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

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ach other in O. Join AO, and produce it to meet BC in F. Then F is

he point where the inscribed circle touches the third side BC. If F be

lot the point of contact, let some other point G be the point of contact.

Through D draw DH parallel to AC, and DK parallel to BC. By the

imilar triangles, CG may be proved equal to CF, or G the point of con-

act coincides with F, the point where the line drawn from A through O

aeets BC.

q2

340 GEOMETRICAL EXERCISES, &C.

23. In the figure, Euc. iv. 5. Let AF bisect the angjle at A, and be

produced to meet the circumference in Gr. Join GB, GO and find the

center H of the circle inscribed in the triangle ABC. The lines GH, GB,

GC are equal to one another.

24. Let ABC be any triangle inscribed in a circle, and let the per-

pendiculars AD, BE, CF intersect in G. Produce AD to meet the cir-

cumference in H, and join BII, CH. Then the triangle BHC may be

shewn to be equal in all respects to the triangle BGC, and the circle

which circumscribes one of the triangles will also circumscribe the other.

Similarly may be shewn by producing BE and CF, &c.

25. First. Prove that the perpendiculars Aa, B6, Cc pass through

the same point O, as Theo. 112, p. 171. Secondly. That the triangles

Ac6, Bca, Cab are equiangular to ABC. Euc. iii. 21. Thirdly. That the

angles of the triangle abc are bisected by the perpendiculars ; and lastly,

by means of Prob. 4, p. 71, that ab + be + ca is a minimum.

26. The equilateral triangle can be proved to be the least triangle

which can be circumscribed about a circle.

27. Through C draw CH parallel to AB and join AH. Then HAC

the difference of the angles at the base is equal to the angle HFC. Euc.

III. 21, and HFC is bisected by FG.

28. LetF, G, (figure, Euc.iv. 5,) be the centers of the circumscribed

and inscribed circles ; join GF, GA, then the angle GAF which is equal to

the difference of the angles GAD, FAD, may be shewn to be equal to

half the difference of the angles ABC and ACB.

29. This Theorem may be stated more generally, as follows :

Let AB be the base of a triangle, AEB the locus of the vertex ; D the

bisection of the remaining arc AD B of the circumscribing circle ; then the

locus of the center of the inscribed circle is another circle whose center is

D and radius DB. For join CD : then P the center of the inscribed

circle is in CD. Join AP, PB ; then these lines bisect the angles CAB,

CBA, and DB, DP, DA may be proved to be equal to one another.

30. Let ABC be a triangle, having C a right angle, and upon AC, BC,

let semicircles be described : bisect the hypotenuse in D, and let fall DE,

DF perpendiculars on AC, BC respectively, and produce them to meet

the circumferences of the semicircles in P, Q ; then DP may be proved

to be equal to DQ.

31. Let the angle BAC be a right angle, fig. Euc. iv. 4. Join AD.

Then Euc. iii. 17, note p. ]o5.

32. Suppose the triangle constructed, then it may be shewn that the

difference between the hypotenuse and the sum of the two sides is equal

to the diameter of the inscribed circle.

33. Let P, Q be the middle points of the arcs AB, AC, and let PQ

be joined, cutting AB, AC in DE ; then AD is equal to AE. Find the

center O and join OP, QO.

34. With the given radius of the circumscribed circle, describe a

circle. Draw BC cutting ofi" the segment BAC containing an angle

equal to the given vertical angle. Bisect BC in D, and draw the diame-

ter EDF : join FB, and with center F and radius FB describe a circle:

this will be the locus of the centers of the inscribed circle (see Theorem

33, supra.) On DE take DG equal to the given radius of the inscribed

circle, and through G draw GH parallel to BC, and meeting the locus of

the centers in H. H is the center of the inscribed circle.

85. This may readily be effected in almost a similar way to the pre-

ceding Problem.

36. With the given radius describe a circle, then by Euc. iii. 34.

ON BOOK IV. 341

37. Let ABC be a triangle on the given base BC and having its ver-

tical angle A equal to the given angle. Then since the angle at A is

constant, A is a point in the arc of a segment of a circle described on EC.

Let D be the center of the circle inscribed in the triangle ABC. Join

DA, DB, DC: then the angles at B, C, A, are bisected. Euc. iv. 4.

Also since the angles of each of the triangles ABC, DBC are equal to two

right angles, it follows that the angle BDC is equal to the angle A and

half the sum of the angles B and C. But the sum of the angles B and C

can be found, because A is given. Hence the angle BDC is known, and

therefore D is the locus of the vertex of a triangle described on the base

BC and having its vertical angle at D double of the angle at A.

