Euclid.

# Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

. (page 33 of 38)
Font size the perpendicular.

29. Since the base and area are given, the altitude of the triangle is
known. Hence the problem is reduced to ; â€” 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. II. 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 Prob-
lem is contained in Theorem 3. p. 114.

34. This may be shewn from Euc. i. 47 ; ii- 5. Cor.

35. Erom C let fall CF pd-pendicular on AB. Then ACE is an ob-
tuse-angled, and BEC an acute-angled triangle. Apply Euc. ii. 12, 13 ,
and by Euc. i. 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. I. 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. i. 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 per-
pendicular to BC. Then by Euc. ii. 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 E ; draw also
EG perpendicular to the base. Then by Euc. i. 47 ; n. 5, Cor.

44. Bisect the angle B by BD meeting the opposite side in D, and
draw BE perpendicular to AC. Then by Euc. i. 47 ; ii. 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 bi-
sected 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

S26 GEOMETRICAL EXERCISES, &C.

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 tra-
pezium 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 pro-
position.

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. 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. i. 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 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 difierence, and the rectangle con~
tained by them is equal to the square on AC,

58. Apply Euc. ii. 14.

GEOMETRICAL EXERCISES ON BOOK III.
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 isos-
celes triangles, of which the parallel chords in the circle are the bases.

ON BOOK III. S21

n. The angles in equal segments are equal, and by Euc. i. 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. From the centre C ot 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. i. 27 ; iii. 28.

17. Take the center O, and join AP, AO, &c. and apply Euc. i. 20.

18. Draw any straight line intersecting two parallel chords and meet-
ing the circumference.

19. Produce the radii to meet the circumference.

20. Join AD, and the first equality follows directly from Euc. in.
20, T. 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. in. 20, p. 155.

21. Let DKE, DBO (fig. Euc. in. 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 in, 21.

23. This is the same as Euc. in. 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 con-
verse to Euc. ITT. 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 de-
scribed 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. in. 26, p. 156.

27. The segment must be described on the opposite side of the pro-
duced chord. By converse of Euc. in. 32.

28. If a circle be described upon the side AC as a diameter, the cir-
cumference will pass through the points D, E. Then Euc. in. 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 de-
scribed on AD will always be of the same magnitude, and the angle EAP
in it, is constant : â€” 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 ex-
tremity 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 f^ntained by
the tangents.

328 GEOMETRICAL EXERCISES, &C.

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.
m. 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 ct-nter. Then
CPD is a right-angled triangle, having the sum of the base PC and hypo-
tenuse 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 EFGII.
Join EO, OF, then EO and OF may be proved to be in the same straight
line, and similarly HO, OK.

Note. â€” This Proposition is equally true if AB, CD be any two chords
v.'hatever. It then becomes equivalent to the following proposition: â€”
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
ambles ; 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 fides, being E, II, 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. i. 29, and the note to iii. 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

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.

I

JN BOOK III. 329

45. It may be shewn that the point required is determined by a per-
pendicular drawn from the center of the circle on the <;ivcn 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. i. 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. liCt the tangents at E, F meet in a point K. Produce HE, RF
to meet the diameter AB produced in S, T. Then RST is a triangle,
and the quadrilateral RFOE maybe 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 re-
quired 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 extremi-
ties 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 pei-pendi-
cular to this line, cutting off a segment containing the given angle. The
circle described passing through one extremity of the chord and touch-
ing 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 tlie same point and terminated by the cir-
cumferences. The greatest line therefore depends on the distance be-
tween the centers ot the two circles.

330 GEOMETRICAL EXERCISES, &C.

59. Apply Euc. m. 27. i. 6.

60. Let two unequal circles cut one another, and let the line ABC
drawn throuj^h B, one of the points of intersection, be the line required,
such that AB is equal to BC. Join O, O' the centers of the circles, and
draw OP, O'P' perpendiculars on ABC, then PB is equal to BP' ; through
O' draw O'D parallel to PP' ; then ODO' is a right-angled triangle, and
a semicircle described on 00' as a diameter will pass through the point
D. Hence the synthesis. If the line ABC be supposed to move round
the point B and its extremities A, C to be in the extremities of the two
circles, it is manifest that ABC admits of a maximum.

61. Suppose the thing done, then it will appear that the line joining
the points of intersection of the two circles is bisected at right angles by
the line joining the centers of the circles. Since the radii are known,
the centers of the two circles may be determined,

62. Let the circles intersect in A, B ; and let CAD, EBF be any
parallels passing through A, B and intercepted by the circles. Join
CE, AB, DF. Then the figure CEFD may be proved to be a parallelo-
gram. Whence CAD is equal to EBF.

63. Complete the circle whose segment is ADB ; AHB being the
other part. Then since the angle ACB is constant, being in a given
segment, the sum of the arcs DE and AHB is constant. But AHB is
given, hence ED is also given and therefore constant.

64. From A suppose ACD drawn, so that when BD, BC are joined,
AD and DB shall together be double of AC and CB together. Then
the angles ACD, ADB are supplementary, and hence the angles BCD,
BDC are equal, and the triangle BCD is isosceles. Also the angles
BCD, BDC are given, hence the triangle BDC is given in species.

Again AD + DB = 2.AC + 2.BC, or CD = AC + BC.

Whence, make the triangle bdc having its angles at c?, c equal to that
in the segment BDA ; and make ca â€” cd â€” ch, and join ah. At A make

65. The line drawn from the point of intersection of the two lines
to the center of the given circle may be shewn to be constant, and the
center of the given circle is a fixed point.

