that is, the vertex of which is in the parallel to GH drawn But if the ratio of the fides be not the fame with this great- Ec3 fquare of ML, and the ftraight line PQ is therefore greater than ML. Draw LR parallel to GP, and from P draw PR at right angles to GP: Becaufe PQ is greater than ML, or PR, the circle defcribed from the centre P, at the diftance PQ, muft neceffarily cut LR in two points; let thefe be O, S, and join OG, OH; SG, SH; each of the triangles OGH, SGH have the things mentioned to be given in the propofition: Join OP, SP; and because as GP to PQ, or PO, fo is PO to PH, the triangle OGP is equiangular to HOP; as, therefore, OG to GP, fo is HO to OP, and, by permutation, as GO to OH, fo is GP to PO, or PQ; and fo is GQ to QH: Therefore the triangle OGH has the ratio of its fides GO, OH the fame with the given ratio of GQ to QH; and the perpendicular has to the base the given ratio of K to GH, because the perpendicular is equal to LM, or K: The like may be fhewn in the fame way of the triangle SGH. This conftruction by which the triangle OGH is found, is fhorter than that which would be deduced from the demon. itration of the datum, by reafon that the bafe GH is given in pofition and magnitude, which was not fuppofed in the demonftration: The fame thing is to be observed in the next propofition. a 4. 6, M. I' F the fides about an angle of a triangle be unequal and have a given ratio to one another, and if the perpendicular from that angle to the bafe divides it into fegments that have a given ratio to one another, the triangle is given in fpecies. Let ABC be a triangle, the fides of which about the angle BAC are unequal, and have a given ratio to one another, and let the perpendicular AD to the base BC divide it into the fegments BD, DC which have a given ratio to one another, the triangle ABC is given in fpecies. Let AB be greater than AC, and make the angle CAE equal to the angle ABC; and because the angle AEB is common to the triangles ABE, CAE, they are equiangular to one another: Therefore as AB to BE, fo is CA to AE, and, a by с B by permutation, as AB to AC, fo BE to to EA was fhewn to be given, therefore the ratio of DE to EA is given And ADE is a right angle, wherefore the triangle e 46. dat. ADE is given in fpecies, and the angle AED given: And the ratio of CE to EA is given, therefore f the triangle AEC is gi- f 44. dat. ven in fpecies, and confequently the angle ACE is given, as also the adjacent angle ACB. In the fame manner, because the ratio of BE to EA is given, the triangle BEA is given in fpecies, and the angle ABE is therefore given: And the angle ACB is given; wherefore the triangle ABC is given in fpe- g 43. dat. cies. But the ratio of the greater fide BA to the other AC muft be less than the ratio of the greater fegment BD to DC: Because the fquare of BA is to the fquare of AC, as the fquares of BD, DA to the fquares of DC, DA; and the fquares of BD, DA have to the fquares of DC, DA a lefs ratio than the square of BD has to the fquare of DC †, because the square of BD is greater than the fquare of DC; therefore the fquate of BA has to the fquare of AC a lefs ratio than the fquare of BD has to that of DC: And confequently the ratio of BA to AC is less than the ratio of BD to DC. This being premifed, a triangle which fhall have the things - mentioned to be given in the propofition, and to which the triangle ABC is fimilar, may be found thus: Take a straight line GH given in pofition and magnitude, and divide it in K, fo that the ratio of GK to KH may be the fame with the given ratio of BA to AC: Divide alfo GH in L, fo that the ratio of GL to LH may be the fame with the given ratio to of BD to DC, and draw LMM at right angles to GH: And because the ratio of the fides of a triangle is less than the ratio of the fegments of the bafe, as has been fhewn, the ratio of GK to KH is less than the ratio of GL to LH; wherefore the point L muft fall betwixt K and H: Alfo make as GK to KH, fo GN to NK, and fo fhall NK be to NH. And from the centre N, at the diftance NK, defcribe a circle, and let its circumference meet LM in O, and join OG, OH; then OGH is the triangle which was, to be defcribed: Because GN is to NK, or NO, as NO to NH, the triangle OGN is equiangular to HON; therefore as OG to GN, fo is HO to ON, and, by permutation, as GO to OH, fo is GN to NO, or NK, that is, as GK to KH, that is, in the given ratio of the fides, and, by the conftruction, GL, LH have to one another the given ratio of the fegments of the bafe. PROP. LXXXII. IF a parallelogram given in fpecies and magnitude be increafed or diminished by a gnomon given in magnitude, the fides of the gnomon are given in magni. tude. Firft, let the parallelogram AB given in fpecies and magnitude be increased by the given gnomon ECBDFG, each of the ftraight lines CE, DF is given. Because AB is given in species and magnitude, and that the gnomon ECBDFG is given, therefore the whole space AG is given in magnitude: But AG is alfo given in fpecies, be cause it is fimilar to AB; therefore the fides of AG are gi ven: Each of the ftraight lines AE, AF is therefore given; and each of the ftraight lines CA, AD is given, therefore each of the remainders EC, DF is given . Next, let the parallelogram AG, given in fpecies and magnitude, be diminished by the given gnomon ECBDFG, each of the ftraight lines CE, DF is given. Because the parallelogram AG is given, as G E C B A H FD alfo its gnomon ECBDFG, the remaining space AB is given in magnitude: 2 2. def. 2. and 24. 5. magnitude: But it is alfo given in fpecies; because it is fimilar to AG; therefore its fides CA, AD are given, and each of the ftraight lines EA, AF is given; therefore EC, DF are each of 60. date them given. b The gnomon and its fides CE, DF may be found thus in the firft cafe. Let H be the given fpace to which the gnomon must be made equal, and find a parallelogram fimilar to AB d 25. 6. and equal to the figures AB and H together, and place its fides AE, AF from the point A, upon the ftraight lines AC, AD, and complete the parallelogram AG, which is about the fame diameter with AB; because therefore AG is equal to e 26, 6. both AB and H, take away the common part AB, the remain. ing gnomon ECBDFG is equal to the remaining figure H; therefore a gnomon equal to H, and its fides CE, DF are found:. And in like manner they may be found in the other case, in which the given figure H muft be less than the figure FE from which it is to be taken. IF PROP. LXXXIII. Fa parallelogram equal to a given space be applied to a given straight line, deficient by a parallelogram given in fpecies, the fides of the defect are given. Let the parallelogram AC equal to a given space be applied to the given ftraight line AB, deficient by the parallelogram BDCL given in fpecies, each of the ftraight lines CD, DB are given. Bifect AB in E; therefore EB is given in magnitude, up a GHF K 58. b 26. 6. on EB defcribe the parallelogram EF fimilar to DL and fimi- a 18. 6. · larly placed; therefore EF is given in fpecies, and is about the fame diameter with DL; let BCG be the diameter, and conftruct the figure; therefore, because the figure EF given in fpecies is defcribed upon the given ftraight line EB, EF is given in magnitude, and the A gnomon ELH is equal to the given d EDB c 56. dat. d 36. and 43. I. figure AC; therefore fince EF is diminished by the given gnomon ELH, the fides EK, FH of the gnomon are given; but e 82. dat. EK is equal to DC, and FH to DB; wherefore CD, DB are each of them given. This |