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157. Apply Euc. I. 47, to express the squares of the three sides in terms of the squares on the perpendiculars and on the segments of AB.

158. By Euc. 1. 47. bearing in mind that the square described on any line is four times the square described upon half the line.

159. The former part is at once manifest by Euc. 1. 47. Let the diagonals of the square be drawn, and the given point be supposed to coincide with the intersection of the diagonals, the minimum is obvious. Find its value in terms of the side.

160. (a). This is obvious from Euc. 1. 13.
(6) Apply Euc. 1. 32, 29.
(C) Apply Euc. 1. 5, 29.

(d) Let AL meet the base BC in P, and let the perpendiculars from F, K meet BC produced in M and N respectively; then the triangles APB, FMB may be proved to be equal in all respects, as also APC, CKN.

(e) Let fall DQ perpendicular on FB produced. Then the triangle DQB may be proved equal to each of the triangles ABC, DBF; whence the triangle DBF is equal to the triangle ABC.

Perhaps however the better method is to prove at once that the triangles ABC, FBD are equal, by shewing that they have two sides equal in each triangle, and the included angles, one the supplement of the other.

(1), If DQ be drawn perpendicular on FB produced, FQ may be proved to be bisected in the point B, and DQ equal to AC. Then the square on FD is found by the right-angled triangle FQD. Similarly, the square on KE is found, and the sum of the squares on FD, EK, GH will be found to be six times the square on the hypotenuse.

(9) Through A draw PAQ parallel to BC and meeting DB, EC produced in P, Q. Then by the right angled triangles.

161. Let any parallelograms be described on any two sides AB, AC of a triangle ABC, and the sides parallel to AB, AC be produced to meet in a point P. Join PA. Then on either side of the base BC, let a parallelogram be described having two sides equal and parallel to AP. Produce AP and it will divide the parallelogram on BC into two parts respectively equal to the parallelograms on the sides. Euc. 1. 35, 36.

162. Let the equilateral triangles ABD, BCE, CAF be described on AB, BC, CA, the sides of the triangle ABC having the right angle at A.

Join DC, AK: then the triangles DBC, ABE are equal. Next draw DG perpendicular to AB and join CG: then the triangles BDG, DAG, DGC are equal to one another. Also draw AH, EK perpendicular to BC; the triangles EKH, EKA are equal. Whence may be shewn that the triangle ABD is equal to the triangle BHE, and in a similar way may be shewn that CAF is equal to CHE.

The restriction is unnecessary: it only brings AD, AE into the same line.


HINTS, &c. 6. See the figure Euc. 11. 5.

7. This Problem is equivalent to the following: construct an isosceles right-angled triangle, having given one of the sides which contains the right angle.

8. In the question for E read D. Construct the square on AB, and the property is obvious.

9. The sum of the squares on the two parts of any line is least when the two parts are equal.

10. A line may be found the square on which is double the square on the given line. The problem is then reduced to:-having given the hypotenuse and the sum of the sides of a right-angled. triangle, construct the triangle.

11. This follows from Euc. 11. 5, Cor.

12. This problem is, in other words, Given the sum of two lines and the sum of their squares, to find the lines. Let AB be the given straight line, at B draw BC at right angles to AB, bisect the angle ABC by BD. On AB take AE equal to the side of the given square, and with center A. and radius AE describe a circle cutting BD in D, from. D draw DF perpendicular to AB, the line AB is divided in F as was required.

13. Let AB be the given line. Produce AB to C making BC equal to three times the square on AB. From BA cut off BD equal to BC; then D is the point of section such that the squares on AB and BD are double of the square on AD.

14. In the fig. Euc. 11. 7. Join BF, and draw FL perpendicular on GD. Half the rectangle DB, BG, may be proved equal to the rectangle AB, BC.

Or, join KA, CD, KD, CK." Then CK is perpendicular to BD. And the triangles CBD, KBD are each equal to the triangle ABK. Hence, twice the triangle ABK is equal to the figure CBKD; but twice the triangle ABK is equal to the rectangle AB, BC; and the figure CBKD is equal to half the rectangle DB and CK, the diagonals of the squares on AB, BC. Wherefore, &c.

15. The difference between the two unequal parts may be shewn to be equal to twice the line between the points of section.

16. This proposition is only another form of stating Euc. II. 7. 17. In the figure, Theo. 7, p. 69, draw PQ,PR, PS perpendiculars on AB, AD, AC respectively: then since the triangle PAC is equal to the two triangles PAB, PAD, it follows that the rectangle contained by PS, AC, is equal to the sum of the rectangles PQ, AB, and PR, AD: When is the rectangle PS, AC equal to the difference of the other two rectangles ?

