192 



THE POPULAR EDUCATOE. 



c A, the sides of the given triangle ABC. Draw a straight line 

 D E equal to A B, and from the points r>, E, as centres, with 

 radii respectively equal to the straight lines A c, B c, describe 

 arcs intersecting each other in the point F ; then join F E, F D ; 

 and the triangle D E F is the triangle required ; that is, it has 

 its three sides equal to the three sides A B, B c, c A of the given 

 triangle ABC; or it is equal to the triangle ABC. 



Fig. 14. 



The mode of construction is the same if it be required to draw 

 a triangle having its sides equal to three given straight lines 

 such as the straight lines A, B, c, in Fig. 14. 



PROBLEM IX. To draw a straight line through a given point, 

 that shall be parallel to a given straight line. 



Let B c (Fig. 15), be the given straight line, and A the given 

 point, through which a straight line parallel to B c is to be 

 j, p drawn. Take any point E in the 



straight line B c', join E A ; and 

 from the point E, as centre, with 



the radius E A, draw the arc A F, 



TD A cutting 3 c in F. Then from the 



Fig. 15. point A, as a centre with radius 



A E, draw the indefinite arc E o ; 



and from the point E as centre, with radius E D equal to the 

 distance A F, describe an arc cutting the arc E o, in the point D; 

 then join A r>, and it will be parallel to B c, as required. 



Another ivay. Another mode of constructing this problem 

 may be inserted here. Let A B (Fig. 16) be the given straight 

 line, and c the given point through which a straight line 

 parallel to A B is to be drawn. Take any point o, at a con- 

 venient distance from the straight line A B, but nearer to it 

 than to the point c ; join o c, and from o as centre, with radius 

 o c, describe the circle c D E G, intersect- B 



ing the straight line A B, in the points 

 D, E ; join c D, and then from E as a 

 centre, with radius or distance equal to 

 D c, describe an arc cutting the circle 

 c D E G in the point F ; and through 

 the points c, F draw the straight line 

 c F. The straight line c F is parallel to 

 the given straight line A B, and it is 

 drawn through the given point c, as pig. IQ_ 



required. 



There are various other ways of drawing a straight line 

 parallel to a given straight line, by means of the single ruler 

 and compasses ; but these are about the easiest. But parallel 

 straight lines are most easily drawn by means of the parallel 

 rulers described in a former lesson. Such instruments, however, 

 are not always at hand ; hence the utility of knowing how to 

 work the preceding problem. 



The only exercises or questions that could be given on the pre- 

 ceding problems, would be simply to desire the student to draw 

 all the figures above described according to the rules of con- 

 struction laid down in the different problems, which we earnestly 

 advise our self-educating students to do accordingly, by means 

 of the single ruler and compasses. 



PROBLEM X. To draw a straight line parallel to a given 

 straight line at a given distance from it. 



. Let A B be the given straight 



I line, and c the given distance 



- at which it is required to draw 

 a straight line parallel to A B. 

 Take any two points, D and E, 

 in the straight line A B, and 

 from these points as centres, 

 Fig. 17. with a radius equal to the given 



distance c, describe the arcs 



T G H, K L M. Draw G L touching these arcs, but not cutting 

 them. The straight line G L is parallel to the given straight 

 line AB. 



Another way. From any two points D E, in the straight line 

 A B, draw the straight lines D G, EL, perpendicular to A B ; 

 then from the same points as centres, with a radius equal to 

 the given distance c, draw the arcs F G H, K L M, cutting the 

 perpendiculars D G, E L in the points G and L. Join G L, and 

 produce it as far as may be required at either end. The straight 

 line G L is parallel to A B. 



PROBLEM XI. To trisect a right angle, or to divide a right 

 angle into three equal parts. 



Let B A c be the right angle that is to be divided into three 

 equal parts. Take any point D in A B, 

 and from the centre A at the distance 

 A D, describe the arc D E, cutting A c 

 in E. Then from the points D, E as 

 centres, with the radius D A or E A, 

 draw arcs, cutting the arc D E in the 

 points F G. Join A F, A G. The right 

 angle B A c is divided into three equal 

 parts by the straight lines A F, A G. 



If the angles B A F, F A G, G A E be 

 bisected by Problem VI., the right 

 angle B A c will be divided into six 

 equal parts, and by continued bisec- 

 tion it may be divided into any num- 



Fig. 18. 



ber of equal parts denoted by the series 6, 12, 24, 48, 96, 

 192, etc. 



PROBLEM XII. To divide a given straight line into any 

 number of equal parts. 



Let A B be the given straight line. From its extremity A 

 draw the straight lino A c, forming with A B the angle CAB, 

 and from the extremity 

 B draw B D parallel to 

 A c, and forming with 

 it the angle DAB, 

 which is equal to the 

 angle CAB. Set off 

 along the straight line 

 A c as many equal parts, 

 less one, as the number 

 of parts into which A B 

 is to be divided: that Fi S- 19 - 



is to say, if A B is to 



be divided into six equal parts, set off five equal parts, 

 A E, E F, F G, G H, H K along the straight line A c, and the 

 same number of equal parts, B L, L M, M N, NO, OP, along 

 the straight line B D. Join the straight lines p E, OF, N G, 

 M H, L K, cutting the straight line A B in the points Q, E, 

 s, T, tr. The parts A Q, Q R, R s, s T, T TJ, u B, into which tho 

 straight line A B is thus divided, are equal to one another, and 

 the straight line A B is divided into tho number of equal parts 

 required. 



PROBLEM XIII. To find a mean proportional between two 



given straight lines. 



Let A and B be the two given straight lines to which it is 



required to find a mean proportional that is to say, if A be the 



shorter of the two lines, a 

 line to which A bears the 

 same proportion as the line 

 required bears to B. Draw 

 the straight line c x, and 

 on c x set off c D equal 

 to A, and D E equal to B ; 

 bisect c E in G, and from 



CAB 



Fig. 20. 



the centre G at the dis- 

 tance G c or G E describe 



the semicircle c F E. From D draw the straight line D F perpen- 

 dicular to c E, and cutting the semicircle c F E in F, the 

 straight line D F is a mean proportional to A and B that is, A 

 is to D F as D F is to B. 



If we know the length of A and B we can find the mean 

 proportional to them by multiplying the numbers representing 

 the length of the lines together and extracting the square root 

 of the product. Thus, if A measure three feet, and B measure 

 twelve feet, the mean proportional to A and B measures six feet, 

 for 3 X 12 = 36 ; and the square root of 36, or the number 

 which when multiplied by itself gives 36, is 6. 



