64. IF PROP. LXXIV. Fa triangle has a given obtufe angle; the excess of the fquare of the fide which fubtends the obtufe angle above the fquares of the fides which contain it, shall have a given ratio to the triangle. Let the triangle ABC have a given obtufe angle ABC; and produce the ftraight line CB, and from the point A draw AD perpendicular to BC. the excefs of the fquare of AC above the a. 12. 2. fquares of AB, BC, that is a the double of the rectangle contained by DB, BC, has a given ratio to the triangle ABC. c. 1. 6. Because the angle ABC is given, the angle ABD is also given; b. 43. Dat. and the angle ADB is given, wherefore the triangle ABD is given in fpecies; and therefore the ratio of AD to DB is given. and as AD to DB, fo is the rectangle AD, BC to the rectangle DB, BC; wherefore the ratio of the rectangle AD, BC to the rectangle DB, BC is given, as also the ratio of twice the rectangle DB, BC to the rectangle AD, BC. but the ratio of the rec I. tangle AD, BC to the triangle ABC is given, A d. 41. 1. because it is doubled of the triangle; there- rectangle DB, BC is the excess of the square D B E H F G And the ratio of this excess to the triangle ABC may be found thus; take a straight line EF given in position and magnitude; and because the angle ABC is given, at the point F of the ftraight line EF make the angle EFG equal to the angle ABC; produce GF, and draw EH perpendicular to FG. then the ratio of the excess of the fquare of AC above the fquares of AB, BC to the triangle ABC is the fame with the ratio of quadruple the straight line HF to HE. Because the angle ABD is equal to the angle EFH, and the angle ADB to EHF, each being a right angle; the triangle ADB is f. 4. 6. equiangular to EHF. therefore f as BD to DA, fo FH to HE; and g. Cor. 4. 5. as quadruple of BD to DA, fo is & quadruple of FH to HE. but as twice BD is to DA, fo is twice the rectangle DB, BC to the rectangle AD, BC; and as DA to the half of it, fo is the rectangle AD, BC to its half the triangle ABC; therefore, ex ae h. C. 5. quali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as quadruple of FH to HE, fo is twice the rectangle DB, BC to the triangle ABC. IF PROP. LXXV. F a triangle has a given acute angle; the space by which the fquare of the fide fubtending the acute angle is less than the fquares of the fides which contain it, fhall have a given ratio to the triangle. 65° Let the triangle ABC have a given acute angle ABC, and draw AD perpendicular to BC; the space by which the fquare of AC is less than the squares of AB, BC, that is the double of the rec- a. 13. 2ą tangle contained by CB, BD, has a given ratio to the triangle ABC. Because the angles ABD, ADB are each of them given, the triangle ABD is given in fpecies; and therefore the ratio of BD to DA is given. and as BD to DA, fo is the rec tangle CB, BD to the rectangle CB, AD; therefore the ratio of these rectangles is given, as also the ratio of twice the rectangle CB, BD to the rectangle CB, AD. but the rectangle CB, AD has a given ratio to its half the triangle ABC, therefore the ratio of twice the rectangle CB, BD to the triangle ABC is given. and twice the rectangle CB, BD is a the space by which the fquare of AC is lefs than the squares of AB, BC; therefore the ratio of this fpace to the triangle ABC is given. and the ratio may be found as in the preceding Propofition. LEMMA. D C F from the vertex A of an Ifofceles triangle ABC, any straight line AD be drawn to the bafe BC; the fquare of the fide AB is equal to the rectangle BD, DC of the fegments of the base together with the fquare of AD. but if AD be drawn to the base produced, the fquare of AD is equal to the rectangle BD, DC together with the fquare of AB. CAS. 1. Bifect the base BC in E, and join AE which will be perpendicular 2 to BC; wherefore the fquare of AB is equal to the fquares of AE, EB. but the fquare of EB is A a. 8. I, b. 47. I. C c. 5. 2. equal to the rectangle BD, DC togetherD BDE with the fquare of DE. therefore the fquare b. 47. I. 07. of AB is equal to the fquares of AE, ED, that is to the fquare of AD, together with the rectangle BD, DC. the other cafe is fhewn in the fame way by 6. 2. Elem. IF PROP. LXXVI. F a triangle have a given angle, the excefs of the fquare of the ftraight line which is equal to the two fides that contain the given angle, above the fquare of the third fide, fhall have a given ratio to the triangle. Let the triangle ABC have the given angle BAC, the excess of the fquare of the straight line which is equal to BA, AC together above the square of BC, shall have a given ratio to the triangle ABC. Produce BA, and take AD equal to AC, join DC and produce it to E, and thro' the point B draw BE parallel to AC; join AE, and draw AF perpendicular to DC. and because AD is equal to AC, BD is equal to BE; and BC is drawn from the vertex B of the Ifofceles triangle DBE, therefore, by the Lemma, the square of BD, that is of BA and AC together, is equal to the rectangle DC, CE together with the fquare of BC; and therefore the square of BA, AC together, that is of BD, is greater than the fquare of BC by the rectangle DC, CE; and this rectangle has a given ratio to the triangle ABC. because the angle BAC is given, the adjacent angle CAD is given; and each of the angles ADC, DCA is given, for each of them is a.5.