38. Suppose the parallelogram to be rectangular and inscribed in the

given triangle and to be equal in area to half the triangle : it may be

shewn that the parallelogram is equal to half the altitude of the triangle,

and that there is a restriction to the magnitude of the angle which two

adjacent sides of the parallelogram make with one another.

39. Let ABC be the given triangle, and A'B'C the other triangle, to

the sides of which the inscribed triangle is required to be parallel.

Through any point a in AB draw a^ parallel to A'B' one side of the given

triangle and through a, b draw ac, be respectively parallel to AC, BC.

Join Ac and produce it to meet BC in D ; through D draw DE, DF,

parallel to ca, c6, respectively, and join EF. Then DEF is the triangle

required.

40. This point will be found to be the intersection of the diagonals

of the given parallelogram.

41. The difference of the two squares is obviously the sum of the four

triangles at the corners of the exterior square.

42. (1) Let ABCD be the given square: join AC, at A in AC,

make the angles CAE, CAF, each equal to one-third of aright angle, and

join EF.

(2) Bisect AB any side in P, and draw PQ parallel to AD or BC,

then at P make the angles as in the former case.

43. Each of the interior angles of a regular octagon may be shewn to

be equal to three-fourths of two right angles, and the exterior angles

made by producing the sides, are each equal to one fourth of two right

angles, or one-half of a right angle.

44. Let the diagonals of the rhombus be drawn ; the center of the

inscribed circle may be shewn to be the point of their intersection.

45. Let ABCD be the required square. Join O, O' the centers of the

circles and draw the diagonal AEC cutting 00' in E. Then E is the

middle point of 00' and the angle AEO is half a right angle.

46. Let the squares be inscribed in, and circumscribed about a circle,

and let the diameters be drawn, the relation of the two squares is manifest.

47. Let one of the diagonals of the square be drawn, then the isos-

celes right-angled triangle which is half the square, may be proved to be

greater than any other right-angled triangle upon the same hypotenuse.

48. Take half of the side of the square inscribed in the given circle,

this will be equal to a side of the required octagon. At the extremities

on the same side of this line make two angles each equal to three-fourths

of two right angles, bisect these angles by two straight lines, the point

at which they meet will be the center of the circle which circumscribes

the octagon, and either of the bisecting lines is the radius of the circle.

49. First shew the possibility of a circle circumscribing such a figure,

and then determine the center of the circle.

50. By constructing the figures and drawing lines from the center of

84:2 GEOMETRICAL EXERCISES, &C.

the circle to the angles of the octagon, the areas of the eight triangles

may be easily shewn to be equal to eight times the rectangle contained

by the radius of the circle, and half the side of the inscribed square.

51. Let AB, AC, AD, be the sides of a square, a regular hexagon and

an octagon respectively inscribed in the circle whose center is O. Pro-

duce AC to E making AE equal to AB ; from E draw EF touching the

circle in F, and prove EF to be equal to AD.

52. Let the circle required touch the given circle in P, and the given

line in Q. Let C be the center of the given circle and C that of the re-

quired circle. Join CC, C'Q, QP ; and let QP produced meet the given

circle in R, join RC and produce it to meet the given line in V. Then

RCV is perpendicular to VQ. Hence the construction.

53. Let A, B be the centers of the given circles and CD the given

straight line. On the side of CD opposite to that on which the circles

are situated, draw a line EF parallel to CD at a distance equal to the

radius of the smaller circle. From A the center of the larger circle de-

scribe a concentric circle GH with radius equal to the difference of the

radii of the two circles. Then the center of the circle touching the

circle GH, the line EF, and passing through the center of the smaller

circle B, may be shewn to be the center of the circle w'hich touches the

circles whose centers are A, B, and the line CD.

51. Let AB, CD be the two lines given in position and E the center

of the given circle. Draw two lines FG, HI parallel to AB, CD respec-

tively and external to them. Describe a circle passing through E and

touching FG, HL Join the centers E, O, and with center O and radius

equal to the difference of the radii of these circles describe a circle ; this

will be the circle required.

55. Let the circle ACF having the center G, be the required circle

touching the given circle whose center is B, in the point A, and cutting

the other given circle in the point C. Join BG, and through A draw a

line perpendicular to BG ; then this line is a common tangent to the

circles whose centers are B, G. Join AC, GC. Hence the construction.

56. liCt C be the give^ point in the given straight line AB, and D

the center of the given circle. Through C draw a line CE perpendicular

to AB ; on the other side of AB, take CE equal to the radius of the given

circle. Draw ED, and at D make the angle EDF equal to the angle

DEC, and produce EC to meet DF. This gives the construction for one

case, when the given line does not cut or touch the other circle.

