GEOMETRICAL EXERCISES ON BOOK VI.

HINTS, &c.

6. In the figure Euc. vi. 23, let the parallelograms be supposed to be rectangular. Then the rectangle AC: the rectangle DG :: BC: CG, Euc. vi. 1. and the rectangle DG: the rectangle CF :: CD: EC,

whence the rectangle AC: the rectangle CF :: BC. CD: CG. EC. In a similar way it may be shewn that the ratio of any two parallelograms is as the ratio compounded of the ratios of their bases and altitudes. 7. Let two sides intersect in O, through O draw POQ parallel to the base AB. Then by similar triangles, PO may be proved equal to OQ: and POFA, QOEB, are parallelograms: whence AE is equal to FB.

8. Apply Euc. vI. 4, v. 7.

9. Let ABC be a scalene triangle, having the vertical angle A, and suppose ADE an equivalent isosceles triangle, of which the side AD is equal to AE. Then Euc. vi. 15, 16, AC.AB AD.AE, or AD'.

Hence AD is a mean proportional between AC, AB. Euc. vI. S.

10. The lines drawn making equal angles with homologous sides, divide the triangles into two corresponding pairs of equiangular triangles; by Euc. vi. 4, the proportions are evident.

11. By constructing the figure, the angles of the two triangles may easily be shewn to be respectively equal.

12. A circle may be described about the four-sided figure ABDC. By Euc. 1. 13; Euc. III. 21, 22. The triangles ABC, ACE may be shewn to be equiangular.

13. Apply Euc. 1. 48; 11. 5. Cor., vi. 16.

14. This property follows as a corollary to Euc. vI. 23: for the two triangles are respectively the halves of the parallelograms, and are therefore in the ratio compounded of the ratios of the sides which contain the same or equal angles: and this ratio is the same as the ratio of the rectangles by the sides.

15. Let ABC be the given triangle, and let the line EGF cut the base BC in G. Join AĞ. Then by Euc. vI. 1, and the preceding theorem (14,) it may be proved that AC is to AB as GE is to GF.

16. The two means and the two extremes form an arithmetic series of four lines whose successive differences are equal; the difference therefore between the first and the fourth, or the extremes, is treble the difference between the first and the second.

17. This may be effected in different ways, one of which is the following. At one extremity A of the given line AB draw AC making any acute angle with AB and join BC; at any point D in BC draw DEF parallel to AC cutting AB in E and such that EF is equal to ED, draw FC cutting AB in G. Then AB is harmonically divided in E, G.

18. In the figure Euc. vi. 13. If E be the middle point of AC; then AE or EC is the arithmetic mean, and DB is the geometric mean, between AB and BC. If DE be joined and BF be drawn perpendicular on DE; then DF may be proved to be the harmonic mean between AB and BC. 19. In the fig. Euc. vI. 13. DB is the geometric mean between AB and BC, and if AC be bisected in E, AE or EC is the Arithmetic mean. The next is the same as- -To find the segments of the hypotenuse of a right-angled triangle made by a perpendicular from the right angle,

S

having given the difference between half the hypotenuse and the perpendicular.

20. Let the line DF drawn from D the bisection of the base of the triangle ABC, meet AB in E, and CA produced in F. Also let AG drawn parallel to BC from the vertex A, meet DF in G. Then by means of the similar triangles; DF, FE, FG, may be shewn to be in harmonic progression.

21. If a triangle be constructed on AB so that the vertical angle is bisected by the line drawn to the point C. By Euc. vi. A, the point required may be determined.

22. Let DB, DE, DCA be the three straight lines, fig. Euc. III. 37: let the points of contact B, E be joined by the straight line BC cutting DA in G. Then BDE is an isosceles triangle, and DG is a line from the vertex to a point G in the base. And two values of the square on BD may be found, one from Theo. 37, p. 118: Euc. 11. 35; 11. 2; and another from Euc. III. 36; II. 1. From these may be deduced, that the rectangle DC, GA, is equal to the rectangle AD, CG. Whence the, &c.

23. Let ABCD be a square and AC its diagonal. On AC take AE equal to the side BC or AB: join BE and at E draw EF perpendicular to AC and meeting BC in F. Then EC, the difference between the diagonal AC and the side AB of the square, is less than AB; and CE, EF, FB may be proved to be equal to one another: also CE, EF are the adjacent sides of a square whose diagonal is FC. On FC take FG equal to CE and join EG. Then, as in the first square, the difference CG between the diagonal FC and the side EC or EF, is less than the side EC. Hence EC, the difference between the diagonal and the side of the given square, is contained twice in the side BC with a remainder CG: and CG is the difference between the side CE and the diagonal CF of another square. By proceeding in a similar way, CG, the difference between the diagonal CF and the side CE, is contained twice in the side CE with a remainder and the same relations may be shewn to exist between the difference of the diagonal and the side of every square of the series which is so constructed. Hence, therefore, as the difference of the side and diagonal of every square of the series is contained twice in the side with a remainder, it follows that there is no line which exactly measures the side and the diagonal of a square.

24. Let the given line AB be divided in C, D. On AD describe a semicircle, and on CB describe another semicircle intersecting the former in P; draw PE perpendicular to AB; then E is the point required.

