TRY. 



100 



U CRY.- VII. 



XIV. To Jlnd a third proportional to two givtn 



\ mid it bo tlio two given straight lino* to which it in re- 

 quired to find a third proportional. Draw two straight lines c v, 



c Q, forming with each other a small angle p c Q. On c P set 

 off c n equal to A, and n F equal to 6, and on c Q set off c K 

 o<[uul to B. Join D E, and through tho point P draw F a parallel 

 to D E, and cutting c Q in o ; the straight line E a is a third 

 proportional to A and B ; that is, A is to B as B is to E o. 



If we know tho length of A and B, we can find the third pro- 

 portional to them by dividing the square of tho length B by the 

 length of A. Thus, if A be three feet, and B be six feet, tho 

 third proportional to A and B measures twelve feet, for the 

 square of 6 divided by 3, or 36 4- 3 = 12. 



PROBLEM XV. To Jlnd a fourth proportional to three given 

 straight lines. 



Let A, B, and c be the three given straight lines to which it is 

 required to find a fourth proportional. Draw two straight lines 

 P, D Q, forming with each other a small angle, p D Q. On D p 



Fig. 22. 



ot off D E equal to A, and E p equal to c, and on D Q set off 

 D o equal to B. Join E G, and through F draw F H parallel to 

 E a, and cutting D Q in H. The straight line H G is a fourth 

 proportional to A, B, and c ; that is, A is to B as c is to H o. 



If we know the length of A, B, and c, we can find the fourth 

 proportional to them by multiplying the length of B and c 

 together, and dividing the product by the length of A. Thus, if 

 A be four feet, B six feet, and c two feet, the fourth propor- 

 tional to A, B, and c measures three feet; for 6 X 2 = 12, and 

 12 H- 4 = 3. 



PROBLEM XVI. To divide a given straight line into any num- 

 ber of parts which shall be to one another in a given proportion. 



Let A B be the given straight line, which it is required to 

 divide into five parts, which are to one another in tho following 

 proportions namely, 5, 2, 3, 1, 4. First draw the straight 



Fig. 23. 



line A o of indefinite length, making a small angle B A c with the 

 given straight line A B. Along A c, from a scale of equal parts, 

 set off in regular succession A D equal to 5 of these equal parts, 

 D K equal to 2, E F equal to 3, p a equal to 1 , and a H equal to 

 4. Join H B, and through the points D, E, p, o draw the straight 

 linos D i, E K, P L, G M, cutting the straight ]iae" A B in the 

 points I, K, L, M. The given straight line A B is now divided 

 into five parts, A I, I K, K L, L M, M B, which are to one another 

 in the required proportions namely, 5, 2, 3, 1, and 4. 



This method of dividing a straight line into any number of 

 parts, which shall be to one another in a given proportion, is 

 based on Problem XII. (page 192). Supposing it had been 

 required to divide A B into 15 equal parts, it is manifestly only 

 requisite to set off along AC 15 equal parts, denoted by the 

 dots on the lino A c, from A to H, and then draw straight lines 

 in succession through each dot on H A, from H to A, parallel to 

 HB. 



The process that has been described in this Problem ensures 



VOL. I. 



Pig. 21. 



an accurate divuiou in COM* where the different parU * 

 represented by fraction* or mixed number* (Me Leak' 



tic, page 160), if we endearoorod to arrive at them by aa 

 arithmetical process. For example, had the line' A B in fig. S3 

 measured 80 inched, we can ao at onee that, as the com of the 

 number* wbieh show the proportion of the lines into whi 

 required to divide it i* equal to 15, the half of 30, we bare 

 only to multiply each number by 2, and mark off A I equal to 10 

 (or 5X2) inches, I K equal to 4 (or 2 X 2) inches and *o on. 

 But supposing A B had measured 29 inches, instead of 30, then 

 A i would be represented numerically by 9, i x by 3tf inches, 

 etc., and lines involving fractions of inches such as y, which 

 are not to be found on an ordinary scale, would be very difficult 

 to mark out without making a special scale for the purpose, or 

 resorting to the plan given above. 



PROBLEM XVII. To draw an equilateral triangle on any yiv* 

 straight line. 



Let A B bo the given straight line on which it is required to 

 draw an equilateral triangle. From the point A as a centre, 

 with A B as a radius, describe the arc B c ; 

 and from the point B as a centre, with 

 B A as a radius, describe the arc A c, cut- 

 ting the arc B c in the point c. Join A c, 

 B c ; the triangle A B c is equilateral or 

 equal-sided (see Definition 19, page 53), 

 and it is drawn on the given straight line 

 A B. 



If the arcs c A, c B be extended to cut 

 each other in the point D below the straight 

 lino A B, by joining D A, D B, we get another 

 equilateral triangle A B D, which is equal 

 to the equilateral triangle ABC, and which 

 is also drawn on the given straight line 

 A B. By taking any straight line as a 

 radius, and from each of its extremities as 

 centres striking arcs intersecting or cutting 

 each other on opposite sides of it, we get, by drawing straight 

 lines from the points in which the arcs cut each other to the 

 extremities of the straight line used as a radius, a regularly- 

 formed diamond-shaped figure, whose four sides and shortest 

 diagonal or diameter are all of equal length, such as A c B D in 

 the above figure. This figure with four equal sides is called a 

 rhombus. (See Definition 30, page 53.) 



The learner should construct Fig. 24 on a large scale by the 

 aid of his compasses and ruler. On applying a parallel ruler to 

 the opposite sides of the figure A c B D, he will find that they are 

 parallel to each other, namely, A c to B D, and BCtoAD; A c B D 

 is therefore a parallelogram, and A B, c D are its diagonals. (See 

 Definition 26, page 53.) From Theorem 5 (page 156) the 

 student learnt that the greatest side of every triangle is opposite 

 the greatest angle, and that the greater the opening of the angle 

 the greater must be the line that subtends or is opposite to it. 

 Now in the triangle A B c, or in any other equilateral triangle, 

 the three stright lines or sides by which it is contained are all 

 equal to one another, and as equal sides must necessarily subtend 

 equal angles, tho three angles of the triangle A B c namely, 

 A B c, B c A, c A B are also all equal to one another. Again, 

 from Theorem 7 (page 156) we have learnt that the three interior 

 angles of any triangle are equal to two right angles. A right 

 angle contains 90 degrees, and as two right angles contain just 

 twice as many, or 180 degrees, each of the equal angles ABC, 

 B c A, c A B, in the interior of the equilateral triangle ABC, 

 contains 180 -r- 3 or 60 degrees. 



Continuing our investigations a little further, we find that 

 each of the angles A c E, B c E is half of the angle A c B, and is 

 Jierefore an angle of 30 degrees. The angles A D E, B D r are 

 also angles of 30 degrees, because each of them is half of the 

 angle A D B, which, like the angle A c B, is an angle of 60 

 degrees. The angle c A D is equal to the angles c A B, D A B, 

 and as each of these equal angles contains 60 degrees, the angle 

 CAD contains 120 degrees. In the same way tho angle c B D 

 also contains 120 degrees. The diagonals of the rhombus 

 A c B D intersect each other at right angles, therefore it wfll 

 be seen that each of the angles CEA, CIB,DEA, ozsia 

 a right angle. 



Fig. 24 teaches us how to draw an angle of 45 degrees with- 

 out the aid of the protractor, as we will proceed to .show. 

 A c E is an angle of 30 degrees, and so is its adjacent angta 



14 