57. This is a particular case of the general problem ; To describe a

circle passing through a given point and touching two straight lines

given in position.

Let A be the given point between the two given lines which when

produced meet in the point B. Bisect the angle at B by BD and through

A draw AD perpendicular to BD and produce it to meet the two given

lines in C, E. Take DF equal to DA, and on CB take CG such that the

rectangle contained by CF, CA is equal to the square on CG. The circle

described through the points F, A, G, will be the circle required. De-

duce the particular case when the given lines are at right angles to one

another, and the given point in the line which bisects the angle at B. If

the lines are parallel, when is the solution possible ?

58. Let A, B, be the centers of the given circles, which touch

externally in E ; and let C be the given point in that whose center is B.

Make CD equal to AE and draw AD ; make the angle DAG equal to

the angle ADG : then G is the center of the circle required, and GC

its radius.

ON BOOK IV. S4.0

59. If the three points be such as when joined by straight lines a

triangle is formed ; the points at which the inscribed circle touches the

sides of the triangle, are the points at which the three circles touch one

another. Euc. iv. 4. Different cases arise from the relative position

of the three points.

60. Bisect the angle contained by the two lines at the point where

the bisecting line meets the circumference, draw a tangent to the circle

and produce the two straight lines to meet it. In this triangle inscribe

a circle.

61. From the given angle draw a line through the center of the circle,

and at the point where the line intersects the circumference, draw a

tangent to the circle, meeting two sides of the triangle. The circle

inscribed within this triangle will be the circle required.

62. Let the diagonal AD cut the arc in P, and let O be the center of

the inscribed circle. Draw OQ perpendicular to AB, Draw PE a

tangent at P meeting AB produced in E : then BE is equal to PD. Join

PQ, PB. Then AB may be proved equal to QE. Hence AQ is equal

to BE or DP.

63. Suppose the center of the required circle to be found, let fall

two perpendiculars from this point upon the radii of the quadrant,

and join the center of the circle with the center of the quadrant and

produce the line to meet the arc of the quadrant. If three tangents be

drawn at the three points thus determined in the two semicircles and

the arc of the quadrant, they form a right-angled triangle which

circumscribes the required circle.

64. Let AB be the base of the given segment, C its middle point.

Let DCE be the required triangle having the sum of the base DE and

perpendicular CF equal to the given line. Produce CF to H making

FH equal to DE. Join HD and produce it, if necessary, to meet AB

produced in K. Then CK is double ofDF. Draw DL perpendicular

toCK.

65. From the vertex of the isosceles triangle let fall a perpendicular

on the base. Then, in each of the triangles so formed, inscribe a circle,

Euc. IV. 4 ; next inscribe a circle so as to touch the two circles and the

two equal sides of the triangle. This gives one solution : the problem

is indeterminate.

C)Q. If BD be shewn to subtend an arc of the larger circle equal to

one-tenth of the whole circumference : â€” then BD is a side of the decagon

in the larger circle. And if the triangle ABD can be shewn to be

inscriptible in the smaller circle, BD will be the side of the inscribed

pentagon.

67. It may be shewn that the angles ABF, BFD stand on two arcs,

one of which is three times as large as the other.

68. It may be proved that the diagonals bisect the angles of the

pentagon , and the five-sided figure formed by their intersection, may be

shewn to be both equiangular and equilateral.

69. The figure ABODE is an irregular pentagon inscribed in a circle ;

it may be shewn that the five angles at the circumference stand upon

arcs whose sum is equal to the whole circumference of the circle ; Euc.

III. 20.

70. If a side CD (figure, Euc. iv. 11) of a regular pentagon be

produced to K, the exterior angle ADK of the inscribed quadrilateral

figure ABCD is equal to the angle ABC, one of the interior angles of the

pentagon. From this a construction may be made for the method of

folding the ribbon.

344 GEOMETRICAL EXERCISES ON BOOK IV.

71. In the figure, Euc. iv. 10, let DC be produced to meet the circum-

ference in F, and join FB. Then FB is the side of a regular pentagon

inscribed in the larger circle, D is the middle of the arc subtended by

the adjacent side of the pentagon. Then the diflference of F.D and Bl)

is equal to the radius AB. Next, it may be shewn, that FD is divided

in the same manner in C as AB, and by Euc. ii. 4, 11, the squares on

FD and DB are three times the square on AB, and the rectangle of FD

and DB is equal to the square on AB.

