PRACTICAL APPLICATIONS. PROBLEM I. To bisect a given straight line. Let AB be a given straight line. From A and B, as centres, with a radius greater than one half of AB, describe arcs intersecting at E and F: join E and F, by the straight line EF. Then EF bisects the given line AB. For, E and F are each equally distant from A and B; and consequently, the line EF bisects AB (B. I., P. XVI., C.). PROBLEM II. To erect a perpendicular to a given straight line, at a given point of that line. Let EF be a given line, and let A be a given point of that line. From A, lay off the equal distances AB and AC; from B and C, as centres, with a radius greater than one half 85 draw the line AD: For, D and A are of BC, describe arcs intersecting at D; PROBLEM III. To draw a perpendicular to a given straight line, from a given point without that line. Let FG be the given line, and A the given point. From A, as a centre, with a radius sufficiently great, describe an arc cutting FG in two points, B and D; with B and D as radius greater than one half of BD, centres, and a describe arcs intersecting at E; draw AE: then AE is the perpendicular required. For, A and E are each equally distant from B and D: hence, AE is perpendicular to BD (B. I., P. XVI., C.). PROBLEM IV. At a point on a given straight line, to construct an angle equal to a given angle. Let A be the given point, AB the given line, and IKL the given angle. From the vertex K as a center, with any radius Kl, describe the arc IL, terminating in the sides of the angle. From A as K A a centre, with a radius AB, equal to Kl, describe the GEOMETRY. indefinite arc BO; then, with a radius equal to the chord LI, from B as a centre, describe an BO in D; draw AD: then BAD is equal to the angle K. For the arcs BD, IL, have equal radii and equal chords: hence, they are equal (P. IV.); arc cutting the arc B therefore, the angles BAD, IKL, measured by them, are also equal (P. XV.). PROBLEM V. To bisect a given arc or a given angle. 1°. Let AEB be a given arc, and C its centre. Draw the chord AB; through C, draw CD perpendicular to AB (Prob. III.): then CD bisects the arc AEB (P. VI.). 2°. Let ACB be a given angle. ID With C as a centre, and any radius CB, describe the arc BA; bisect it by the line CD, as just explained: then CD bisects the angle ACB. For, the arcs AE and EB are equal, from what was just shown; consequently, the angles ACE and ECB are also equal (P. XV.). Scholium If each half of an are or angle is bisected, the original arc or angle is divided into four equal parts; and if each of these is bisected, the original arc or angle is divided into eight equal parts; and so on. BOOK III. PROBLEM VI. Through a given point, to draw a straight line parallel to a given straight line. B Let A be a given point, and BC a given line. From the point A as a centre, with a radius AE, greater than the shortest distance from A to BC, describe an indefinite arc EO; from E as a centre, with the same radius, describe the arc AF; lay off ED equal to AF, and draw AD: then AD is the parallel required. are equal (P. XV.); therefore, the lines AD, EF are parallel (B. I., P. XIX., C. 1). For, drawing AE, the angles AEF, EAD, PROBLEM VII. Given, two angles of a triangle, to construct the third angle. Let A and B be given angles of a triangle. Draw a line DF, and at some point of it, as E, construct the angle FEH equal to A, and HEC equal to B. Then, CED is equal to the required angle. B E H For, the sum of the three angles at E is equal to two right angles (B. I., P. I., C. 3), as is also the sum of the three angles of a triangle (B. I., P. XXV.). Consequently, the third angle CED must be equal to the third angle of the triangle. PROBLEM VIII. Given, two sides and the included angle of a triangle, to construct the triangle. Let B and C denote the given sides, and A the given angle. Draw the indefinite line DF, and at D construct an angle FDE, equal to the angle A; on DF, lay off DH equal to the side C, and on DE, lay off DG equal to the side B; draw GH: then DGH is the required triangle (B. I., P. V.). B E PROBLEM IX. Given, one side and two angles of a triangle, to construct the triangle. The two angles may be either both adjacent to the given side, or one may be adjacent and the other opposite to it. In the latter case, construct the third angle by Problem VII. We shall then have two angles and their included side. Draw a straight line, and on it lay off DE equal to the given side; at D construct an angle equal to one of the adjacent angles, and at E construct an angle equal to the other adjacent angle; F G produce the sides DF and EG till they intersect at H: then DEH is the triangle required (B. I., P. VI.). |