Solution. Draw the indefinite line DEF; at the point E make the angle DECA, and the angle CEH = B; the remaining angle HEF will be the third angle required; for these three angles are together equal to two right angles. PROBLEM. 141. Two sides of a triangle B and C (fig. 77) being given, and Fig. 77. the angle A contained by them, to construct the triangle. Solution. Draw the indefinite line DE, and make at the point D the angle EDF equal to the given angle A; then take DG = B, DHC, and draw GH; DGH will be the triangle required (36). PROBLEM. 142. One side and two angles of a triangle being given, to construct the triangle. Solution. The two given angles will be either both adjacent to the given side, or one adjacent and the other opposite. In this last case, find the third angle (140); we shall thus have the two adjacent angles. Then draw the straight line DE (fig. 78) Fig. 78. equal to the given side, at the point D make the angle EDF equal to one of the adjacent angles, and at the point E the angle DEG equal to the other; the two lines DF, EG, will cut each other in H, and DEH will be the triangle required (38). PROBLEM. 143. The three sides A, B, C (fig. 79), of a triangle being given, Fig. 79. to construct the triangle. Solution. 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 the point D, as a centre, with a radius equal to the third side C, describe another arc cutting the former in F; draw DF, EF, and DEF will be the triangle required (41). 144. Scholium. If one of the sides be greater than the sum of the other two, the arcs will not cut each other; but the solution will always be possible, when each side is less than the sum of other two. PROBLEM. 145. Two sides A and B of a triangle being given with the angle C opposite to the side B, to construct the triangle. Solution. The problem admits of two cases. 1. If the angle Fig. 80. C (fig. 80) is a right angle, or an obtuse angle, make the angle EDF equal to the angle C; take DE = A, from the point E, as a centre, and with a radius equal to the given side B, describe an arc cutting the line DF in F; draw EF, and DEF will be the triangle required. Fig. 81. Fig. 82. Fig. 83. It is necessary, in this case, that the side B should be greater than A, for the angle C being a right or an obtuse angle, it is the greatest of the angles of the triangle, and the side opposite must consequently be the greatest of the sides. 2. If the angle C (fig. 81) is acute, and B greater than A, the construction is the same, and DEF is the triangle required. But if, while C (fig. 82) is acute, the side B is less than A, then the arc described from the centre E with the radius EF = B, will cut the side DF in two points F and G situated on the same side of D; there are therefore two triangles DEF, DEG, which equally answer the conditions of the problem. 146. Scholium. The problem would be in every case impossible, if the side B were less than the perpendicular let fall from E upon the line DF. PROBLEM. 147. The adjacent sides A and B (fig. 83) of a parallelogram being given together with the included angle C, to construct the parallelogram. = Solution. Draw the line DEA; make the angle FDE = C, and take DFB; describe two arcs, one from the point F, as a centre, with the radius FG DE, and the other from the point E, as a centre, with the radius EG = DF; to the point G, where the two arcs cut each other, draw FG, EG; and DEGF will be the parallelogram required. For, by construction, the opposite sides are equal, therefore the figure is a parallelogram (86), and it is formed with the given adjacent sides and included angle. 148. Corollary. If the given angle be a right angle, the figure will be a rectangle; and, if the adjacent sides are also equal, the figure will be a square. PROBLEM. 149. To find the centre of a given circle, or of a gwen arc. Solution. Take at pleasure three points A, B, C (fig. 84), in Fig. 84 the circumference of the circle or in the given arc; join AB and BC, and bisect them by the perpendiculars DE, FG; the point O, in which these perpendiculars meet, is the centre sought. 150. Scholium. By the same construction a circle may be found, the circumference of which will pass through three given points A, B, C, or in which a given triangle ABC may be inscribed. PROBLEM. 151. Through a given point, to draw a tangent to a given circle. Solution. If the given point A (fig. 85) be in the circumfer- Fig. 85. ence, draw the radius CA, and through A draw AD perpendicular to CA, then AD will be the tangent sought (110). If the point A (fig. 86) be without the circle, join the point A and the centre Fig 86, by the straight line AC; bisect AC in O, and from O, as a centre, with the radius OC, describe an arc cutting the given circle in the point B; draw AB, and AB will be the tangent required. For, if we draw CB, the angle CBA inscribed in a semicircle is a right angle (128); therefore AB, being a perpendicular at the extremity of the radius CB, is a tangent. 