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may be proved to have a certain ratio, and the triangles A.CB, CPM in the same way, may be proved to have the same ratio.
111. Let BA, AC be the bounding radii, and D a point in the arc of a quadrant. Bisect BAC by AE, and draw through D, the line HDGP perpendicular to AE at G, and meeting AB, AC, produced in H, P. From H draw HM to touch the circle of which BC is a quadrantal arc ; produce AH, making HL equal to HM, also on HA, take HK equal to HM. Then K, L, are the points of contact of two circles through D which touch the bounding radii, AB, AC.
Join DA. Then, since BAC is a right angle, AK is equal to the radius of the circle which touches BA, BC in K, K'; and similarly, AL is the radius of the circle which touches them in L, L'. Also, HAP being an isosceles triangle, and AD drawn to the base, AD® is shewn to be equal to AK. KL. Euc. 111. 36 ; 11. 5, Cor.
112. Let E, F, G be the centers of the circles inscribed in the triangles ABC, ADB, ACD. Draw EH, FK, GL perpendiculars on BC, BA, AC respectively, and join CE, EB; BF, FA; CĠ, GA. Then the relation between R, r,r, or EH, FK, GL may be found from the similar triangles, and the property of right-angled triangles.
113. The two hexagons consist each of six equilateral triangles, and the ratio of the hexagons is the same as the ratio of their equilateral triangles.
114. The area of the inscribed equilateral triangle may be proved to be equal to half of the inscribed hexagon, and the circumscribed triangle equal to four times the inscribed triangle.
115. The pentagons are similar figures, and can be divided into the same number of similar triangles. Euc. vi. 19.
116. Let the sides AB, BC, CA of the equilateral triangle ABC touch the circle in the points D, E, F, respectively. Draw AE cutting the circumference in G; and take o the center of the circle and draw OD: draw also HGK touching the circle in G. The property may then be shewn by the similar triangles AHG, AOD.
GEOMETRICAL EXERCISES ON BOOK XI.
3. Let AD, BE be two parallel straight lines, and let two planes ADFC, BEFC pass through AD, BE, and let CF be their common intersection, fig. Euc. XI, 10. Then CF may be proved parallel to BE and AD.
4. This theorem is analogous to Euc. XI. 8. Let two parallel lines AC, BD meet a plane in the points A, B. Take AC equal to BD and draw CE, DF, perpendiculars on the plane, and join AE, BF. Then the angles CAE, DBF, are the inclinations of AC, BD to the plane, Euc. XI. def. 5, and these angles may be proved to be equal.
5. Let AB, CD be parallel straight lines, and let perpendiculars be drawn from the extremities of AB, CD on any plane, and meet it in the points A', B', C', D'. Draw A'B', C'D'; these are the projections of AB, CD on the plane, and may be proved to be parallel.
6. Draw the figure, the proof offers no difficulty. 7. Let AB, AC drawn from the point A, and A'B', A'C' drawn from
the point A', in two parallel planes, make equal angles with a plane EF passing through AA', and perpendicular to the planes BAC, B'A'C'. Let AB in the plane ABC be parallel to A'B' in the plane A'B'C': then AC may be proved to be parallel to A'C'.
8. The plane must be drawn through the given line so that the plane and the other given line may be equally inclined to a third plane.
9. The required plane must be drawn through the given point so as to have the same inclination to a third plane, as the plane which passes through the two given lines.
10. From the point A let AB be drawn perpendicular to a plane, and AC perpendicular to a given line CD in a plane : join BC, then BC is at right angles to CD. For AB, BC, CD may be considered as three consecutive edges of a rectangular parallelopiped, and AC the diagonal of one face.
11. In the triangle BCD in which BE is drawn from the vertex to a point E in the base CD; it may be proved that the difference of the squares on the sides BC, BD is equal to the difference of the squares on the segments CE, ED of the base. By the converse of Theo. 149, p. 83.
12. Let BC be the common intersection of the two planes ABCD, EFGH which are inclined to each other at any angle. From K at any point in the plane ABCD, let KL be drawn perpendicular to the plane EFGH, and KM perpendicular to BC, the line of intersection of the two planes. Join LMT, and prove that the plane which passes through KL, KM is perpendicular to the line BC.
13. About the given line let a plane be made to revolve, till it passes through the given point. The perpendicular drawn in this plane from the given point upon the given line is the distance required.
14. Through any point in the first line draw a line parallel to the second ; the plane through these is parallel to the second line. Through the second line draw a plane perpendicular to the fore-named plane cutting the first line in a point. Through this point draw a perpendicular in the second plane to the first, and it will be perpendicular to both lines.
