the given angle CAD, and take GK equal to GH, join KH, and COR. And if a triangle have a given angle, the space by which the fquare of the ftraight line which is the difference of the fides which contain the given angle is less than the fquare of the third fide, hall have a given ratio to the triangle. this is demonstrated the fame way as the preceding Propofition, by help of the fecond cafe of the Lemma. 4. 6. 222.58 F the perpendicular drawn from a given angle of a See N. triangle to the oppofite fide, or bafe, 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 bafe BC, have a given ratio to it; the triangle ABC is given in fpecies. If ABC be an Ifofceles triangle, it is evident that if any one of a.5.& 32.1, its angles be given, the reft are alfo given; and therefore the tri angle is given in fpecies, without the confideration of the ratio of the perpendicular to the bafe, which in this cafe is given by Prop. 50, But when ABC is not an Ifofceles 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 alfo are the triangles MLE, DAB, as ML to LE, fo 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 becaufe 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 c alfo is given. complete the parallelogram LMFK, and because LM is given, FK is given in magnitude; it is alfo given in position, c. 30. Dat. and the point F is given, and confequently the point K; and be- caufe thro' the ftraight line KL is drawn parallel to EF which d. 31. Dat. is given in pofition, therefore KL is given in pofition; and the circumference ELF is given in pofition, therefore the point Lis e. 18. Dat. given. and because the points L, E, F are given, the ftraight lines f. 29. Dat. LE, EF, FL are given f in magnitude; therefore, the triangle LEF g. 41. Dat. is given in fpecies 8. and the triangle ABC is fimilar to LEF, wherefore alfo ABC is given in fpecies. Becaufe LM is lefs 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 ftraight line, in a fegment of a circle containing an angle equal to the given angle, that bifects the bafe 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 bafe BC, as the perpendicular LM to the bafe EF; the triangles ABC, LEF are fimilar. Defcribe 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 to the alternate angle FNL, that is to the angle FEL in the fame fegment, therefore the triangle NEF is fimr to LEF. and in the fegment EGF there can be no other triangle upon the bafe EF which has the ratio of its perpendicular to that base the fame with the ratio of LM or NO to EF, because the perpendicular must be greater or lefs than LM or NO. but, as has been fhewn in the preceding demonftration, a triangle fimilar to ABC can be defcribed in the fegment EGF upon the bafe EF, and the ratio of its perpendicular to the bafe is the fame, as was there fhewn, 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 fimilar 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 oppofite fide BC, in a given angle ARC, has a given ratio to BC; the triangle ABC is given in fpecies. Draw AD perpendicular to BC; therefore the triangle ARD is given in fpecies; wherefore the ratio of AD to AR is given; and the b 423 ratio of AR to BC is given, and confequently the ratio of AD h. 9. Dat. to BC is given; and the triangle ABC is therefore given in fpecies 1. i. 77. Dat. COR. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if ftraight lines drawn from thefe angles to the bafes, making with them given and equal angles, have the fame ratio to the bafes, each to each; then the triangles are fimilar. for, having drawn perpendiculars to the bafes from the equal angles, as one perpendicular is to its bafe, fo is the other to its bafe k, wherefore, by Cor. 1. the triangles are fimilar. A triangle fimilar to ABC may be found thus; having described the fegment EGF and drawn the ftraight line GH as was directed in the Propofition, 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 becaufe, as has been fhewn, the ratio of AD to BC, that is of FK to EF, muft be less than the ratio of GH to EF; therefore FK is less than GH; and confequently the parallel to EF drawn through the point K must meet the circumference of the fegment in two points. let 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 AD is to BC, as KF, that is LM to EF, the triangle ABC is fimilar to triangle LEF. by Cor. 1. k. 4. 6. 22. 5. 80. 2. 41. z. IF PROP. LXXVIII. a triangle have one angle given, and if the ratio of the rectangle of the fides which contain the given angle to the fquare 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 fquare of BC be given; the triangle ABC, given in fpecies. From the point A draw AD perpendicular to BC; the rectangle AD, BC has a given ratio to its half the triangle ABC. and because the angle BAC is given, the ratio of the triangle ABC b. Cor. 62. to the rectangle BA, AC is given ; and, by the hypothefis, the ratio of the rectangle BA, AC to the fquare of BC is given. therec. 9. Dat. fore the ratio of the rectangle AD, BC to the fquare of BC, that is the ratio of the ftraight line AD to BC is given. wherec. 77. Dat. fore the triangle ABC is given in fpecies. d. 1. Dat. 6. c A triangle fimilar to ABC may be found thus; take a straight line EF given in pofition 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 fimilar. and the rectan M to it, is to the rectangle BA, let the given ratio of the BD N rectangle BA, AC to the fquare of BC be the fame with the ratio of the ftraight line EF to FL; therefore, ex aequali, the ratio of the rectangle AD, BC to the fquare of BC, that is the ratio of the ftraight 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 fegment of the circle defcribed about the triangle ABC, which bifects BC at right angles; the ratio of AD to BC, that is of HF to FL, muft not be greater than the ratio of MN to BC. let it be fo, and by the 77. Dat. find a triangle OPQ which has one of its angles POQ equal to the given angle BAC, and the ratio of the perpendicular OR, drawn from that angle, to the bafe PQ the fame with the ratio of HF to FL. then the triangle ABC is fimilar to OPQ. because, as has been fhewn, the ratio of AD to BC is the fame with the ratio of (HF to FL, that is, by the conftruction, with the ratio of) OR to PQ; and the angle BAC is equal to the angle POQ. therefore the triangle ABC is fimilar f to the triangle POQ. f. 1. Cor. Otherwise, Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the fquare of BC be given; the triangle ABC is given in fpecies. 77. Dat. Because the angle BAC is given, the excefs of the fquare of both the fides BA, AC together above the fquare of the third fide BC has a given ratio to the triangle ABC. let the figure D a. 76. Dat. be equal to this excefs; 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 b, because BAC is a given angle; and the rec- b. Cor. 62. tangle BA, AC has a given ratio to the fquare of BC; wherefore the ratio of Dat. A C. Io. Dat. D d. 7. Dat. D to the fquare of BC is given. and, by compofition, the ratio of the space D together with the fquare of BC to theB fquare of BC is given. but D together with the fquare of BC is equal to the fquare of both BA and AC together; therefore the ratio of the fquare of BA, AC together to the fquare of BC is given; and the ratio of BA, AC together to BC is therefore given. c. 59. Dat. and the angle BAC is given, wherefore the triangle ABC is gi- f. 48. Dat. ven in fpecies. The compofition of this which depends upon thofe of the 76. and 48. Propofitions is more complex than the preceding compofition which depends upon that of Prop. 77. which is eafy." IF F a triangle have a given angle, and if the straight See N. line drawn from that angle to the bafe, making a given angle with it, divides the base into fegments which have a given ratio to one another; the triangle is given in fpecies. Let the triangle ABC have the given angle BAC, and let the ftraight line AD drawn to the base BC making the given angle ADB, divide BC into the fegments BD, DC which have a given ratio to one another; the triangle ABC is given in fpecies. |