25. Let AB be equal to a side of the given square. On AB describe a semicircle; at A draw AC perpendicular to AB and equal to a fourth proportional to AB and the two sides of the given rectangle. Draw CD parallel to AB meeting the circumference in D. Join AD, BD, which are the required lines.

26. Let the two given lines meet when produced in A. At A draw AD perpendicular to AB, and AE to AC, and such that AD is to AE in the given ratio. Through D, E, draw DF, EF, respectively parallel to AB, AC and meeting each other in F. Join AF and produce it, and the perpendiculars drawn from any point of this line on the two given lines will always be in the given ratio.

27. The angles made by the four lines at the point of their divergence, remain constant. See Note on Euc. vi. A, p. 295.

28. Let AB be the given line from which it is required to cut off a part BC such that BC shall be a mean proportional between the remainder AC and another given line. Produce AB to D, making BD

equal to the other given line. On AD describe a semicircle, at B draw BE perpendicular to AD. Bisect BD in O, and with center O and radius OB describe a semicircle, join OE cutting the semicircle on BD in F, at F draw FC perpendicular to OE and meeting AB in C. C is the point of division, such that BC is a mean proportional between AC and BD.

29. Find two squares in the given ratio, and if BF be the given line (figure, Euc. vI. 4), draw BE at right angles to BF, and take BC, CE respectively equal to the sides of the squares which are in the given ratio. Join EF, and draw CA parallel to EF: then BF is divided in A as required.

30. Produce one side of the triangle through the vertex and make the part produced equal to the other side. Bisect this line, and with the vertex of the triangle as center and radius equal to half the sum of the sides, describe a circle cutting the base of the triangle.

31. If a circle be described about the given triangle, and another circle upon the radius drawn from the vertex of the triangle to the center of the circle, as a diameter, this circle will cut the base in two points, and give two solutions of the problem. Give the Analysis.

32. This Problem is analogous to the preceding.

33. Apply Euc. vi. 8, Cor.; 17.

34. Describe a circle about the triangle, and draw the diameter through the vertex A, draw a line touching the circle at A, and meeting the base BC produced in D. Then AD shall be a mean proportional between DC and DB. Euc. II. 36.

35. In BC produced take CE a third proportional to BC and AC; on CE describe a circle, the center being O; draw the tangent EF at E equal to AC; draw FO cutting the circle in T and T; and lastly draw tangents at T, T′ meeting BC in P and P'. These points fulfil the conditions of the problem.

By combining the proportion in the construction with that from the similar triangles ABC, DBP, and Euc. III. 36, 37: it may be proved that CA.PD CP". The demonstration is similar for P'D'.

=

36. This property may be immediately deduced from Euc. vr. 8, Cor. 37. Let ABC be the triangle, right-angled at C, and let AE on AB be equal to AC, also let the line bisecting the angle A, meet BC in D. Join DE. Then the triangles ACD, AED are equal, and the triangles ACB, DEB equiangular.

38. The segments cut off from the sides are to be measured from the right angle, and by similar triangles are proved to be equal; also by similar triangles, either of them is proved to be a mean proportional between the remaining segments of the two sides.

39. First prove AC: AD :: BC: 2. BD : then 2. AC: AD2 :: BC: BD, whence 2. AC2 AD: AD2:: BC - BD: BD,

2. AC2 - (AC2 + DC2) = AC2 — CD3,

the property is immediately deduced.

40. The construction is suggested by Euc. I. 47, and Euc. vi. 31. 41. See Note Euc. vi. A. p. 295. The bases of the triangles CBD, ACD, ABC, CDE may be shewn to be respectively equal to DB, 2.BD, 3.BD, 4.BD.

42. (1) Let ABC be the triangle which is to be bisected by a line drawn parallel to the base BC. Describe a semicircle on AB, from the center D draw DE perpendicular to AB meeting the circumference in E, join EA, and with center A and radius AE describe a circle cutting AB in F, the line drawn fron F parallel to BC, bisects the triangle. The

proof depends on Euc. vi. 19; 20, Cor. 2. (2) Let ABC be the triangle, BC being the base. Draw AD at right angles to BA meeting the base produced in D. Bisect BC in E, and on ED describe a semicircle, from B draw BP to touch the semicircle in P. From BA cut off BF equal to BP, and from F draw FG perpendicular to BC. The line FG bisects the triangle. Then it may be proved that BFG: BAD :: BE: BD, and that BAD: BAC :: BD : BC; whence it follows that BFG: BAC :: BE: BC or as 1 : 2.

43. Let ABC be the given triangle which is to be divided into two parts having a given ratio, by a line parallel to BC. Describe a semicircle on AB and divide AB in D in the given ratio; at D draw DE perpendicular to AB and meeting the circumference in E; with center A and radius AE describe a circle cutting AB in F: the line drawn through F parallel to BC is the line required. In the same manner a triangle may be divided into three or more parts having any given ratio to one another by lines drawn parallel to one of the sides of the triangle. 44. Let these points be taken, one on each side, and straight lines be drawn to them; it may then be proved that these points severally bisect the sides of the triangle.

