PART THIRD. PROBLEMS. 134. REMARKS. 1. The only instruments required for the following Problems are a Ruler for drawing straight lines, and Compasses or Dividers for taking or laying off distances, and for drawing arcs of circles. The beginner may sometimes need directions from his teacher in regard to the best method of holding or applying these instruments. If the Ruler is not at hand, a straight edge may be obtained by doubling over a piece of paper upon itself; and by doubling again this edge upon itself we obtain a right angle (§ 25). A string or a small piece of wood with two pins in it may sometimes serve as a rude substitute for the Compasses or Dividers. § 135. 2. As a means of determining distance, it is often necessary to draw arcs of circles. A CIRCLE is a plane figure, bounded by a line which is everywhere equally distant from a point within called the centre. This line is called the circumference, and parts of it are called arcs. A line drawn from the centre to the circumference is called a radius. All radii, from the very definition of the circle, are equal. 3. Every problem should not only be performed, but strictly demonstrated. In the actual construction of geometrical figures, from the imperfection of our senses and instruments, absolute accuracy is of course impossible (§ 5. g). Still we should make this our standard, and approach as near to it as may be. K PROBLEM I. 136. To bisect a given straight line AB. From A and B as centres, with a radius greater than half of AB, describe arcs cutting each other in C and D. Draw the straight line CED. AB is bisected in E. If it is not convenient to obtain two points on opposite sides of AB, they may be obtained on the same side (as C and F) by using different radii. As the points C and D (or F) are C LIV. A B D each equally distant from A and B, the line passing through them must be a perpendicular upon the middle of AB (§ 92). PROBLEM II. 137. To erect a perpendicular to a straight line AB, at a given point C. From C lay off CD and CE equal to each other. From D and E as centres, describe arcs cutting each other in F. Draw FC, which will be the perpendicular required. As F and C are each equally distant from D and E, FC must be DE (§ 92). A LV. F B D C E § 138. If the given point C is at or near the end of the line, one of the following methods may be more convenient. THIRD METHOD. E LVII. § 139. From C as a centre, with any convenient radius CD, describe the arc DEF. In this arc take the points E and F, so that the distances DE and EF shall be = DC. From E and F as centres, with the same radius as before (or with any radius greater than half of EF), describe arcs cutting each other in G. Draw GC, which will be the perpendicular required. If straight lines were drawn joining DE, EF, and FC, as they would be all equal to DC, the figure DEFC would be a parallelogram (§ 108). Therefore, EF || AB. But, as G and C are each equally distant from E and F, GC (§ 92), and therefore AB (§ 46. a). PROBLEM III. A D C B EF 140. From a given point A without a straight line BC, to let fall a perpendicular upon that line. From A as a centre, with a radius greater than the distance to BC, describe an arc cutting BC in D and E. From D and E as centres, describe arcs cutting each other in F. Draw AF, which will be the perpendicular required. As A and F are each equally distant from D and E, AF DE (§ 92). SECOND METHOD. § 141. If A is nearly opposite the point C, draw AD obliquely to BC, and bisect it in P. From P as a centre, with the radius PA, describe an arc cutting BC in E. Draw AE, which will be the perpendicular required. LVIII. A E B BD LVI. Because AP PD = PE, aAED = 90° (§ 131). A E THIRD METHOD. § 142. In BC take any two points B and D; and from these as centres, with the radii BA and DA, describe arcs cutting each other in E. Draw AE, which will be the perpendicular required. As B and D are each equally distant from A and E, therefore BC – AE (§ 92). LIX. A C NOTE. In drawing perpendiculars, the instrument called a square is convenient. PROBLEM IV. $143. Through a given point A, to draw a parallel to a given straight line BC. B F LX. A E C D From A as a centre, with a radius greater than the distance to BC, describe the arc DE. From D as a centre, with the same radius, describe the arc AF. Take DE AF, and draw AE, which will be the parallel required. For, if AD be joined, as DF: AE, and straight lines joining DE and AF would be equal, therefore aDAE the alternate aADF (§ 98). .. AE || BC (§ 49. c). = AD = NOTE. In drawing parallels, the instrument called a parallel ruler is convenient. PROBLEM V. § 144. At a given point A in a line AB, to make an angle equal to a given angle D. From D as a centre, with any radius, describe an arc cutting the sides of the angle in E and F. From A as a centre, with the same radius, describe an arc GH. In this arc, take H at the same distance from G, that E is from F; and draw AH. the angle required. As AH AG= HAB is D DE = DF, and as straight lines joining HG and EF would be equal, aA = aD (§ 98). PROBLEM V1. § 145. To bisect a given angle A. From A as a centre, with any radius, LXII. A describe an arc cutting the sides of the angle in B and C. From B and C as centres, with a radius greater than half the distance between B and C, describe arcs cutting each other in D. which bisects the arc. B Draw AD, If BD and CD be joined, we have two triangles which agree in their three sides. Therefore, aBAD =aCAD (§ 98). PROBLEM VII. § 146. Two angles of a triangle, A and B, being given, to describe the third angle. Take any point E in any line DEF. and make aDEC = aA, and aCEG = aB. Then, as the three angles of a triangle are equal to 180°, aGEF must be equal to the third angle (§ 54. b). PROBLEM VIII. LXIII. C G D S 147. To describe a triangle, two sides and the in cluded angle being given. 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