111. PROP. XIV. To construct á right-angled triangle with given parts. 1st. When each of the two sides forming the right angle are given. The construction in this case is too obvious to require even to be stated. 2nd. When one side, and the adjacent acute angle, are given. From one end of the given side, draw a straight line at right angles to it, and from the other end, draw another straight line making with the given side an angle equal to the given angle (105). These two lines, with the given line, will form the required triangle. 3rd. When one side, and the opposite angle is given. See fig. 2 in (110); let DE be the given side; from D and E draw DF, EG at right angles to DE; make <GEF the given angle; DEF is the required triangle. For DF is parallel to EG, .. ¿DFE=¿FEG=the given angle. = 4th. When one side, and the hypothenuse, are given. Upon the hypothenuse as a diameter describe a semicircle. From one end draw a chord equal to the given side; and join the other end with the end of that chord. 112. PROP. XV. From the vertex of one of the angles of a given triangle to draw a perpendicular upon the opposite side. Let ABC be the given triangle. It is required to draw from the point C a perpendicular to AB. Bisect AC in D; with centre D and radius DA or DC describe the semi-circle AEC, intersecting AB in E. Join CE, and it is the perpendicular required. A D B For CEA is an 'angle in a semi-circle', and is therefore a right angle (54). The same thing may, of course, be done by any of the methods given in (103) for drawing a straight line perpendicular to a given straight line from a given point without it. 113. PROP. XVI. To bisect a given angle, that is, to divide it into two equal angles. (1) Let BAC be the given angle. With centre A, and any convenient radius (the L = L D (2) Most persons, in practice, knowing that in the same circle equal arcs subtend equal angles at the centre, instead of bisecting the chord DE, would bisect the arc DE, and join the point of bisection and the point A. And the arc DE would mostly be bisected by trial with the compasses, since equal chords subtend equal arcs. (3) The following is an expeditious method of bisecting an angle: With centre A and any convenient radius describe the arc DE; then with the same radius and centres D and E describe arcs intersecting in F; and join AF. AF bisects the angle BAC. с The accuracy of the work may also readily be tested. For, diminishing the opening of the_compasses, if with centres D and E another pair of intersecting arcs be drawn, their point of intersection ought to be in AF. A D B (4) Another Method by means of the Parallel-Ruler. Take any point D in AB; through D draw a straight line DF parallel to AC, and make DF-AD, and join AF. The angle BAC is bisected by AF (26 and 34). 114. PROP. XVII. To bisect the angle between two given straight lines, when the vertex of the angle is not given, and cannot conveniently be determined. Let AB, CD be the two given straight lines, which cannot conveniently be produced to meet. From any point E in one of them, CD, draw EF parallel to the other, AB. Bisect DEF by the straight line EG (113), meeting AB in G. With centres E and G, and any radius, 2 draw two pairs of intersecting arcs, and draw HI joining the points of intersection; then HI is the line required. For, by construction, DEG F = FEG. Also FEG=LEGB, since FE is parallel to AB (34); .. 4 DEG =4 EGB, and .. the D K triangle which would be form ed by producing the given lines to meet would be isos- A celes upon base EG. Also HI bisects that base at right angles (101), and every straight line which bisects the base of an isosceles triangle at right angles will, if produced, pass through the vertex of the opposite angle, as may easily be proved. 115. PROP. XVIII. To trisect a right angle, that is, to divide it into three equal angles. Let BAC be the given right angle; with centre A, and any radius AD, a part of AB, (the larger the better), describe an arc of a circle meeting E AB in D, and AC in E; with centre D, and the same radius as before, draw a small arc to intersect the former in F; and, again, with centre E, and radius as before, another small are to intersect the first in G. Join AF, AG; and the thing required is done. = A DB For, joining DF, and EG, AFD, and AEG are both equilateral triangles; DAF one-third of two right angles (26 Cor. and 37,) that is, two-thirds of one right angle, and.. EAF, which makes up the right angle, must be one-third of a right angle. Similarly, since AEG is an equilateral triangle, 4 EAG= two-thirds of a right angle, and .. ‹ DAG= one-third of a right angle. Hence FAG which, added to these two, makes up the right angle, must also be one-third of a right angle. COR. Hence also a quadrant, or the arc of a quadrant, may be divided into three equal parts. 116. PROP. XIX. To find the point which is at certain given distances from two given straight lines in the same plane. C Let AB, and CD be the two given straight lines. Take any point E in AB; draw EF at right angles to AB, and equal to the given distance from AB. Also from any point G in CD draw GH at right angles to CD, and H D F Ι equal to the other given dis tance. Through F draw FI parallel to AB, and through A H draw HI parallel to CD; and the point of intersection, I, of these parallels is the point required. For, drawing through I two straight lines parallel to EF, GH, the proof is obvious by (40). 117. PROP. XX. To construct a square with each of its sides equal to a given straight line. D This is done in (42); but in practice it will most frequently be done thus:-Take AB equal to the given length or line; from A and B, by means of the "Square', or the Triangle', draw AC, BD, at right angles to AB; and make each of them equal to AB. Then join CD; and ACDB is the square re quired. The correctness of the work always be easily tested by applying the compasses or rule to the opposite corners, for in a true square the diagonals must be equal to one another, as well as the sides. 118. PROP. XXI. To construct a square with its diagonal equal to a given straight line or length. Let AB be taken equal to the given straight line; with centres A and B, and any convenient A radius greater than the half of AB, draw two pairs of intersect D B ing arcs on opposite sides of AB; join the points of intersection by E a straight line cutting AB in C. In this line take CD, CE, each equal to CA or CB; join AD, BD, AE, BE, and ADBE is the square required. For, if with centre C and radius CA a circle be described, it will pass through the points A, D, B, E; and each of the angles of the fig. ADEB will be the 'angle in a semi-circle', and .. a right angle (54). Also the sides will be the chords of equal arcs, and .. will be equal (58). 119. PROP. XXII. To construct a square which shall be equal to the sum of two given squares. Draw the straight line AB equal to a side of one of the given squares, and BC, at right angles to AB, equal to a side of the other; and join AC. Then construct a square whose side is equal to AC (117), and it will be the square required. For, since ABC is a right angle, the square of AC = square of A AB+ square of BC (43). 120. PROP. XXIII. C B To construct a square which shall be equal to the difference of two given squares. Draw the straight line AB equal to a side of the greater of the given squares; bisect it in C; with centre C and radius CA describe a semi-circle. With centre B, and radius equal to a side of the other square, describe a small arc intersecting the semi-circle in D. Join AD. Then construct a square whose side is equal to AD (117), and it will be the square required. A C B For, joining BD, ADB is a right-angled triangle (54), square of AB=square of AD + square of BD; and taking from these equals the square of BD, we have the difference of the squares of AB and BD= the square of AD. 121. PROP. XXIV. To construct a square which shall be half of a given square. |