45. Let ABC be any triangle and D be the given point in BC, from which lines are to be drawn which shall divide the triangle into any number (suppose five) equal parts. Divide BC into five equal parts in E, F, G, H, and draw AE, AF, AG, AH, AD, and through E, F, G, H draw EL, FM, GN, HO parallel to AD, and join DL, DM, DN, DO; these lines divide the triangle into five equal parts.

By a similar process, a triangle may be divided into any number of parts which have a given ratio to one another.

46. Let ABC be the larger, abc the smaller triangle, it is required to draw a line DE parallel to AC cutting off the triangle DBE equal to the triangle abc. On BC take BG equal to be, and on BG describe the triangle BGH equal to the triangle abc. Draw HK parallel to BC, join KG; then the triangle BGK is equal to the triangle abc. On BA, BC take BD to BE in the ratio of BA to BC, and such that the rectangle contained by BD, BE shall be equal to the rectangle contained by BK, BG. Join DE, then DE is parallel to AC, and the triangle BDE is equal to abc.

47. Let ABCD be any rectangle, contained by AB, BC,

Then AB2: AB.BC::AB: BC,

and AB.BC: BC:: AB: BC,

whence AB2: AB. BC :: AB.BC: BC2,

or the rectangle contained by two adjacent sides of a rectangle, is a mean proportional between their squares.

48. In a straight line at any point A, make Ac equal to Ad in the given ratio. At A draw AB perpendicular to cAd, and equal to a side of the given square. On cd describe a semicircle cutting AB in b; and join bc, bd; from B draw BC parallel to be, and BD parallel to bd: then AC, AD are the adjacent sides of the rectangle. For, CA is to AD as cA to Ad, Euc. vi. 2; and CA.AD = AB2, ČBD being a right-angled triangle.

49. From one of the given points two straight lines are to be drawn perpendicular, one to each of any two adjacent sides of the parallelogram; and from the other point, two lines perpendicular in the same manner to each of the two remaining sides. When these four lines are drawn to intersect one another, the figure so formed may be shewn to be equi angular to the given parallelogram.

50. It is manifest that this is the general case of Prop. 4, p. 197. If the rectangle to be cut off be two-thirds of the given rectangle ABCD. Produce BC to E so that BE may be equal to a side of that square which is equal to the rectangle required to be cut off; in this case, equal to two-thirds of the rectangle ABCD. On AB take AF equal to AD or BC; bisect FB in G, and with center G and radius GE, describe a semicircle meeting AB, and AB produced, in H and K. On CB take CL equal to AH and draw HM, LM parallel to the sides, and HBLM is two-thirds of the rectangle ABCD.

51. Let ABCD be the parallelogram, and CD be cut in P and BC produced in Q. By means of the similar triangles formed, the property may be proved.

52. The intersection of the diagonals is the common vertex of two triangles which have the parallel sides of the trapezium for their bases.

53. Let AB be the given straight line, and C the center of the given circle; through C draw the diameter DCE perpendicular to AB. Place in the circle a line FG which has to AB the given ratio; bisect FG in H, join CH, and on the diameter DCE, take CK, CL each equal to CH; either of the lines drawn through K, L, and parallel to AB is the line required.

54. Let C be the center of the circle, CA, CB two radii at right angles to each other; and let DEFG be the line required which is trisected in the points E, F. Draw CG perpendicular to DH and produce it to meet the circumference in K; draw a tangent to the circle at K: draw CG, and produce CB, CG to meet the tangent in L, M, then MK may be shewn to be treble of LK.

55. The triangles ACD, BCE are similar, and CF is a mean proportional between AC and CB.

56. Let any tangent to the circle at E be terminated by AD, BC tangents at the extremity of the diameter AB. Take O the center of the circle and join OC, OD, OE; then ODC is a right-angled triangle and OE is the perpendicular from the right angle upon the hypotenuse.

57. This problem only differs from problem 59, infra, in having the given point without the given circle.

58. Let A be the given point in the circumference of the circle, C its center. Draw the diameter ACB, and produce AB to D, taking AB to BD in the given ratio: from D draw a line to touch the circle in E, which is the point required. From A draw AF perpendicular to DE, and cutting the circle in G.

59. Let A be the given point within the circle whose center is C, and let BAD be the line required, so that BA is to AD in the given ratio. Join AC and produce it to meet the circumference in E, F. Then EF is a diameter. Draw BG, DH perpendicular on EF: then the triangles BGA, DHA are equiangular. Hence the construction.

60. Through E one extremity of the chord EF, let a line be drawn parallel to one diameter, and intersecting the other. Then the three angles of the two triangles may be shewn to be respectively equal to one another.

61. Let AB be that diameter of the given circle which when produced is perpendicular to the given line CD, and let it meet that line in C; and let P be the given point: it is required to find D in CD, so that DB may be equal to the tangent DF. Make BC: CQ:: CQ: CA, and join PQ; bisect PQ in E, and draw ED perpendicular to PQ meeting CD in D; then D is the point required. Let O be the center of the circle, draw the tangent DF; and join OF, OD, QD, PD. Then QD may be shewn