66. This is at once obvious from Euc. iii. 36.

67. This follows directly from Euc. iii. 36.

68. Each of the lines CE, DF may be proved parallel to the common
chord AB.

69. By constructing the figure and joining AC and AD, by Euc.
III. 27, it may be proved that the line BC falls on BD.

70. By constructing the figure and applying Euc i. 8, 4, the truth
is manifest.

71. The bisecting line is a common chord to the two circles ; join the
other extremities of the chord and the diameter in each ch'cle, and the
angles in the two segments may be proved to be equal.

72. Apply Euc. m. 27 ; i. 32, 6.

73. Draw a common tangent at C the point of contact of the circles,
and prove AC and CB to be in the same straight line.

74. Let A, B, be the centers, and C the point of contact of the two
circles ; D, E the points of contact of the circles with the common tangent
DE, and CF a tangent common to the two circles at C, meeting DF in E.
Join DC, CE. Then DF, FC, FE may be shewn to be equal, and FC
to be at right angles to AB.

75. The line must be drawn to the extremities of the diameters which
are on opposite sides of the line joining the centers.

ON BOOK III. 331

76. The sum of the distances of the center of the third circle from
the centers of the two given circles, is equal to the sum of the radii of
the given circles, which is constant.

77. Let the circles touch at C either externally or internally, and
their diameters AC, BC through the point of contact will either coincide
or be in the same straight line. CDE any line through C will cut off
similar segments from the two circles. For joining AD, BE, the angles
in the segments DAC, EBC are proved to be equal.

The remaining segments are also similar, since they contain angles
which are supplementary to the angles DAC, EBC.

78. Let the line which joins the centers of the two circles be pro-
duced to meet the circumferences, and let the extremities of this line
and any other line from the point of contact be joined. From the center
of the larger circle draw perpendiculars on the sides of the right- angled
triangle inscribed within it.

79. Li general, the locus of a point in the circumference of a circle
which rolls within the circumference of another, is a curve called the
Hypocycloid ; but to this there is one exception, in which the radius of
one of the circles is double that of the other: in this case, the locus is
a straight line, as may be easily shewn from the figure.

80. Let A, B be the centers of the circles. Draw AB cutting the
circumferences in C, D. On AB take CE, DF each equal to the radius
of the required circle : the two circles described with centers A, B, and
radii AE, BF, respectively, will cut one another, and the point of inter-
section will be the center of the required circle.

81. Apply Euc. III. 31.

82. Apply Euc. iii. 21.

83. (1) When the tangent is on the same side of the two circles.
Join C, C their centers, and on CC describe a semicircle. With center
C and radius equal to the difference of the radii of the two circles, describe
another circle cutting the semicircle in D : join DC and produce it to
meet the circumference of the given circle in B. Through C draw CA
parallel to DB and join BA ; this line touches the two circles.

(2) When the tangent is on the alternate sides. Having joined C,
C ; on CC describe a semicircle ; with center C, and radius equal to the
sum of the radii of the two circles describe another circle cutting the
semicircle in D, join CD cutting the circumference in A, through C
draw CB parallel to CA and join AB.

84. The possibility is obvious. The point of bisection of the segment
intercepted between the convex circumferences will be the center of one
of the circles : and the center of a second circle will be found to be the
point of intersection of two circles described from the centers of the
given circles with their radii increased by the radius of the second circle.
The line passing through the centers of these two circles will be the locus
of the centers of all the circles which touch the two given circles.

85. At any points P, R in the circumferences of the circles, whose
centers are A, B, draw PQ, RS, tangents equal to the given lines, and
join AQ, BS. These being made the sides of a triangle of which AB
is the base, the vertex of the triangle is the point required.

86. In each circle draw a chord of the given length, describe circles
concentric with the given circles touching these chords, and then draw
a straight line touching these circles.

87. Within one of the circles draw a chord cutting off a segment
equal to the given segment, and describe a concentric circle touching
the chord : then draw a straight line touching this latter circle and the
other given circle.

8S2 GEOMETRICAL EXERCISES, &C.

88. The tangent may intersect the line joining the centers, or the line
produced. Prove that the angle in the segment of one circle is equal to
the angle in the corresponding segment of the other circle.

89. Join the centers A, B ; at C the point of contact draw a tangent,
and at A draw AF cutting the tangent in F, and making with OF an
angle equal to one-fourth of the given angle. From F draw tangents
to the circles.

90. Let C be the center of the given circle, and D the given point in
the given line AB. At D draw any line DE at right angles to AB, then
the center of the circle required is in the line AE. Through C draw a
diameter FG parallel to DE, the circle described passing through the
points E, F, G will be the circle required.

91. Apply Euc. III. 18.

92. Let A, B, be the two given points, and C the center of the given
circle. Join AC, and at C draw the diameter DCE perpendicular to AC,
and through the points A, D, E describe a circle, and produce AC to
meet the circumference in F. Bisect AF in G, and AB in H, and draw
GK, HK, perpendiculars to AF, AB respectively and intersecting in K.
Then K is the center of the circle which passes through the points A, B,
and bisects the circumference of the circle whose center is C.

93. Let D be the given point and EF the given straight line. (fig.
Euc. III. 32.) Draw DB to make the angle DBF equal to that contained
in the alternate segment. Draw BA at right angles to EF, and DA at
right angles to DB and meeting BA in A. Then AB is the diameter of
the circle.

94. Let A, B be the given points, and CD the given line. From E
the middle of the line AB, draw EM perpendicular to AB, meeting CD
in M, and draw MA. In EM take any point F ; draw FII to make the
given angle with CD ; and draw FG equal to FH, and meeting MA
produced in G. Through A draw AP parallel to FG, and CPK parallel
to FH. Then P is the center, and C the third defining point of the
circle required : and AP mav be proved equal to CP by means of the
triangles GMF, AMP; and HMF, CMP, Euc. vi. 2. Also CPK the
diameter makes with CD the angle KCD equal to FHD, that is, to the