* 76 Dat. Dat. * 10 Dat. *7 Dat. square of the third side BC has a given * ratio to the triangle ABC. Let the figure D be equal to this ex, cess; therefore the ratio of D to the triangle ABC is given: and the ratio of the triangle ABC to the rect* Cor. 62 angle BA, AC, is given *, because BAC is a given an gle; and the rectangle BA, square D B ther with the square of BC, to the square of BC is given ; but D, together with the square of BC, is equal to the square of both BA and AC together; therefore the ratio of the square of BA, AC, together, to the square of BC, is given; and the ratio of BA, AC, together, to BC is therefore given*; and the angle BAC is given, wherefore* the triangle ABC is given in species. The composition of this, which depends upon those of the 76th and 48th Propositions, is more complex than the preceding composition, which depends upon that of Prop. 77. which is easy. * 59 Dat. PROP. LXXIX. See N. . If a triangle have a given angle, and if the straight line drawn from that angle to the base, making a given angle with it, divide the base into segments which have a given ratio to one another ; the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the straight line AD, drawn to the base BC, making the given angle ADB, divide BC into the segments BD, DC, which have a given ratio to one another; the triangle ABC is given in species. Describe* the circle BAC about the triangle, and from its centre E, draw EA, EB, EC, ED; because the angle BAC is given, the angle BEC at the centre, which is the double * of it, is given. And the ratio of BE to EC is given, because they are equal to one an* 41 Dat. Other; therefore * the triangle BEC is given in species, and the ratio of EB to BC is given; also the ratio of CB to BD is given *; because the ratio of BD to DC is given, therefore the ratio of EB to BD is given *, and the angle EBC is given, wherefore the triangle EBD is given * in * 5. 4. * 20. 3. *7 Dat. 9 Dat. * 44 Dat. * * 44 Dat. * 43 Dat. species, and the ratio of EB, that is, of EA, to ED, is there- A B В which are equal, is given; and the triangle AEC is therefore given in species, and the angle ECA is given; and the angle ECB is given, wherefore the angle ACB is given; and the angle BAC is also given; therefore* the triangle ABC is given in species. A triangle similar to ABC may be found, by taking a straight line given in position and magnitude, and dividing it in the given ratio which the segments BD, DC, are required to have to one another; then, if upon that straight line a segment of a circle be described containing an angle equal to the given angle BAC, and a straight line be drawn from the point of division in an angle equal to the given angle ADB, and from the point where it meets the circumference, straight lines be drawn to the extremity of the first line, these together with the first line, shall contain a triangle similar to ABC, as may easily be shewn. The demonstration may be also made in the manner of that of the 77th Prop., and that of the 77th may be made in the manner of this. If the sides about an angle of a triangle have a given ra tio to one another, and if the perpendicular drawn from that angle to the base have a given ratio to the O base, the triangle is given in species. Let the sides BA, AC, about the angle BAC of the triangle ABC have a given ratio to one another, and let the perpendicular AĎ have a given ratio to the base BC: the triangle ABC is given in species. First, let the sides AB, AC, be equal to A one another, therefore the perpendicular AD bisects * the base BC; and the ratio 1 of AD to BC, and therefore to its balf B D DB is given; and the angle ADB is 26. 1. * 44. Dat. * 9 Dat. * 6 Dat. * 9 Dat. 2. * 43 Vat. given; wherefore the triangle * ABD, and consequently the triangle ABC, is given * in species. Mano But let the sides be unequal, and BA be greater than a AC; and make the angle CAE equal to the angle de ABC; because the angle AEB is common to the tri- d) angles AEB, CEA, they are similar; therefore as AB 09 to BE, so is CA to AE, and, by permutation, as BA16 to AC, so is BE to EA, and so is EA to EC; and the A ratio of BA to AC is given, therefore the ratio of a BE to EA, and the ratio of EA to EC, as also the ra- ja tio of BE to EC, is given *; wherefore the ratio of EB to BC is given *; and the ratio of AD to BC is given by the hypothesis, therefore * the ratio of AD A to BE is given; and the ratio of BE to EX was shewn to be given; wherefore the ratio of AD to EA B FC ED 2 is given; and ADE is a right an* 46 Dat. gle, therefore the triangle ADE is given * in species, and the angle AEB given; the ratio of BE to EA is * 44 Dat. likewise given, therefore * the triangle ABE is given in species, and consequently the angle EAB, as also the angle ABE, that is, the angle CAE, is given : therefore the angle BAC is given, and the angle ABC being also 11 * 43 Dat. given, the triangle ABC is given * in species. How to find a triangle which shall have the things which are mentioned to be given in the proposition, is evident in the first case; and to find it the more easily straight line EF equal to EA be placed in EB towards in that, if the B, the point F