33. Bisect the base by a line drawn in the given direction, whether parallel to a given line, or tending to a given point. The plane drawn through the bisecting line and the vertex of the pyramid, gives the solution of the problem.

34. Through each line draw a plane parallel to the other; these planes will be parallel, and obviously form two of the faces of the parallelopiped. Through each line and one extremity of the other, draw a plane; and a second plane parallel to it through the remaining extremity. This will complete the figure; but there will be four varieties of cases according as the extremities are situated.

35. From the vertex A draw a line to any point B in the base of the pyramid, and meeting the given section in B'. From the angular points of the base draw lines to the point B; also from the angular points of the given section to the point B'. Then any triangle in the section, may be shewn to be similar to the corresponding triangle in the base. Euc. vi. 20.

36. Let AB be at right angles to the plane BCED, and let the perpendiculars from AB intersect the plane GHKL in the line MN, and let HNK be the common intersection of the planes CBDE, GHKL. Join AM, BN, and prove MN to be a straight line perpendicular to HK. 37. Draw the necessary lines, and by Euc. I. 47.

38. Let AE meet the straight lines BE, DE, in the plane BED, fig. Euc. XI. 6, and let the angle AEB measure the inclination of AE to the plane BDE; then the angle AEB is less than the angle AED. Draw AB perpendicular to the plane, make ED equal to EB and join BD, AD. Euc. 1. 18, 19.

39. Let HM be the common section of the two planes MN, MQ; and let AB be drawn from a point A in HM perpendicular to the plane MN: then, if planes be drawn through AB to cut the planes MN, MQ in lines which make the angles CAD, EAF with each other, and that the plane BACD is perpendicular both to MN and MQ; the angle CAD will be greater than EAF. Shew that the angle BAD is less than the angle BAF, and it follows that CAD is greater than EAF.

40. Let GH be the edge of the wall, A, B the two points, and let the line joining A, B, meet the edge of the wall GH in E. If the points AE, BE make equal angles with GH, then AE, EB may be proved to be less than any other two lines drawn from A, B, to meet GH in any other point E'.

41. Let A, B, be the given points, and GH the given straight line; draw AC, BD, perpendicular on GH, and in the plane AGH produced, draw DB' perpendicular to GH, and equal to DB; join AB', meeting GH in E, and draw EB. Then AE + EB is the minimum. For the triangles EDB, EB'D are equal, being right-angled at D, and having one side common, and the others equal. Whence the angle BEH is equal to GEA, each being equal to B'EH. The conclusion follows from the demonstration of the preceding theorem.

42. Let AB, A'B' be any portions of the two straight lines. At B' draw B'C' parallel to AB, and B'C perpendicular to the plane passing through A'B'C'. Let the plane passing through A'B'C intersect the line AB in the point A. In the plane A'B'C, from A draw AA' perpendicular to A'B', and AC perpendicular to AA'. Then the plane CAB passing through the line AB may be shewn to be parallel to the plane A'B'C passing through the line A'B', and that no other parallel planes can be drawn through AB, A'B'. Also AA' is the perpendicular distance between the two planes, and that AA' is less than any other line which can be drawn between the two planes.

GEOMETRICAL EXERCISES ON BOOK XII.

HINTS, &c.

5. Apply Euc. XII. 2.

6. First, to bisect a circle by a concentric circle. Let C be its center, AC any radius. On AC describe a semicircle, bisect AC in B, draw BD perpendicular to AC, and meeting the semicircle in D; join CD, and with center C, and radius CD, describe a circle; its circumference shall bisect the given circle. Join AD. Then by Euc. vi. 20, Cor. 2, the square on AC is to the square on CD as AC is to CB; and Euc. XII. 2. In the same way, if the radius AC be trisected, and perpendiculars be drawn from the points of trisection to meet the semicircle in D, E, the two circles described from C with radii CD, CE shall trisect the circle. And generally, a circle may be divided into any number of equal parts. NOTE. By a similar process a circle may be divided into any number of parts which shall have to each other any given ratios.

7. To divide the circle into two equal parts. Let any diameter ACB be drawn, and two semicircles be described, one on each side of the two radii AC, CB: these semicircles divide the circle into two equal parts which have their perimeters equal. In a similar way a circle may be divided into three equal parts, by dividing the diameter into three equal parts, AB, BC, CD, and describing semicircles upon AB, AC on one side of the diameter, and then semicircles upon DC, DB on the other side of the diameter.

8. By Euc. XII. 2, the area of the quadrant ADBEA is equal to the area of the semicircle ABCA.

9. By Euc. XII. 2. The squares on the radii of the two circles may be shewn to be in the ratio of 3 to 1.

10. By reference to Theorem 2, p. 346 and Euc. XII. 2, the parts of the diameter may be proved to bear to each other the ratio of 1 to 2. 11. Apply Euc. XII. 2.

12. If the circles whose centers are B and C touch each other in S, the problem may mean:-to find the point R, so that the figure between the three circles (see fig. Theo. 2, p. 346) may be bisected by the line RS; or it may mean, if two chords be drawn from P, Q, to R, the portions of the lunes bounded by parts of these chords and portions of the circles may be equal.

