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 same with the ratio of the rectangle DC, CE to the triangle ABC. because the angles HGK, DAC at the vertices of the Isosceles triangles GHK, ADC are equal to one another, these triangles are similar, and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, so is h (DC to AF, and so is) the rect- h. § 4.6. angle DC, CE to the rectangle AF, CE; but as GL to its half, so is the rectangle AF, CE to its half which is the triangle ACE, or the triangle A BC; 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 . s which contain the given angle is less than the square of the third side, shall have a given ratio to the triangle. this is demonstrated the fame way as the preceeding Proposition, by help of the second cafe of the Lemma. PROP. LXXVII, I. F the perpendicular drawn from a given angle of a tri- See N. angle to the opposite side, or base, has a given ratio to the base; the triangle is given in species. 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 Isosceles triangle, it is evident a that if any one of a s.& 32.5. G L K B RDC E O H M F its angles be given, the rest are also given ; and therefore the tij. angle is given in species, without the consideration of the ratio of the perpendicular to the base, which in this case is given by Prop 50. But when ABC is not an Ifofceles triangle, take any straight lino EF given in position and magnitude, and upon it describe the f«g ment ment of a circle EGF containing an angle equal to the given angl. BAC; draw GH bisecting EF at right angles, and join EG, GF. then since the angle EGF is equal to the angle BAC, and that EGF is an Isosceles 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, fo is DA to AB; and as LE to EF, so is AB to BC; wherefore, ex aequali, as LM to EF, so is AD to BC. and because the ratio of AD to BC is given, therefore Þ. 3. Dat. the ratio of LM to EF is given; and EF is given, wherefore b LM also is given. complete the parallelogram LMFK, and because LM is given, FK is given in magnitude; it is also given in position, and ç. 30. Dat. the point F is given, and consequently the point K; and because throʻK the straight line KL is drawn parallel to EF which is given in &. 31. Dat. position, therefore d KL is given in polition; and the circumference G K B RDC E O H M F e. 18. Dat. ELF is given in position, therefore the point L is given and be. cause the points L, E, F are given, the straight lines LE, EF, FL f. 29. Dat. are given fin magnitude; therefore the triangle LEF is given in 8. 43. Dat. species, and the triangle ABC is funilar to LEF, wherefore also ABC is given in species. 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 bisects the base of the segment at right angles, has unto the base. 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 similar. 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 equal equal to the alternate angle FNL, that is to the angle FEL in the same segment, therefore the triangle NEF is similar to LEF. and in the segment EGF there can be no other triangle upon the base EF which has the ratio of its perpendicular to that base the same with the ratio of LM or NO to EF, because the perpendicular must be greater or less than LM or NO. but, as has been shewn in the preceeding demonstration, a triangle fimilar to ABC can be described in the regment EGF upon the base EF, and the ratio of its perpendicular to the base is the same, as was there thewn, with the ratio of AD to BC, that is of LM to EF. therefore that triangle must be either LEF, or NEF, which therefore are similar to the triangle ABC. Cor. 2. If a triangle ABC has a given angle BAC, and if the ftraight line AR drawn from the given angle to the opposite side BC, in a given angle ARC, has a given ratio to BC; the triangle ABC is given in species. Draw AD perpendicular to BC; therefore the triangle ARD is given in species; wherefore the ratio of AD to AR is given; and the ratio of AR to BC is given, and consequently the ratio of AD to h. 9. Dat. BC is given; and the triangle ABC is therefore given in species i. i. 77. Dat. Cor. