AB. Q draw QM parallel to AB. Let HA be drawn, and the solution of the following problem: Two points A, B, produced till it meet DE in E", and draw BE", meet(fig. 4.) and also three straight lines DE, FK, KL, be- ing QM in M; through M draw KML parallel to ing given in position, together with two points H and M BE', then is KML the line and M the point which in two of these lines, to inflect from A and B to a point were to be found. There are two lines which will anin the third line, two lines that shall cut off from KF swer the conditions of this porism; for if in QB, produand KL two segments, adjacent to the given points H ced on the other side of B, there be taken BG=BQ, and M, having to one another the given ratio of a to B. and if q m be drawn parallel to AB, cutting MB in m; Now, to find whether a porism be connected with this and if ma be drawn parallel to BQ, the part m n, cut VOL. XVII, Part I. + Dd off Fig. 4. LT LA't · LB: + Porism. off by EB produced, will be equal to MN, and bave the first, second, and tifth of these propositions, it is ma- Poriss to HG the ratio required. It is plain, that whatever nifest that •LB'++ LD'=ABXLETEKxGH LA therefore DB X DG=ADB'. CL last example, that the magnitudes required may all, or And because a part of them, be found by considering the extreme CL: LB::(LA: LE :: DA: DH::)DA': DAXDH, cases; but for the discovery of the relation between LB them, and the indefinite magnitudes, we must have re therefore DA DA X DH= -DA'. From the result of CL LA ·DB’=DA X DH + BD X DG; CL CL but DAXDII= twice trian. ADH, and DBX DG= For this purpose Dr Simson frequently employs two twice trian. BDG, and therefore DAX DH+DBX statements of the general hypothesis , which he compares DG=2 (trian. A DH+trian. BDG)= 2 (trian. AEB together. As for instance, in his analysis of the last po. +trian. AEG)=AB XLE+EKXHG. Now it has he only in LB but also another point o, anywhere in the same line, been proved, that D.1XDII+DBX DG=DA' CL to both of which he supposes lines to be inflected from LA LB the points A, B. This double statement, however, AB -= cannot be made without rendering the investigation long CL and complicated; nor is it even necessary, for it may be LA LB --AD + LD’, therefore CL LA LA AB -BD’culty, but with considerable advantage both with re CL CL CL gard to the simplicity and shortness of the demonstration. was to be demonstrated. It will be proper to premise the following lemma. Let Por ism. Let there be three straight lines AB, AC, AB (fig. 7.) be a straight line, and D, L any two points CB given in position (fig. 5.); and from any point in it, one of which D is between A and B; also let whatever in one of them, a: D, let perpendiculars be CL be any straight line. Then shall drawn to the other twn, as DF, DE, a point G may be LB LA LB LA AB found, such, that if GD be dirawn from it to the point CL the square of that line shall have a given ratio to the sum of the squares of the perpendiculars DF and DE, For place CL perpendicular to AB, and through the which ratio is to be found. points A, C, B describe a circle, and let CL meet the Draw AH, BK perpendicular to BC and AC; and circle again in E, and join AE, BE. Also draw DG Also draw DG in AB take L, so that AL : LB :: AH' : BK: :: parallel to CE, meeting AE and BE in H and G, and AC: CB*. The point L is therefore given; and if draw EK parallel to AB. Then, from the elements of a line N be taken, so as to have to AL the same ratio geometry, that AB' has to AH', N will be given in magnitude. CL : LB :: (LA: LE ::) LA: LAXLE, Also, since All': BK: ::AL: LB, and AH': AB': AL : N, ex equo, BK*: AB: :: LB : N. Draw LO, LB and hence LBXLE= -·LA. LM perpendicular to AC, CB; LO, LM are thereCL fore given in magnitude. Now, because AB' : BK :: LB LA AL N LB LB AD'+ -.BD'and CL : LA :: LB : LE :: EK or LD : KG, by the preceding lemma, N N therefore, (Geom. Sect. III. Theor. 8.) AL AB •AL’+ * BL+ -DL"; that is, DE+DFCL: AB :: (LD: GH ::) LD': EK XGH, N AB LOʻ+LA+W.DL. Join LG, then by hspoele- sis LO'+LM has to LG', the same ratio as DF+ From the three equations which we liare deduced from DE' bas to DG’; let it be that of R to N, then LOʻ --AL' + Fig. 7. D LV N 1 R R The same porism also assists in the solution of another Pori-m. Prism LM'=LGʻ; and therefore DE'+DF=NIG?+ problem. For if it were required to find D such that DE: N would therefore be given ; whence the solution is obvi- DL'S DG’and ·DL=-(DG- The connection of this porism with the impossible case of consequence the impossible cases of problems. The fore AB be divided in I, so that AL : LB :: AH': sible, and admits of just one solution ; but if I be in 1: BK”; and if LG, a mean proportional between AL DH, the problem admits of no solution at all, the point ble to be assigned. There is, however, this exception, to one another, some one be supposed to vary, while the inconsistency can take place between them. There are, Dd 2 however, |