13. This will be found by Theorem 1, p. 346.

14. Produce CD to meet the arc of the quadrant in E. Then the sector ACE is half of the quadrant: also the semicircle CDA may be shewn to be equal to half the quadrant. The segments on CD and DA are similar and equal, if the figure bounded by DĂ, AC, and the arc CD be added to each, the remaining part of the semicircle on AC is equal to the triangle ACD which is a right-angled isosceles triangle.

15. The area of the circle of which the quadrant is given, is to the area of the circle which touches the three circles, as 36 is to 1. And the quadrant is one-fourth of the area of the circle. Hence the quadrant is to the circle as 9 to 1.

1

16. The circles on BA, AC are as the squares on BA, AC; Euc. XII. 2. and the square on BA is equal to the rectangle BC, BD, also the square on AC is equal to the rectangle CB, CD; whence it follows that the circles are as BD, CD.

17. Let ABC be the right-angled trianglé, BC being the hypotenuse, and let semicircles be described on AB, AC as diameters. Bisect AB, AC, BC, in E, F, G; from G draw perpendiculars on AB, AC, meeting the semicircles in H, K, and shew that GH is equal to GK. By Euc. XII. 2. the difference is found.

18. Let AB, A'B' be arcs of concentric circles whose center is C and adii CA, CA', and such that the sector ACB is equal to the sector A CB'. Assuming that the area of a sector is equal to half the rectangle contained by the radius and the included arc: the arc AB is to the arc A'B' as the radius A'C is to the radius AC. Let the radii AC, BC be cut by the interior circle in A', D. Then the arc A'D is to the arc AB, as A'C is to AC; because the sectors A'CD, ACB are similar: and the arc AB' is to the arc AD, as the angle ACB' is to the angle ACD, or the angle ACB. Euc. vi. 33. From these proportions may be deduced the proportion:- -as the angle ACB is to the angle A'CB', so is the square on the radius A'C to the square on the radius AC. And by Euc. XII. 2, the property is manifest.

19. Let AB, A'B' be arcs of two concentric circles, whose center is C. ACB, A'CB' two sectors such that the angle ACB is to the angle ACB', as A'C2 is to AC. If AC, BC be cut by the interior circle in A', D; then the arc A'B' is to the arc A'D, as the angle A'CB' is to the angle A CD, or ACB. Euc. vi. 33. And the arc A'D is to the arc AB, as the radius A/C is to the radius AC, by similar sectors. By means of these two proportions and the given proportion, the rectangle contained by the arc AB and the radius AC, may be proved equal to the rectangle contained by the arc A'B' and the radius A'C.

20. Let the arc of a semicircle on the diameter AB be trisected in the points D, E; C being the center; join AD, AE, CD, CE; then the difference of the segments on AD and AE, may be proved to be equal to the sector ACD or DCE.

21. Assuming that the area of a sector of a circle is equal to half the rectangle contained by the radius and the arc, the sector AOC is shewn to be equal to AOB.

22. Let POQ be any quadrant, O being the center of the circle, and let BG, DH be drawn perpendicular to_the_radius PO, and OB, OD be joined. The triangle GBO is equal to DHO.

23. The radii of the circles may be proved to be proportional to the two sides of the original triangle. Then by Euc. XII. 2; vI. 19.

24. The triangles CEA, CEB are equal, and the difference of the two segments is equal to the difference of the parts of the semicircle made by CE. The difference of the same parts may also be shewn to be equal to double the sector DEC.

25. Let AB be the hypotenuse of the right-angled triangle ABC, and let the semicircles described upon the sides AC, BC, intersect the hypotenuse in D. Join AD. AD is perpendicular to AB. The segments on AC, AD, and on one side of CD are similar; and the segments on AC may be shewn to be equal to the segments on AD, CD. Also the segment on BC may be shewn to be equal to the segments on BD, and the other side of CD. If Euc. vi. 31 be true for all similar figures, the conclusions above stated follow at once.

26. The area of the triangle ABC is equal to the quadrant ABD. From these equals take the figure AEDB.

27. The segments on BC, BA, AC may be shewn to be similar. And similar segments of circles may be proved to be proportional to the squares on their radii, Euc. XII. 2, and to the squares on the chords on which they stand, Euc. vI. 6.

If Euc. vi. 31 be extended to any similar figures, the equality follows directly.

28. This is shewn from Euc. XII. 2; 1. 47; v. 18.

29. The sum of the squares on the segments of the diagonals, is equal to the sum of the squares on each pair of opposite sides of the quadrilateral figure. Hence by Euc. XII. 2; I. 47; v. 18, the property is proved.

30. The squares on the four segments, are together equal to the square on the diameter. Theorem 6, p. 163. Then by Euc. XII. 2. 31. This is shewn by Euc. I. 47; XII. 2; v. 18.

32. Apply Theorem 1, p. 346.

33. Is analogous to Euc. III. 14.

34. The arc of a circle being considered as the measure of an angle which the arc subtends; the angle between the planes of two great circles can be shewn to be equal to the angle between the two radii of that great circle which bisects the two planes at right angles.