& 32.1. the half of the given angle BAC; thereb. 43. Dat. fore the triangle ADC is given in fpe d. 1. 6. e. 41. I. B D C H L E K cies; and AF is drawn from its vertex to the base in a given angle, c. 50. Dat. wherefore the ratio of AF to the base CD is given . and as CD to AF, fo is the rectangle DC, CE to the rectangle AF, CE; and the ratio of the rectangle AF, CE to its half the triangle ACE is given; therefore the ratio of the rectangle DC, CE to the triangle ACE, that is to the triangle ABC is givens. and the rectangle DC, CE is the excefs of the fquare of BA, AC together above the fquare of BC; therefore the ratio of this excess to the triangle ABC is given. f. 37. I. The ratio which the rectangle DC, CE has to the triangle ABC is found thus. take the ftraight line GH given in pofition and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take GK equal to GH, join KH, and draw GL perpendicular to it. then the ratio of HK to the half of GL is the fame with the ratio of the rectangle DC, CE to the triangle ABC. because the angles HGK, DAC at the vertices of the Ifofceles triangles GHK, ADC are equal to one another, these triangles are fimilar, and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, fo is h (DC to AF, and so is) the h. rectangle DC, CE to the rectangle AF, CE; but as GL to its half, fo is the rectangle AF, CE to its half which is the triangle ACE, or the triangle ABC; therefore, ex aequali, HK is to the half of the straight line GL, as the rectangle DC, CE is to the triangle ABC. COR. And if a triangle have a given angle, the space by which the square of the straight line which is the difference of the sides which contain the given angle is less than the fquare of the third fide, fhall have a given ratio to the triangle. this is demonftrated the fame way as the preceding Propofition, by help of the second cafe of the Lemma. IF $4.6. 222.5. F the perpendicular drawn from a given angle of a See N. triangle to the oppofite fide, or base, has a given ra tio to the bafe; the triangle is given in fpecies. Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the base BC, have a given ratio to it; the triangle ABC is given in species. If ABC be an Ifofceles triangle, it is evident a that if any one of 2.5.& 32.8. its angles be given, the rest are also given; and therefore the tri angle is given in fpecies, without the confideration of the ratio of the perpendicular to the base, which in this case is given by Prop. 50. But when ABC is not an Isofceles triangle, take any straight line EF given in pofition and magnitude, and upon it defcribe the feg ment of a circle EGF containing an angle equal to the given angle BAC; draw GH bifecting EF at right angles, and join EG, GF. then fince the angle EGF is equal to the angle BAC, and that EGF is an Ifofceles triangle and ABC is not, the angle FEG is not equal to the angle CBA. draw EL making the angle FEL equal to the angle CBA, join FL, and draw LM perpendicular to EF. then because the triangles ELF, BAC are equiangular, as also are the triangles MLE, DAB, as ML to LE, so is DA to AB; and as LE to EF, fo is AB to BC; wherefore, ex aequali, as LM to EF, fo is AD to BC, and because the ratio of AD to BC is given, therefore b. 2. Dat. the ratio of LM to EF is given; and EF is given, wherefore ↳ LM also is given. complete the parallelogram LMFK, and because LM is given, FK is given in magnitude; it is also given in pofition, c. 30. Dat. and the point F is given, and consequently the point K; and be cause thro' K the straight line KL is drawn parallel to EF which d. 31. Dat. is given in position, therefore KL is given in pofition; and the B RDC E N OHM F K circumference ELF is given in position, therefore the point L is e. 28. Dat. given a. and because the points L, E, F are given, the straight lines f. 29. Dat. LE, EF, FL are given fin magnitude; therefore, the triangle LEF 5. 42. Dat. is given in species %. and the triangle ABC is fimilar to LEF, wherefore alfo ABC is given in fpecies. Because LM is less than GH, the ratio of LM to EF, that is the given ratio of AD to BC must be less than the ratio of GH to EF which the straight line, in a segment of a circle containing an angle equal to the given angle, that bifects the base of the fegment at right angles, has unto the bafe. COR. 1. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if the perpendicular AD be to the base BC, as the perpendicular LM to the base EF; the triangles ABC, LEF are fimilar. Describe the circle EGF about the triangle ELF, and draw LN parallel to EF, join EN, NF, and draw NO perpendicular to EF. because the angles ENF, ELF are equal, and that the angle EFN is |