. Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson. AE being produced, will also passthrough H; and the point H, and the angle GHA and its equalHAG are given. Hence also the center G and the circle aregiven, and the method of solution is plain. Prop. XVIH.—Prob.— Given the base of a triangle^ the ver-tical angle, and the straight line bisecting that angle y to con-struct the
. Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson. AE being produced, will also passthrough H; and the point H, and the angle GHA and its equalHAG are given. Hence also the center G and the circle aregiven, and the method of solution is plain. Prop. XVIH.—Prob.— Given the base of a triangle^ the ver-tical angle, and the straight line bisecting that angle y to con-struct the triangle. On the given base BC describe (HI. 19) the segment BACcapable of containing an angle equal to the given vertical an-gle, and complete the circle; bisect the arc BEC in E, and joinEC; perpendicular to this, draw CF equal to half the linebisecting the vertical angle, and from F as center, with FC asradius, describe the circle CGII, cutting the straight line pass-in through E and F in G and H; make ED equal to EG, anddraw EDA; lastly, join AB, AC, and ABC is the requiredtriangle. 238 EXEKCI8E8. For the triangles CEA, CED are equiangular, the angle CEAbeing tomnion, and BCE, CAE being each equal to AE : EC :: EC : ED, and (V. 9, cor.) EC=. But (ITI. 16, cor. 3, and 21) or therefore ^; whence AE=:HE, and (I. ) AD=GH = 2CF. AD is therefore equal to the given bisect-ing line, and it bisects the angle BAC. Hence ABC is the re-quired triangle. Method of Computation. Draw EL perpendicular to BC,and join CH. Then BCE is equal to BAE, half the verticalangle A; and therefore, to the radius EC ; CL is the cosine of?§ A, and CF is the tangent of CEF to the same radius ; where-fore, to any radius, CL : CF, or BC : AD : : cos ^ A : tanCEF or cot EFC; and hence the angle H, being half of EEC,is known. Also ECU is the complement of II, because ECF isa right angle, and FCH equal to H. But (Tgig. 2) EC : EH orEA : : sin II : sin E
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Keywords: ., bo, bookcentury1800, booksubjectgeometry, booksubjecttrigonometry