72. If one of the diagonals be drawn, this line with three sides of the

pentagon forms a quadrilateral figure of which three consecutive sides

are equal. The problem is reduced to the inscription of a quadrilateral

in a square.

73. This may be deduced from Euc. iv. 11.

74. The angle at A the center of the circle (fig. Euc. iv. 10.) is one-

tenth of four right angles, the arc BD is therefore one-tenth of the

circumference, and the chord BD is the side of a regular decagon

inscribed in the larger circle. Produce DC to meet the circumference

in F and join BF, then BF is the side of the inscribed pentagon, and AB

is the side of the mscribed hexagon. Join FA. Then FCA may be

proved to be an isosceles triangle and FB is a line drawn from the

vertex meeting the base produced. If a perpendicular be drawn from

F on BC, the difference of the squares on FB, FC may be shewn to be

equal to the rectangle AB, BC, (Euc. i. 47 ; ii. o. Cor.) ; or the square

on AC.

75. Divide the circle into three equal sectors, and draw tangents to

the middle points of the arcs, the problem is then reduced to the

inscription of a circle in a triangle.

76. Let the inscribed circles whose centers are A, B touch each

other in G, and the circle whose center is C, in the points D, E ; join

A, D ; A, E ; at D, draw DF perpendicular to DA, and EF to EB,

meeting in F. Let F, G be joined, and FG be proved to touch the two

circles in G whose centers are A and B.

77. The problem is the same as to find how many equal circles may

be placed round a circle of the same radius, touching this circle and

each other. The number is six.

78. This is obvious from Euc. iv. 7, the side of a square circum-

scribing a circle being equal to the diameter of the circle.

79. Each of the vertical angles of the triangles so formed, may be

proved to be equal to the difi'erence between the exterior and interior

angle of the heptagon.

80. Every regular polygon can be divided into equal isosceles tri-

angles by drawing lines from the center of the inscribed or circumscribed

circle to the angular points of the figure, and the number of triangles

will be equal to the number of sides of the polygon. If a perpendicular

FG be let fall from F (figure, Euc. iv. 14) the center on the base CD of

FCD, one of these triangles, and if GF be produced to H till FH be

equal to FG, and HC, HD be joined, an isosceles triangle is formed,

such that the angle at H is half "the angle at F. Bisect HC, HD in K,

L, and join KL; then the triangle HKL may be placed round the

vertex H, twice as many times as the triangle CFD round the vertex F.

81. The sum of the arcs on which stand the 1st, 3rd, oth, &c. angles,

is equal to the sum of the arcs on which stand the 2nd, 4th, 6th, &c.

angles.

82. The proof of this property depends on the fact, that an isosceles

triangle has a greater area than any scalene triangle of the same perimeter.

I

GEOMETRICAL EXERCISES ON BOOK VI.

HINTS. &c.

6. In the figure Euc. vi. 23, let the parallelograms be supposed to be

rectangular.

Then the rectangle AC : the rectangle DG :: BC : CG, Euc. vi. 1.

and the rectangle DG : the rectangle CF :: CD : EC,

whence the rectangle AC : the rectangle CF : : BC . CD : CG . EC.

In a similar way it may be shewn that the ratio of any two parallelo-

grams is as the ratio compounded of the ratios of their bases and altitudes

7. Let two sides intersect in O, through O draw POQ parallel to

the base AB. Then by similar triangles, PO may be proved equal to

OQ : and POFA, QOEB, are parallelograms : whence AE is equal

toFB.

8. Apply Euc. VI. 4, v. 7.

9. Let ABC be a scalene triangle, having the vertical angle A, and

suppose ADE an equivalent isosceles triangle, of which the side AD is

equal to AE. Then Euc. vi. 15, 16, AC.AB=AD.AE, or ADÂ«,

Hence AD is a mean proportional between AC, AB. Euc. vi. 8.

10. The lines drawn making equal angles with homologous sides,

divide the triangles into two corresponding pairs of equiangular triangles ;

by Euc. VI. 4, the proportions are evident.

11. By constructing the figure, the angles of the two triangles may

easily be shewn to be respectively equal.

12. A circle may be described about the four-sided figure ABDC.

By Euc. I. 13 ; Euc. iii. 21, 22. The triangles ABC, ACE may be

shewn to be equiangular.

13. Apply Euc. I. 48 ; ii. 5. Cor., vi. 16.

14. This property follows as a corollary to Euc. vi. 23 : for the two

triangles are respectively the halves of the parallelograms, and are

therefore in the ratio compounded of the ratios of the sides which contain

the same or equal angles : and this ratio is the same as the ratio of the

rectangles by the sides.