15. Through any point draw perpendiculars to both planes ; the plane passing through these two lines will fulfil the conditions required.
16. From the points where the lines meet the planes, draw two lines perpendicular to the intersection of the planes.
17. Let AB, AC in one of the planes make equal angles with DE the line of the intersection of the planes. Let AB be equal to AC. Draw BF, CG perpendiculars on the other plane, and draw FA, GA in that plane, and prove the angle BAF equal to the angle CAG.
18. If the intersecting plane be perpendicular to the three straight lines ; by joining the points of their intersection with the plane, the figure formed will be an equilateral triangle. If the plane be not perpendicular, the triangle will be isosceles.
19. Let the straight lines intersect in A, and let a plane be drawn cutting the three given lines in the points B, C, D, and the fourth in E.
20. This will appear from Euc. I. 19.
21. Let S be the proposed solid angle, in which the three plane angles ASB, ASC, BSC are known, it is required to find the angle contained by two of these planes, such as ASB, ASC. On a plane make the angles B'SA, ASC, B"SC equal to the angles BSA, ASC, BSC in the solid figure; take B'S and B”S each equal to BS in the solid figure; from the points B', and B" at right angles to SA and SC draw B'A and B"C, which will intersect each other at the point 0. From O as a center, with radius
AB' describe the semicircle B'WE ; at the point 0, erect Ob perpendicular to B'E and meeting the circumference in h; join Ab: the angle EAb will be the required inclination of the two planes ASC, ASB in the solid angle. (Legendre's Geometry, translated by Sir David Brewster, pp. 125, &c.)
22. Let ASC, ASB (same figure as in 21) be the two given plane angles ; and suppose for a moment that CSB"is the third angle required; then employing the same construction as in the foregoing problem, the angle included between the planes of the two first, the inclination of these planes would be EAb. Now as EAb can be determined by means of CSB", the other two being given, so likewise may CSB" be determined by means of EAb, which is just what the problem requires.
Having taken SB' at pleasure, upon SA let fall the indefinite perpendicular B’E; make the angle EAV equal to the inclination of the two given planes ; from the point b, where the side Ab meets the circle described from the center A with the radius AB', draw 60 perpendicular to AE; from the point 0, at right angles to SC draw the indefinite line OCB"; make SB” equal to SB'; the angle CSB" will be the third plane angle required. (Legendre's Geometry, translated by Sir David Brewster, pp. 127, &c.)
23. Let the three lines meet in the point A, and let a plane intersect them in the points B, C, D, so that AB, AC, AD are equal to one another. Describe a circle about the triangle BCD, and let O be the center; the line AO is perpendicular to the plane BCD.
24. This may be readily proved by Euc. xi. 17.
25. Construct the figure, and it will be found that the angle between the diagonal and one side of the cube measures the inclination of the two planes.
26. The diagonal plane of a cube is at right angles to two of the faces of the cube, and makes angles, each equal to half a right angle with the other four faces.
27. Let a rectangular parallelogram ABCD, be formed by four squares, each equal to a face of the given cube, and let EF, GH, KL, be the lines of division of the four squares. Let BD the diagonal of ABCD, cut EF in M; the square on BM to the square on AB is as 17 to 16. Let BG the diagonal of ABHG cut EF in N; the square on BN is to the square on AB, as 20 is to 16; hence there is some square between that on BM and BN which bears to the square on AB, the ratio of 18 to 16, or of 9 to 8.
The following addition may be easily proved. If six edges of a cube taken in order round the figure, be bisected, and the points of bisection be joined in succession, these six lines will form a regular hexagon.
28. From the six points out of the perpendicular, draw perpendiculars to the plane, and join the points where the perpendiculars meet the plane.
29. This is to shew that the square on the diagonal of a rectangular parallelopiped is equal to the sum of the squares on its three edges.
30. This theorem is analogous to the corresponding theorem respecting a rectangular parallelogram.
The axis of a parallelopiped must not be confounded with its diagonal,
31. Let the figure be described in a similar manner to that of Theorem 2, page 337: by employing Euc. 11, 12, 13, instead of Euc. I. 47, the truth of the theorem may be proved. .
32. Describe a circle passing through the three given points, and from the center draw a line perpendicular to its plane. Then every point in this perpendicular fulfils the conditions required.
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 X 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 Bʻ. 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. 1. 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'Č, 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.
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. xi. 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. XI. 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.