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if straight lines drawn from these angles to the bases, making with them given and equal angles, have the same ratio to the bases, each to each; then the triangles are similar. for, having drawn perpendiculars to the bases from the equal angles, as one perpendicular is to its base, fo is the other to its basek. where- k. $ 22.5. fore, by Cor. 1. the triangles are similar. A triangle similar to ABC may be found thus; having described the segment EGF and drawn the straight line GH as was directed in the Proposition, find FK which has to EF the given ratio of AD to BC; and place FK at right angles to EF from the point F. then bäcause, as has been shewn, the ratio of AD to BC, that is of FK to EF, must be less than the ratio of GH to EF; therefore FK is less than GH; and consequently the parallel to EF drawn thro' the point K ment meet the circumference of the segment in two points. lct L be either of them, and join EL, LF, and draw LM perpendicular to EF. then because the angle BAC is equal to the angle ELF, and that AÐ is to BC, as KF, that is LM to EF, the triangle ABC is similar to triangle LEF. by Cor 1. PROP. 4.6. 80. IF с Dat. c. 9. Dat. PRO P. LXXVIII. the rectangle of the fides which contain the given angle to the square of the third fide be given; the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the square of BC be given; the triangle ABC is given in species. From the point A draw A D perpendicular to BC; the recta. 41. 1. angle AD, BC has a given ratio to its half a the triangle ABC. and because the angle BAC is given, the ratio of the triangle ABC to 5. Cor. 62. the rectangle BA, AC is given b; and, by the hypothesis, the ratio of the rectangle BA, AC to the square of BC is given. therefore d. 1. 6. the ratio of the rectangle AD, BC to the square of BC, that is the ratio of the straight line AD to BC is given. wherefore the tric. 77. Dat. angle ABC is given in species, A triangle fimilar to ABC may be found thus; take a straight line EF given in position and magnitude, and make the angle FEG equal to the given angle BAC, and draw FH perpendicular to EG, and BK perpendicular to AC; therefore the triangles ABK, EFH are similar and the rect M 0 angle AD, BC, or the rect. K angle BK, AC, which is e E L R BD N is as FH to FE; let the G gi C F ven ratio of the rectangle BA, AC to the square of BC be the same with the ratio of the straight line EF to FL; therefore, ex aequali, the ratio of the rectangle AD, BC to the square of BC, that is the ratio of the straight line AD to BC, is the fame with the ratio of HF to FL. and because AD is not greater than the straight line MN in the feginent of the circle described about the triangle ABC, which bisects BC at right angles; the ratio of AD to BC, that is of HF to FL, must not be greater than the ratio of MN to BC. let it be so, and by the 77. Dat. find a triangle OPQwhich has one of its angles PO equal to the given angle BAC, and the ratio of the perpendicular OR, drawn from that angle, to the base PQ the same with the ratio of HF to FL. then the triangle ABC is similar to OPQ. 1 77. Dat. OPQ. because, as has been shewn, the ratio of AD to BC is the same with the ratio of (HF to FL, that is, by the construction, with the ratio of) OR to PQ_; and the angle BAC is equal to the angle f. 1. Cor. POQ. therefore the triangle ABC is similar to the triangle POQ. Otherwise, Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the square of BC be given; the triangle ABC is given in species. Because the angle BAC is given, the excess of the square of both the sides BA, AC together above the square of the third fide BC has a given a ratio to the triangle ABC. let the figure D be equal to a. 96. Dat. this excess; therefore the ratio of D to the triangle ABC is given; and the ratio of the triangle ABC to the rectangle BA, AC is given, because BAC is a given angle; and the rectangle BA, b. Cor. 63. AC has a given ratio to the square of A BC; wherefore the ratio of D to the square of BC is given. and, by compo D sition 4, the ratio of the space D together d. 7. Dat. with the square of BC to the square B С of BC is given. but D together with the square of BC is equal to the square of both B A 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 e. 59. Dat. the angle BAC is given, wherefore the triangle ABC is given in f. 48. Dat. species. The composition of this which depends upon those of the 76. and 48. Propositions is more complex than the preceeding compofition which depends upon that of Prop. 77. which is easy. Dat. c. 10. Dat. PRO P. LXXIX. K. IF a triangle have a given angle, and if the straight line See N. drawn from that angle to the base, making a given angle with it, divides 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 B C 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 |