18. Through E draw EG parallel to AB, and through F, draw FHK parallel to BC and cutting EG in H. Then the area of the rectangle is made up of the areas of four triangles ; whence it may be readily shewn that twice the area of the triangle AFE, and the figure AGHK is equal to the area ABCD.

19. Apply Euc. 11. 11.

20. The vertical angles at L may be proved to be equal, and each of them a right angle.

21. Apply Euc. II. 4, 11. I. 47. 22. Produce FG, DB to meet in L, and draw the other diagonal LHC, which passes through H, because the complements AG, BẮ are equal. Then LH may be shewn to be equal to Ff, and to Dd.

23. The common intersection of the three lines divides each into two parts, one of which is double of the other, and this point is the vertex of three triangles which have lines drawn from it to the bisection of the bases. Apply Euc. 11. 12, 13.

24. Apply Theorem: 3, p. 104, and Euc. 1..47.

25. This will be found to be that particular case of Euc. ir. 12, in which the distance of the obtuse angle from the foot of the perpendicular,

is half of the side subtended by the right angle made by the perpendicular and the base produced.

26. (1) Let the triangle be acute-angled, (Euc. 11. 13, fig. 1.)

Let AC be bisected in E, and BE be joined; also EF be drawn perpendicular to BC. EF is equal to FC. Then the square on BE may be proved to be equal to the square on BC and the rectangle BD, BC.

(2) If the triangle be obtuse-angled, the perpendicular EF falls within or without the base according as the bisecting line is drawn from the obtuse or the acute angle at the base.

27. This may be shewn from theorem 3. p. 114.

28. Let the perpendicular AD be drawn from A on the base BC. It may be shewn that the base BC must be produced to a point E, such that CE is equal to the difference of the segments of the base made by the perpendicular,

29. Since the base and area are given, the altitude of the triangle is known. Hence the problem is reduced to ;-Given the base and altitude of a triangle, and the line drawn from the vertex to the bisection of the base, construct the triangle.

30. This follows immediately from Euc. I. 47. 31. Apply Euc. 11. 13.

32. The truth of this property depends on the fact that the rectangle contained by AC, CB is equal to that contained by AB, CD.

33. Let P the required point in the base AB be supposed to be known. Join CP. It may then be shewn that the property stated in the Problem is contained in Theorem 3. p. 114.

34. This may be shewn from Euc, 1. 47; 11. 5. Cor.

35. From C let fall CF perpendicular on AB. Then ACE is an obtuse-angled, and BEC an acute-angled triangle. Apply Euc. II, 12, 13; and by Euc. 1. 47, the squares on AC and CB are equal to the square on AB.

36. Apply Euc. I. 47, 11. 4; and the note p. 102, on Euc. II. 4,

37. Draw a perpendicular from the vertex to the base, and apply Euc. 1. 47 ; 11. 5, Cor. Enunciate and prove the proposition when the straight line drawn from the vertex meets the base produced.

38. This follows directly from Euc. II. 13, Case 1.
39. The truth of this proposition may be shewn from Euc. I. 47; 11. 4.

40. Let the square on the base of the isosceles triangle be described. Draw the diagonals of the square, and the proof is obvious.

41. Let ABC be the triangle required, such that the square on AB is three times the square on AC or BC. Produce BC and draw AD perpendicular to BC. Then by Euc. 11. 12, CD may be shewn to be equal: to one half of BC. Hence the construction.

42. Apply Euc. 11. 12, and Theorem 38, p. 118. 43. Draw EF parallel to AB and meeting the base in F; draw also EG perpendicular to the base. Then by Euc. 1. 47; 11, 5, Cor.

44. Bisect the angle B by BD meeting the opposite side in D, and draw BE perpendicular to AČ. Then by Euc. 1. 47; 11. 5, Cor.

45. This follows directly from Theorem 3, p. 114.

46. Draw the diagonals intersecting each other in P, and join OP. By Theo. 3, p. 114.

47. Draw from any two opposite angles, straight lines to meet in the bisection of the diagonal joining the other angles. Then by Euc. 11. 12, 13.

48. Draw two lines from the point of bisection of either of the bisected sides to the extremities of the opposite side; and three triangles will be formed, two on one of the bisected sides and one on the other, ini each of which is a line drawn from the vertex to the bisection of the base. Then by Theo. 3, p, 114.

49. 'If the extremities of the two lines which bisect the opposite sides of the trapezium be joined, the figure formed is a parallelogram which has its sides respectively parallel to, and equal to, half the diagonals of the trapezium. The sum of the squares on the two diagonals of the trapezium may be easily shewn to be equal to the sum of the squares on the four sides of the parallelogram.

50. Draw perpendiculars from the extremities of one of the parallel sides, meeting the other side produced, if necessary. Then from the four right-angled triangles thus formed, may be shewn the truth of the proposition.

