PROBLEM VI. Through a given point A, to draw a parallel to a given straight line BC. From the point A as a centre, with a radius sufficiently great, describe the indefinite arc EO; from the point E as a centre, with the same radius, describe the arc AF; take EO=AF, and draw AD: this will be the parallel required. B For, joining AE, the alternate angles AEF, EAD are evidently equal; therefore (23. I.) the lines AD, EF are parallel. Two angles A and B of a triangle being given, to find the third. Draw the indefinite line DEF; at the point E, make the angle DEC A, and the angle CEH CB: the remaining angle HEF will be the third angle required; because those three angles are together equal to two right angles. D H PROBLEM VIII. Two sides B and C of a triangle, and the angle A which they contain, being given, to describe the triangle. Having drawn the indefinite line TB TC then take DG-B, DH-C, and draw H PROBLEM IX. A side and two angles of a triangle being given, to describe the triangle. H The two angles will either be both adjacent to the given side, or the one adjacent and the other opposite: in the latter case, find the third angle (Prob. 7.); and the two adjacent angles will thus be known: in the former case, draw the straight line DE equal to the given side; at the point D, make an angle EDF equal to one of the adjacent angles, and at E, an angle DEG equal to the other; the two lines DF, EG, will cut each other in H; and DEH will be the triangle required. PROBLEM X. The three sides A, B, C of a triangle being given, to describe the triangle. Draw DE equal to the side A; from the point E as a centre with a radius equal to the second side B, describe an arc; from D as a centre with a radius equal to the third side C, describe another arc intersecting D the former in F; draw DF, EF; and DEF will be the triangle required. A. B Scholium. If one of the sides were C greater than the sum of the other two, the arcs would not intersect each other: but the solution will always be possible, when the sum of two sides, anyhow taken, is greater than the third. PROBLEM XI. Two sides A and B of a triangle, and the angle C opposite the side B, being given, to describe the triangle. There are two cases. First. When the angle C is right or take DE=A; from this point as agus ba shil obtuse, make the angle EDF C; neory centre, with a radius equal to the E given side B, describe an arc cutting DF in F; draw EF: then DEF will be the triangle required. In this first case, the side B must be greater than A; for the angle C, being right or obtuse, is the greatest angle of the triangle, and the side opposite to it must, therefore, also be the greatest. Secondly. If the angle C is acute, and B greater than A, the same construction will again apply, and DEF will be the triangle required. Scholium. The problem would be impossible in all cases, if the side B were less than the perpendicular let fall from E on the line DF. PROBLEM XII. The adjacent sides A and B of a parallelogram, with the angle C which they contain, being given, to describe the parallelogram. Draw the line DEA; at the point D, make the angle FDE=C; take DF-B; describe two arcs, the one from F as a centre with a radius FG-DE, the other D from E as a centre with a radius EG-DF; to the point G, where these arcs intersect each other, B draw FG, EG; DEGF will be the parallelogram required. A E For the opposite sides are equal, by construction; hence the figure is a parallelogram (30. I,) and it is formed with the given sides and the given angle. Cor. If the given angle is right, the figure will be a rectangle; if, in addition to this, the sides are equal, it will be a square. PROBLEM XIII. To find the centre of a given circle or arc. Take three points, A, B, C, anywhere in the circumference, or the arc; join AB, BC, or suppose them to be joined; bisect those two lines by the perpendiculars DE, FG: the point O, where these perpendiculars meet, will be the centre sought. Scholium. The same construction serves for making a circumference pass through three given points A, B, C; and also for describing a cir E cumference in which a given triangle ABC shall be inscribed. PROBLEM XIV. Through a given point to draw a tangent to a given circle. If the given point A lies in the circumference, draw the radius CA, and erect AD perpendicular to it: AD (9. II.) will be the tangent required. If the point A lies without the circle, join A and the centre by the straight line CA; bisect CA in O; from O as a centre, with the radius OC, describe a circle intersecting the given circumference in B; join AB: this will be the tangent required. For, drawing CB, the angle CBA being inscribed in a semicircle is a right angle (18. II.); therefore AB is a perpendicular at the extremity of the radius CB; therefore it is a tangent. D B Scholium. When the point A lies without the circle, there will evidently be always two equal tangents AB, AD, passing through the point A they are equal, because the right-angled triangles CBA, CDA have the hypotenuse CA common, and the side CB-CD; hence they are equal; hence AD is equal to AB, and also the angle CAD to CAB. PROBLEM XV. To inscribe a circle in a given triangle ABC. the right angle ADO=AFO; hence the third angle AOD is equal to the third AOF. Moreover, the side AO is common to the two triangles AOD, AOF; and the angles adjacent to the equal side are equal: hence the triangles themselves are equal; and DO is equal to OF. In the same man |