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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. 1. 47.
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. 11. 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, II. 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; II. 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. 1. 47; II. 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. II. 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.
Bisect the angle B by BD meeting the opposite side in D, and draw BE perpendicular to AC. 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. II. 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, in
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. Let AD be parallel to BC in the figure ABCD. 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.
52. 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. II. 12, 13.
54. Convert the triangle into a rectangle, then Euc. I. 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. I. p. 113.
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 O 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.
GEOMETRICAL EXERCISES ON BOOK III.
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. It 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. From the centre C of the circle, draw CA, CB at right angles to each other meeting the circumference; join AB, and draw CD perpendicular to AB.
16. Join the extremities of the chords, then Euc. 1. 27; III. 28.
17. Take the center O, 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, 1. 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. 11. 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. 1. 15, and г. 21.
23. This is the same as Euc. 111. 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. 11. 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. III. 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, AC. 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 O, 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 intersection of two tangents to a circle, bisects the angle contained by the tangents.
33. Draw two radii containing an angle equal to the supplement of the given angle; the tangents drawn at the extremities of these radii will contain the given angle.
34. Since the circle is to touch two parallel lines drawn from two given points in a third line, the radius of the circle is determined by the distance between the two given points.
35. It is sufficient to suggest that the angle between a chord and a tangent is equal to the angle in the alternate segment of the circle. Euc. III. 32.
36. Let AB be the given chord of the circle whose center is O. Draw DE touching the circle at any point E and equal to the given line; join DO, and with center O and radius DO describe a circle: produce the chord AB to meet the circumference of this circle in F: then F is the point required.
37. Let D be the point required in the diameter BA produced, such that the tangent DP is half of DB. Join CP, C being the center. Then CPD is a right-angled triangle, having the sum of the base PC and hypotenuse CD double of the perpendicular PD.
38. If BE intersect DF in K (fig. Euc. III. 37). Join FB, FE, then by means of the triangles, BE is shewn to be bisected in K at right angles.
39. Let AB, CD be any two diameters of a circle, O the center, and let the tangents at their extremities form the quadrilateral figure EFGH. Join EO, OF, then EO and OF may be proved to be in the same straight line, and similarly HO, OK.
NOTE.-This Proposition is equally true if AB, CD be any two chords whatever. It then becomes equivalent to the following proposition :— The diagonals of the circumscribed and inscribed quadrilaterals, intersect in the same point, the points of contact of the former being the angles of the latter figure.
40. Let C be the point without the circle from which the tangents CA, CB are drawn, and let DE be any diameter, also let AE, BD_be joined, intersecting in P, then if CP be joined and produced to meet DE in G: CG is perpendicular to DE. Join DA, EB, and produce them to meet in F.
Then the angles DAE, EBD being angles in a semicircle, are right angles; or DB, EA are drawn perpendicular to the sides of the triangle DEF: whence the line drawn from F through P is perpendicular to the
third side DE.
41. Let the chord AB, of which P is its middle point, be produced both ways to C, D, so that AC is equal to BD. From C, D, draw the tangents to the circle forming the tangential quadrilateral CKDR, the points of contact of the sides, being E, H, F, G. Let O be the center of the circle. Join EH, GF, CO, GO, FO, DO. Then EH and GF may be proved each parallel to CD, they are therefore parallel to one another. Whence is proved that both EF and DG bisect AB.
42. This is obvious from Euc. 1. 29, and the note to г. 22. p. 156. 43. From any point A in the circumference, let any chord AB and tangent AC be drawn. Bisect the arc AB in D, and from D draw DE, DC perpendiculars on the chord AB and tangent AC. Join AD, the triangles ADE, ADC may be shewn to be equal.
44. Let A, B, be the given points. Join AB, and upon it describe a segment of a circle which shall contain an angle equal to the given angle. If the circle cut the given line, there will be two points; if it only touch the line, there will be one; and if it neither cut nor touch the line, the problem is impossible.
45. It may be shewn that the point required is determined by a perpendicular drawn from the center of the circle on the given line.
46. Let two lines AP, BP be drawn from the given points A, B, making equal angles with the tangent to the circle at the point of contact P, take any other point Q in the convex circumference, and join QA, QB: then by Prob. 4, p. 71, and Euc. 1. 21.
47. Let C be the center of the circle, and E the point of contact of DF with the circle. Join DC, CE, CF.
48. Let the tangents at E, F meet in a point R. Produce RE, RF to meet the diameter AB produced in S, T. Then RST is a triangle, and the quadrilateral RFOE may be circumscribed by a circle, and RPO may be proved to be one of the diagonals.
49. Let C be the middle point of the chord of contact: produce AC, BC to meet the circumference in B', A', and join AA', BB'.
50. Let A be the given point, and B the given point in the given line CD. At B draw BE at right angles to CD, join AB and bisect it in F, and from F draw FE perpendicular to AB and meeting BE in E. E is the center of the required circle.
51. Let O be the center of the given circle. Draw OA perpendicular to the given straight line; at O in OA make the angle AOP equal to the given angle, produce PO to meet the circumference again in Q. Then P, Q are two points from which tangents may be drawn fulfilling the required condition.
52. Let C be the center of the given circle, B the given point in the circumference, and A the other given point through which the required circle is to be made to pass. Join CB, the center of the circle is a point in CB produced. The center itself may be found in three ways.
53. Euc. III. 11 suggests the construction.
54. Let AB, AC be the two given lines which meet at A, and let D be the given point. Bisect the angle BAC by AE, the center of the circle is in AE. Through D draw DF perpendicular to AE, and produce DF to G, making FG equal to FD. Then DG is a chord of the circle, and the circle which passes through D and touches AB, will also pass through G and touch AC.
55. As the center is given, the line joining this point and center of the given circle, is perpendicular to that diameter, through the extremities of which the required circle is to pass.
56. Let AB be the given line and D the given point in it, through which the circle is required to pass, and AC the line which the circle is to touch. From D draw DE perpendicular to AB and meeting AC in C. Suppose O a point in AD to be the centre of the required circle. Draw OE perpendicular to AC, and join OC, then it may be shown that CO bisects the angle ACD.
57. Let the given circle be described. Draw a line through the center and intersection of the two lines. Next draw a chord perpendicular to this line, cutting off a segment containing the given angle. The circle described passing through one extremity of the chord and touching one of the straight lines, shall also pass through the other extremity of the chord and touch the other line.
58. The line drawn through the point of intersection of the two circles parallel to the line which joins their centers, may be shewn to be double of the line which joins their centers, and greater than any other straight line drawn through the same point and terminated by the circumferences. The greatest line therefore depends on the distance between the centers of the two circles.