152. Scholium. The point A being without the circle, it is evident that there are always two equal tangents AB, AD, which pass through the point A; they are equal (56) because the right-angled triangles CBA, CDA, have the hypothenuse CA common, and the side CB = CD; therefore AD = AB, and at the same time the angle CAD CAB. = PROBLEM. 153. To inscribe a circle in a given triangle ABC (fig. 87). Fig. 87. Bisect the angles A and B of the triangle by the straight lines AO and BO, which will meet each other in O; from the point draw the perpendiculars OD, OE, OF, to the three sides of the triangle; these lines will be equal to each other. For, by construction, the angle DAO = OAF, and the right angle ADO = AFO ; consequently the third angle AOD is equal to the third AOF. Besides, the side AO is common to the two triangles AOD, AOF; therefore, a side and the adjacent angles of the one being respectively equal to a side and the adjacent angles of the other, the two triangles are equal; hence DO = OF. It may be shown, in like manner, that the two triangles BOD, BOE, are equal; consequently OD OE; therefore the three perpendiculars OD, OE, OF, are equal to each other. Now, if from the point O, as a centre, and with the radius OD, we describe a circle, it is evident that this circle will be inscribed in the triangle ABC; for the side AB, perpendicular to the radius at its extremity, is a tangent. The same may be said of the sides BC, AC. 154. Scholium. The three lines, which bisect the three angles of a triangle, meet in the same point. PROBLEM. Fig. 88, 155. Upon a given straight line AB (fig. 88, 89) to describe a segment capable of containing a given angle C, that is a segment 89. such, that each of the angles, which may be inscribed in it, shall be equal to a given angle C. Solution. Produce AB toward D, make at the point B the angle DBE = C, draw BO perpendicular to BE, and GO perpendicular to AB, G being the middle of AB; from the point of meeting O, as a centre, and with the radius OB, describe a circle; the segment required will be AMB. For, since BF is perpendicular to the radius at its extremity, BF is a tangent, and the angle ABF has for its measure the half of the arc AKB (131); besides, the angle AMB, as an inscribed angle, has also for its measure the half of the arc AKB; consequently the angle AMB = ABF = EBD = C ; therefore each of the angles inscribed in the segment AMB is equal to the given angle C. 156. Scholium. If the given angle were a right angle, the segment sought would be a semicircle described upon the diameter AB. PROBLEM. 157. To find the numerical ratio of two given straight lines AB, CD (fig. 90), provided, however, these two lines have a common Fig. 90. measure. Solution. Apply the smaller CD to the greater AB, as many times as it will admit of, for example, twice with a remainder BE. Apply the remainder BE to the line CD, as many times as it will admit of, for example, once with a remainder DF. Apply the second remainder DF to the first BE, as many times as it will admit of, once, for example, with a remainder BG. Apply the third remainder BG to the second DF, as many times as it will admit of. Proceed thus, till a remainder arises, which is exactly contained a certain number of times in the preceding. This last remainder will be the common measure of the two proposed lines; and, by regarding it as unity, the values of the preceding remainders are easily found, and, at length, those of the proposed lines from which their ratio in numbers is deduced. If, for example, we find that GB is contained exactly twice in FD, GB will be the common measure of the two proposed lines. Let GB = 1, we have FD = 2; but EB contains FD once plus GB; therefore EB =3; CD contains EB once plus FD; therefore CD = 5; AB contains CD twice plus EB; therefore AB = 13; consequently the ratio of the two lines AB, CD, is as 13 to 5. If the line CD be considered as unity, the line AB would be 13; and, if the line AB be considered as unity, the line CD would be โร 158. Scholium. The method, now explained, is the same as that given in arithmetic for finding the common divisor of two numbers (Arith. 61), and does not require another demonstration. It is possible, that, however far we continue the operation, we may never arrive at a remainder, which shall be exactly contained a certain number of times in the preceding. In this case the two lines have no common measure, and they are said to be incommensurable. We shall see, hereafter, an example of this in the ratio of the diagonal to the side of a square. But, although the exact ratio cannot be found in numbers, by neglecting the last remainder we may find an approximate ratio to a greater Geom. 6 |