divides the base BC into the segments BF, FC, which have to one another the ratio of the sides BA, CA, because BE, EA, or EF, and EC, were shewn to be proportionals, therefore * BF is to FC, as BE to EF or EA, that is, as BA to AC; and AE cannot be less than the altitude of the triangle ABC, but it may be equal to it, which if it be, the triangle, in this case, as also the ratio of the sides, may be thus found: having given the ratio of the perpendicular to the base, take the straight line GH, given in position and magnitude, for the base of the triangle to be found; and let the given ratio of the perpendicular to the base be that of the straight line K to GH, that is, let K be equal to the perpendicular; and suppose GLH to be the triangle which is to be found, therefore having made the angle HLM equal to LGH, it is required ! * 19. 5. 6. 2. * 6. 6. that LM be perpendicular to GM, and equal to K; and because GM, ML, MH, are proportionals, as was. shewn of BE, EA, EC, the rectangle GMH is equal to the square of ML. Add the common square of NH (having bisected GH in N), and the square of NM is equal to the squares of the given straight lines NH * 6. %. and ML, or K: therefore the square of NM, and its side NM, is given, as also the point M, viz. by taking the straight line NM,the square of which is equal to the squares of NH, ML. Draw ML equal to K, at right angles to GM; and because ML is given in position and magnitude, therefore the point L is given: join LG, LH; then the triangle LGH is that which was to be found; for the square of NM is equal to the squares of NH and ML, and taking away the common square of NH, the rectangle GMH is equal * to the К. squareof ML; there I. R fore as GM to ML, so is ML to MH; 3. and the triangle LGM is * therefore equiangular to HLM, and the angle Ġ NO H MĚ HLM equal to the angle LGM, and the straight line LM, drawn from the vertex of the triangle, making the angle HLM equal to. LGH, is perpendicular to the base, and equal to the given straight line K, as was required; and the ratio of the sides GL, LH, is the same with the ratio of GM to ML, that is, with the ratio of the straight line which is made up of GN, the half of the given base, and of NM, the square of which is equal to the squares of GN and K, to the straight line K. And whether this ratio of GM to ML be greater or less than the ratio of the sides of any other triangle upon the base GH, and of which the altitude is equal to the straight line K, that is, the vertex of which is in the parallel to GH drawn through the point L, may be thus found. Let OGH be any such triangle, and draw OP, making the anglé HOP equal to the angle OGH; therefore, as before, GP, PO, PH, are proportionals, and PO cannot be equal to LM, because the rectangle GPH would be equal to the rectangle GMH, which is impossible; al 2 Cor. 20. 6. for the point P cannot fall upon M, because O would then fall on L; nor can PO be less than LM, therefore it is greater; and consequently the rectangle GPH is greater than the rectangle GMH, and the straight line GP greater than GM: therefore the ratio of GM tejto MH is greater than the ratio of GP to PH, and the ratio of the square of GM to the square of ML is therefore * greater than the ratio of the square of GP to the square of PO, and the ratio of the straight line GM to ML greater than the ratio of GP to PO. But as GM to ML, so is GL to LH; and as GP to PO, so is GO to OH; therefore the ratio of GL to LH is greater than the ratio of GO to OH; wherefore the ratio of GL to LH is the greatest of all others; and consequently the given ratio of the greater side to the less must not be greater than this ratio. But if the ratio of the sides be not the same with this greatest ratio of GM to ML, it must necessarily be less than it: let any less ratio be given, and the same things being supposed, viz. that G# is the base, and K equal to the altitude of the triangle, it may be found as follows: divide GH in the point Q, so that the ratio of GQ to QH may be the same with the given ratio of the sides; and as GQ to QH, so make GP to PQ, and so will * PQ be to PH; wherefore the square of GP is to the square of PQ, as the straight line GP to PH: and because GM, ML, MH, are proportionals, the square of GM is to the square of ML, as * the straight line GM to MH: but the ratio of GQ to QH, that is, the ratio of GP to PQ, is less than the ratio of GM to ML; and therefore the ratio of the square of GP to the square of PQ is less than the ratio of the square of GM to that of ML; and consequently the ratio of the straight line GP to PH is less than the ratio of GM to MH; and, by division, the ratio of GH to HP is less than that of GH to HM; wherefore * the straight line HP is greater than HM, and the rectangle GPH, that is, the square of PQ, greater than the rectangle GMH, that is, than the square of ML, and the straight line PQ is therefore greater than ML. Draw LR parallel to GP, and from P draw PR at right angles to GP. Because PQ is greater than ML, or PR, the circle described from the centre P, at the distance PQ, must necessarily cut LR in two points; let these be O, S, and join OG, * 19. 5. * % Cor. 20. 6. 20. 6. * 10. 5. |