51. In the problem, for triangle read rectangle. Let ABCD be any trapezium having the side AD parallel to BC. Draw the diagonal AC, then the sum of the triangles ABC, ADC may be shewn to be equal to the rectangle contained by the altitude and half the sum of AD and BC.

62. Let ABCD be the trapezium, having the sides AB, CD, parallel, and AD, BC equal. Join AC and draw AE perpendicular to DC. Then by Euc. II. 13.

53. Let ABC be any triangle; AHKB, AGFC, BDEC, the squares upon their sides; EF, GH, KL the lines joining the angles of the squares. Produce GA, KB, EC, and draw HN, DQ, FR perpendiculars upon them respectively: also draw AP, BM, CS perpendiculars on the sides of the triangle. Then AN may be proved to be equal to AM; CR to CP; and BQ to BS; and by Euc. 11. 12, 13.

54. Convert the triangle into a rectangle, then Euc. II. 14. 55. Find a rectangle equal to the two figures, and apply Euc. II. 14.

56. Find the side of a square which shall be equal to the given rectangle. See Prob. 1. p. 113.

57. On any line PQ take AB equal to the given difference of the sides of the rectangle, at A draw AC at right angles to AB, and equal to the side of the given square; bisect AB in 0 and join OC; with center O and radius OC describe a semicircle meeting PQ in D and E. Then the lines AD, AE have AB for their difference, and the rectangle contained by them is equal to the square on AC.

58. Apply Euc. II. 14.


HINTS, &c.

7. Euc. III. 3, suggests the construction.

8. The given point may be either within or without the circle. Find the center of the circle, and join the given point and the center, and upon this line describe a semicircle, a line equal to the given distance may be drawn from the given point to meet the arc of the semicircle. When the point is without the circle, the given distance may meet the diameter produced.

9. This may be easily shewn to be a straight line passing through the center of the circle.

10. The two chords form by their intersections the sides of two isosceles triangles, of which the parallel chords in the circle are the bases.

11. The angles in equal segments are equal, and by Euc. 1. 29. If the chords are equally distant from the center, the lines intersect the diameter in the center of the circle.

12. Construct the figure and the arc BC may be proved equal to the arc B'C'.

13. The point determined by the lines drawn from the bisections of the chords and at right angles to them respectively, will be the center of the required circle.

14. Construct the figures : the proof offers no difficulty.

15. On any radius construct an isosceles right-angled triangle, and produce the side which meets the circumference.

16. Join the extremities of the chords, then Euc. I. 27 ; II. 28. 17. Take the center 0, and join AP, AO, &c. and apply Euc. 1. 20.

18. Draw any straight line intersecting two parallel chords and meeting the circumference.

19. Produce the radii to meet the circumference.

20. Join AD, and the first equality follows directly from Euc. III. 20, I. 32. Also by joining AC, the second equality may be proved in a similar way. If however the line AD do not fall on the same side of the center O as E, it will be found that the difference, not the sum of the two angles, is equal to 2. AED. See note to Euc. III. 20, p. 155.

21. Let DKE, DBO (fig. Euc. III. 8) be two lines equally inclined to DA, then KE may be proved to be equal to BO, and the segments cut off by equal straight lines in the same circle, as well as in equal circles, are equal to one another.

22. Apply Euc. I. 15, and 111, 21.

23. This is the same as Euc. III, 34, with the condition, that the line must pass through a given point.

24, Let the segments AHB, AKC be externally described on the given lines AB, AC, to contain angles equal to BAC. Then by the converse to Euc. III. 32, AB touches the circle AKC, and AC the circle AHB.

25. Let ABC be a triangle of which the base or longest side is BC, and let a segment of a circle be described on BC, Produce BA, CA to meet the arc of the segment in D, E, and join BD, CE. If circles be described about the triangles ABD, ACE, the sides AB, AC shall cut off segments similar to the segment described upon the base BC.

26. This is obvious from the note to Euc.I11. 26, p. 156.

27. The segment must be described on the opposite side of the produced chord. By converse of Euc. III. 32.

28. If a circle be described upon the side AC as a diameter, the circumference will pass through the points D, E. Then Euc. IIl. 21. 29. Let AB,

AC be the bounding radii, and D any point in the arc BC, and DE, DF, perpendiculars from D on AB, AČ. The circle described on AD will always be of the same magnitude, and the angle EAF in it, is constant :-whence the arc EDF is constant, and therefore its chord EF.

30. Construct the figure, and let the circle with center 0, described on AH as a diameter, intersect the given circle in P, Q, join OP, PE, and prove EP at right angles to OP.

31. If the tangent be required to be perpendicular to a given line: draw the diameter parallel to this line, and the tangent drawn at the extremity of this diameter will be perpendicular to the given line.

32. The straight line which joins the center and passes through the ntersection of two tangents to a circle, bisects the angle contained by the tangents.

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