a. 5. 4. 44 Dat. Defcribe the circle BAC about the triangle, and from its center E draw EA, EB, EC, ED. because the angle BAC is given, the b. 20. 3. angle BEC at the center, which is the double of it, is given. and the ratio of BE to EC is given, because they are equal to one another; therefore the triangle BEC is given in fpecies, and the ratio d. 7. Dat. 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 e. 9. Dat. BD is given. and the angle EBC is given, wherefore the triangle EBD is given in fpecies, and the ratio of EB, that is of EA to ED is therefore given. and the angle EDA is given, becaufe each of the f. 47. Dat. angles BDE, BDA is given. therefore the triangle AED is given' in fpecies, and the angle AED given; alfo the angle DEC is given, becaufe each of the angles BED, BEC is given; therefore the angle AEC is given. and the ratio of EA to EC, which are equal, is given; and the triangle AEC is therefore given in fpecies, B C and the angle ECA given. and the angle C D ECB is given, wherefore the angle ACB is given. and the angie 43. Dat. BAC is alfo given; therefore the triangle ABC is given in fpecies. L. A triangle fimilar to ABC may be found, by taking a straight line given in pofition and magnitude, and dividing it in the given ratio which the fegments BD, DC are required to have to one another; then if upon that ftraight line a fegment of a circle be defcribed containing an angle equal to the given angle BAC, and a ftraight line be drawn from the point of divifion in an angle equal to the given angle ADB, and from the point where it meets the circumference, ftraight lines be drawn to the extremity of the first line, thefe together with the first line shall contain a triangle fimilar to ABC, as may easily be fhewn. The Demonftration may be alfo made in the manner of that of the 77. Prop. and that of the 77. may be made in the manner of this. F the fides about an angle of a triangle have a given ratio to one another, and if the perpendicular drawn from that angle to the bafe has a given ratio to the bafe; the triangle is given in species. Let the fides BA, AC, about the angle BAC of the triangle ABC have a given ratio to one another, and let the perpendicular AD have a given ratio to the bafe BC; the triangle ABC is given in species. First, let the fides AB, AC be equal to one another, therefore the perpendicular AD bifects the bafe BC. and 2 the ratio of AD to BC, and therefore to its half A a. 26. I. *.43. Dat. b. 44. Dat. A d. 6. Dat. BD But let the fides be unequal, and BA be greater than AC; and make the angle CAE equal to the angle ABC. because the angle AEB is common to the triangles AEB, CEA, they are fimilar; therefore as AB to BE, fo is CA to AE, and, by permutation, as BA to AC, fo is BE to EA, and fo is EA to EC. and the ratio of BA to AC is given, therefore the ratio of BE to EA, and the ratio of EA to EC, as alfo the ratio of BE to EC is given; wherefore c. 9. Dat. 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 to BE is given; and the ratio of BE to EA was fhewn to be given;. wherefore the ratio of AD to AE is given. andB F C E D ADE is a right angle, therefore the triangle ADE is given in fpe- e. 46. Dat. cies, and the angle AEB given; the ratio of BE to EA is likewife given, therefore the triangle ABE is given in fpecies, and confequently the angle EAB, as alfo the angle ABE, that is the angle CAE is given; therefore the angle BAC is given, and the angle ABC being alfo given, the triangle ABC is given f in species. 2 e f. 43. Dat. How to find a triangle which fhall have the things which are mentioned to be givch in the Propofition, is evident in the first cafe. and to find it the more eafily in the other cafe, it is to be obferved that if the ftraight line EF equal to EA be placed in EB towards B, the point F divides the bafe BC into the fegments BF, FC which have to one another the ratio of the fides BA, AC. because BE, EA, or EF, and EC were fhewn to be proportionals, therefore* BF is . 19. 