35. First, shew that all the lines drawn in the plane of the section, from that point where the diameter of the sphere meets the section, to the surface of the sphere, are equal. The second part is analogous to Euc. III. 14.

36. This may be proved indirectly as in Euc. III. 18.

37. Let D be the given point, and from D let DA be drawn through the center E, and meeting the surface in C, A. Let DB be a line from D touching the sphere at B. Join BE. Then the triangle DBE (fig. Euc. 111. 36) is in a plane passing through D, and E the centre of the sphere, and the distances DE, EB are always the same. Hence it follows that BD is always of the same length. Euc. 1. 47.

The sphere which touches the six edges of any tetrahedron, has four circular sections touching the sides of the four triangles which form the surface of it.

38. Let the circle ADB cut the circle AEB in the diameter AB at

any angle, C being their common center. Next let the plane perpendicular to AB cut the circumference of the circle ADB in D, F, and the circumference of AEB in E, G. Then E, D, G, F may be proved to be in the circumference of a circle.

39. Let AB, CD, EF be three lines meeting the surface and intersecting each other at right angles in the point G within a sphere whose centre is O. Join OG and produce it to meet the surface of the sphere in H, K; then HK is a diameter. From O draw OL, OM, ON perpendicular on AB, CD, EF respectively, then these three lines are bisected in L, M, N. Next draw OP perpendicular to the plane of AB, EF, and join PL; PL is perpendicular to the line AB; also in the same plane join PN; PN is also perpendicular to EF. Join also OA, OC, OF. Then Euc. II. 9, the squares on AG, BG, are equal to double the squares AL, LG. Similarly for the lines CD and EF; and by Euc. 1. 48, II. 5. Cor. it may be proved that the squares on AG, GB, ČG, GD, EG, GF, are together equal to the square on HK and twice the rectangle HG, GK. 40. Take a point A on the spherical surface of the fragment as a center, and with any radius AB describe a circle upon it. Take two other

points C, D in the circumference of this circle, and describe a plane triangle A'B'C' having its sides equal to the distances AB, BC, CA, respectively. Describe a circle about the triangle A'B'C', and draw the diameter A'D'; with centers A', D' and the radius equal to AB, describe circles intersecting each other in E', and through the points A', D', E' describe a circle; the diameter of this circle will be equal to that of the sphere of which the fragment is given.

41. All the sections may be proved to be equilateral triangles.

42. From the vertex A draw the line AE perpendicular on BCD the base of the tetrahedron, and from E draw the line EF perpendicular on the plane ABC; the angle between the perpendiculars is equal to the inclination of two planes of the tetrahedron. It will be found that in the triangle AEF, the side AE is three times EF. The inclination may also be found as in Prob. 21, p. 339.

43. The two lines drawn from two angles to bisect the opposite sides of the base of the tetrahedron, are at right angles to the sides of the triangular base.

44. Draw BO and produce it to meet DC in E. Then Euc. 1. 47. 45. First, let ABCD be a tetrahedron; bisect the opposite edges, AB in E, and CD in F; join EF, and prove EF perpendicular to AB, CD. Then conversely.

46. If FE be the shortest distance of the opposite sides AB, CD; join CE, DE, and shew that the square on EF is one-fourth of the square on CD. 47. First prove the direct proposition, then the converse of it.

48. Let ABCD be a tetrahedron and let the line EF joining the bisections E, F of the two opposite sides AB, CD, be bisected in G; the line AO drawn from the vertex A to the plane of the base BCD passes through G. Draw the necessary lines. Euc. VI. 4.

49. The joining lines in the theorem, are the lines joining the centers of the circles inscribed in the four faces of the given tetrahedron.

50. From the vertex A of a tetrahedron draw AO to the point O, the center of the circle which circumscribes the face BCD, and prove AO perpendicular to the plane BCD; then conversely.

61. Let ABCD be a regular tetrahedron. From A in the plane ABC draw AE perpendicular to BC, and join DE in the plane BCD, also from A draw AG perpendicular to the line DE. Then the angle AEG is the inclination of the two faces ABC, DBC of the tetrahedron, and the base EG is one-third of the hypotenuse AE in the right-angled triangle AGE.

Let a bedef be a regular octahedron whose faces are equal to those of the tetrahedron. Join af, two opposite vertices. Draw a g in the plane a b c perpendicular to b c, and g e perpendicular to af. Draw fg in the plane fbc, and from f draw fh perpendicular to a g produced.

Then a gf is the inclination of two faces of the octahedron. Also in the right-angled triangle fhg, g h may be proved to be one-third of fg, and fg is equal to AE. Hence the triangles fgh, AEF are equal in all respects. Therefore the angle f g h is equal to the angle AEB. Hence the angle AFF is the supplement of the angle agf, or the inclination of two contiguous faces of a tetrahedron, is the supplement of the inclination of two contiguous faces of an octahedron.

52. It may be shewn that the diameter of the sphere which circum. scribes a regular octahedron will be to an edge as the diagonal is to the side of a square.