PROP. XII. As the tangent of the vertical angle of a plane triangle ABC (fig. 28.) is to radius, so is the base AB to a fourth proportional; and as the said fourth proportional is to the sum of the semi-base and the line CD bisecting the base, so is the difference of these two to the perpendicular height of the triangle. Let a circle be described about the triangle, and from O, the centre thereof, let OA, OC, and OD be drawn; also let CF, parallel to AB, be drawn, meeting DO, produced (if need be) in F. It is evident that DF will be perpendicular to AB, and equal to the height of the triangle. But DC2 = OC2 + OD2 + 20D × OF (by 12. 2.) = OA2 +-' OD2 + 20D × DF — OD OA2 - OD2 +20D×DF= AD2+20D×DF (by 47. 1.); whence, by taking away AD2 from the first and last of these equal quantities, we have. DC2- AD2 = 20D x DF; or DC + AD x DC-AD =20D X DF (by Cor. to 5. 2.), and therefore 2OD: DC +AD: : DC = AD: DF; but (by Theor. 2.) as the tang. AOD=ACB: radius (: : AD : OD) : : AB : 20D. 2. E. D. PROP. XIII. As twice the rectangle under the base and perpendicular of a plane triangle ABC (fig. 29.) is to the rectangle under the sum and difference of the base and sum of the two sides, so is radius to the co-tangent of half the vertical angle. Let HG, perpendicular to AB, be the diameter of a circle described about the triangle; and let HD and Hd be perpendicular to the two sides of the triangle; also let CF be parallel to AB, and let HA, HB, and HC be drawn. Since the diameter HG is perpendicular to AB, therefore is AE BE (by 3. 3.), AH = BH, and the angle ACH BCH (by 28. and 27. 3.); whence, also, CD = Cd, and HD Hd (by 15. 1.) Therefore the right-angled triangles HAD and HBd, having AH HB and HD= = Hd, have, likewise, AD = Bd (by 26. 1.) From whence it is manifest, that CD will be equal to half the sum, and AD equal to half the difference of the two sides of the triangle. Moreover, because of the similar triangles AEH and HCD, it will be, AE2: CD2 :: HA2= HE × HG (by Cor. to 8. 6.) : HC2 (HF × HG) : : HE: NF (by 1. 6.) Whence, by division, &c. AE2: CD2-AE: HE EF: HEX AE: EFX AE; therefore, by inversion and alternation, EF x AE CD2— AE2 (CD + AE × CD2. CD-AE) :: HEX AE AE. HE AE radius : co-tang, EAH (ACH), by Theor. 2.: whence the truth of the proposition is manifest. PROP. XIV. As twice the rectangle of the base and perpendicular.of a plane triangle ABC (fig. 29.) is to the rectangle under the sum and difference of the base, and the difference of the two sides, so is radius to the tangent of half the vertical angle. Let the preceding construction be retained, and let AG and CG be drawn. The triangles AHD and GHC (being right-angled at D and C, and having HAD = HGC, by 21. 3.) are equiangular; and so AD: GC:: AH : HG :: AE AG (by 8. 6.); whence, alternately, AD : AE :: GC: AG, and AD2: AE2:: GC2 (GF × HG) : AG2 (GE × GH) : : GF: GE; therefore, by division, &c. AE AD : AE2: : EF: GE :: EF × AE "GEX AE; whence, again, by alternation, &c. EF × AE : AE - X AD2 (AE + AD × AE — AD) :: GE × AE: AE2 ::GE: AE :: radius: tang. AGE (by Theor. 2.); from which the truth of the proposition is manifest. PROP. XV. If the relation of three right lines a, b, and x (fig. 30.) be such, that ax x2 = b2, then it will be, as i̟a: b :: radius to the sine of an angle, and as radius to the tangent (or cotangent) of half this angle, so is b: x. Make AB equal to a, upon which let a semicircle ADB be described; also let CD, equal to b, be perpendicular to AB, and meet the periphery in D (for it cannot exceed the radius of the circle when the proposition is possible): moreover, let AD, BD, and the radius QD be drawn. Because AC x CBCD2b2 (by Cor. to 8. 6.), it is plain that AC xa-AC, or BC xa-BC is also = 2; and, therefore, xx a—x being = b2, it is manifest that x may be equal, either to AC or to BC. Now (by Theor. 1.) OD (ja) : CD (b) : : radius : sine DOC, whose half is equal to A or BDC (by 20. 3.) But as radius: tang. BDC :: DC (b): BC; or, as radius co-tang. BDC (tang. CDA):: DC (6): AC. 2. E. D. PROP. XVI. If the relation of three lines a, b, and x (fig. 31.) be such, that x2± ax = b2, then it will be, as a : b:: radius to the tangent of an angle, and as radius is to the tangent or cotangent of half this angle (according as the sign of ax is positive or negative) :: b: x. Make ABb, and AC perpendicular to AB, equal to a; about the latter of which, as a diameter, let a circle be described; and through O, the centre thereof, let BD be drawn, meeting the periphery in E and D; also let A, E and C, E be joined, and draw BF parallel to AC, meeting AE, produced, in F. Then, since (by 36. 3.) BE × BD (= BE × BE +a=BD × BD — a) = AB2 (b2) and x x xab2, by supposition, it is manifest, that BE will be =x, when xxx+a=b2; and BD = x, when x x x - CL = b2. = Furthermore, because the angle F = OAE (by 29. 1.) = OEA (by 5. 1.) BEF (by 15. 1.), it is evident that BF= BE (by 6. 1.), and that the angles BAF and C (being the complements of the equal angles F and OAE) are likewise equal. Now (by Theor. 2.) AO (a): AB (b) : : radius: tang. AOB; whose half is equal to C or BAF. But, as radius : tang. BAF :: AB (6): BF (BE), the value of x in the co-tang. first case, where x2 + axb2. Again, radius BAF (tang. F) : : BF (BE): AB (by Theor. 2.); and BE :AB:: AB: BD (by Cor. to 36. 3. and 16. 6.); whence, by equality, radius: co-tang. BAF:: AB (b): BD; which is the value of x in the second case; where x2 2. E. D. ax = b2. PROP. XVII. In any plane triangle ABC (fig. 32.), it will be, as the line CE, bisecting the vertical angle, is to the base AB, so is the secant of half the vertical angle ACB to the tangent of an angle; and, as the tangent of half this angle is to radius, so is the sign of half the vertical angle to the sign of either angle, which the bisecting line makes with the base. Let ACBD be a circle described about the triangle, and let CE be produced to meet the periphery thereof in D; moreover, let AD and BD be drawn, and likewise DF, perpendicular to the base AB; which will also bisect it, because (BCD being = ACD) the subtenses BD and AD are equal (by 26. and 29. 3.) Moreover, since the angle DBE = ACD (by 21. 3.) = DCB, the triangles DEB and DBC (having D common) are equiangular, and therefore DE X DC = DB2 (by 4. and 17. 6.), or, which is the same, DE2 + DE CE DB2. Therefore (by Prop. 16.) CE : DB :: radius: tangent of an angle (which we will call Q); and as radius: tang. Q: DB : DE. = : : But DB: BF (AB) secant FBD (BCE): radius; therefore, by compounding this with the first proportion, K |