5. to FC, as BE to EF, or EA, that is as BA to AC. and AE cannot be lefs than the altitude of the triangle ABC, but it may be equal to it; which if it be, the triangle, in this cafe, as alfo the ratio of the fides, may be thus found, having given the ratio of the perpendicular to the bafe. take the ftraight line GH given in pofition and magnitude, for the bafe of the triangle to be found; and let the g. 6.2. given ratio of the perpendicular to the bafe be that of the straight line K to GH, that is, let K be equal to the perpendicular; and fuppofe GLH to be the triangle which is to be found. therefore having made the angle HLM equal to LGH, it is required that LM be perpendicular to GM and equal to K. and because GM. ML, MH are proportionals, as was fhewn of BE, EA, EC, the rectangle GMH is equal to the fquare of ML. add the common fquare of NH, (having bifected GH in N) and the fquare of NM is equal to the fquares of the given ftraight lines NH aud ML, or K. therefore the fquare of NM, and its fide NM, is given, as alfo the point M, viz. by taking the straight line NM the fquare of which is equal to the fquares of NH, ML. draw ML equal to K, at right angles to GM. and because ML is given in pofition and magnitude, therefore the point L is given; join LG, LH, then the triangle LGH is that which was to be fouad. for the fquare of NM is equal to the fquares of NH and ML, and taking away the common fquare of NH, the rectangle GMH is the triangle LGM ish there fore equiangular to HLM, 3 and the angle HLM equalNQH MP to the angle LGM, and the ftraight line LM, drawn from the vertex of the triangle making the angle HLM equal to LGH, is perpendicular to the bafe and equal to the given ftraight line K, as was required. and the ratio of the fides GL, LH is the fame 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 bafe and of NM the fquare of which is equal to the fquares of GN and K, to the straight line K. And whether this ratio of GM to ML is greater or less than the ratio of the fides of any other triangle upon the bafe GH, and of which the altitude is equal to the ftraight line K, that is, the vertex of which is in the parallel to GH drawn thro' the point L, may be thus found. Let OGH be any fuch triangle, and draw OP making the angle 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 impoffible, for the point P cannot fall upon M, because O would 20.6. then fall on L; nor can PO be lefs than LM, therefore it is greater; and confequently the rectangle GPH is greater than the rectangle GMH, and the straight line GP greater than GM. therefore the ratio of GM to MH is greater than the ratio of CP to PH, and the ratio of the fquare of GM to the fquare of ML is therefore i greater i. 2. Cor. than the ratio of the fquare of GP to the fquare of PO, and the ratio of the ftraight line GM to ML, greater than the ratio of GP to PO. but as GM to ML, fo is GL to LH; and as GP to PO, fo 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 confequently the given ratio of the greater fide to the lefs must not be greater than this ratio. i But if the ratio of the fides be not the fame with this greateft ratio of GM to ML, it must neceffarily be lefs than it. Let any lefs ratio be given, and the fame things being fuppofed, viz. that GH is the bafe, and K equal to the altitude of the triangle, it may be found as follows. Divide GH in the point Q, fo that the ratio of GQ to QH may be the fame with the given ratio of the fides; and as GQ to QH, fo make GP to PQ, and fo will f PQ be to PII; where- f. 19. 5. fore the fquare of GP is to the fquare of PQ, as i the straight line GP to PH. and becaufe GM, ML, MH are proportionals, the fquare of GM is to the fquare of ML, as i the ftraight 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 fquare of GP to the fquare of PQ is lefs than the ratio of the square of GM to that of ML; and confequently the ratio of the straight line GP to PH is lefs than the ratio of GM to MH, and, by divifion, the ratio of GH to HP is less than that of GH to HM; whereforek k. 10. 5. the straight line HP is greater than HM, and the rectangle GPH, that is the fquare of PQ, greater than the rectangle GMH, that is than the 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. because PQ_is greater than ML, or PR, the circle defcribed from the center P, at the diftance PQ, muft neceffarily cut LR in two points; let these 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 becaufe 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, so 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 bafe 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 Demonstration 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 obferved in the next Propofition. IF 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 species. 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 bafe BC divide it into the fegments BD, DC which have a given ratio to one another; the triangle ABC is given in fpecies. a A Let AB be greater than AC, and make the angle CAE equal to d DC E KLH N CD is given, and confequently the ratio of DE to EC. and the ratio of EC to EA was fhewn to be given, therefore the ratio e. 46. Dat. of DE to EA is given. and ADE is a right angle, wherefore the triangle ADE is given in fpecies, and the angle AED given. and |