15. Let ABC be the given triangle, and let the line EGF cut the

base BC in G. Join AG. Then by Euc. vi. 1, and the preceding

theorem (14,) it may be proved that AC is to AB as GE is to GF.

16. The two means and the two extremes form an arithmetic series

of four lines whose successive differences are equal ; the difference therefore

between the first and the fourth, or the extremes, is treble the difference

between the first and the second.

17. This may be effected in different ways, one of which is the

following. At one extremity A of the given line AB draw AC making

any acute angle with AB and join BC ; at any point D in BC draw DEF

paialiel to AC cutting AB in E and such that EF is equal to ED, draw

FC cutting AB in G. Then AB is harmonically divided in E, G.

18. In the figure Euc. vi. 13. If E be the middle point of AC ; then

AE or EC is the arithmetic mean, and DB is the geometric mean, between

AB and BC. If DE be joined and BF be drawn perpendicular on DE ;

then DF may be proved to be the harmonic mean between AB and BC.

19. In the fig. Euc. vi. 13. DB is the geometric mean between AB

and BC, and if AC be bisected in E, AE or EC is the Arithmetic mean.

The next is the same as â€” To find the segments of the hypotenuse of a

right-angled triangle made by a perpendicular from the right angle,

q5

346 GEOMETRICAL EXERCISES, &C.

having given the diflFerence between half the hypotenuse and the

perpendicular.

20. Let the line DF drawn from D the bisection of the base of the

triangle ABC, meet AB in E, and CA produced in F. Also let AG

drawn parallel to BC from the vertex A, meet DF in G. Then by means

of the similar triangles ; DF, FE, FG, may be shewn to be in harmonic

progression.

21. If a triangle be constructed on AB so that the vertical angle is

bisected by the line drawn to the point C. By Euc. vi. A, the point

required may be determined.

22. Let DB, DE, DCA be the three straight lines, fig. Euc. iii. 37 ;

let the points of contact B, E be joined by the straight line BC cutting

DA in G. Then BDE is an isosceles triangle, and DG is a line from the

vertex to a point G in the base. And two values of the square of BD

may be found, one from Theo. 37, p. 118: Euc. iii. 35; ii. 2; and

another from Euc. iii. 36; ii. 1. From these may be deduced, that

the rectangle DC, GA, is equal to the rectangle AD, CG. Whence

the, &c.

23. Let ABCD be a square and AC its diagonal. On AC take AE

equal to the side BC or AB : join BE and at E draw EF perpendicular

to AC and meeting BC in F. Then EC, the difference between the

diagonal AC and the side AB of the square, is less than AB ; and CE,

EF, FB may be proved to be equal to one another : also CE, EF are the

adjacent sides of a square whose diagonal is FC. On FC take FG equal

to CE and join EG. Then, as in the first square, the difference CG

between the diagonal FC and the side EC or EF, is less than the side EC.

Hence EC, the difference between the cUagonal and the side of the given

square, is contained twice in the side BC with a remainder CG : and CG

is the difference between the side CE and the diagonal CF of another

square. By proceeding in a similar way, CG, the difference between the

diagonal CF and the side CE, is contained twice m the side CE with a

remainder : and the same relations may be shewn to exist between the

difference of the diagonal and the side of every square of the series which

is so constructed. Hence, therefore, as the difference of the side and

diagonal of every square of the series is contained twice in the side with

a remainder, it follows that there is no line which exactly measures the

side and the diagonal of a square.

24. Let the given line AB be divided in C, D. On AD describe a

semicircle, and on CB describe another semicircle intersecting the former

in P ; draw PE perpendicular to AB ; then E is the point required.

25. Let AB be equal to a side of the given square. On AB describe

a semicircle ; at A draw AC perpendicular to AB and equal to a fourth

proportional to AB and the two sides of the given rectangle. Draw CD

parallel to AB meeting the circumference in D. Join AD, BD, which

are the required lines.

26. Let the two given lines meet when produced in A. At A draw

AD perpendicular to AB, and AE to AC, and such that AD is to AE in

the given ratio. Through D, E, draw DF, EF, respectively parallel to

AB, AC and meeting each other in F. Join AF and produce it, and

the perpendiculars drawn from any point of this line on the two given

lines will always be in the given ratio.

27. The angles made by the four lines at the point of their divergence,

remain constant. See Note on Euc. vi. A, p. 295.

28. Let AB be the given line from which it is required to cut off a

part BC such that BC shall be a mean proportional between the

remainder AC and another given line. Produce AB to D, making BD

ON BOOK VI.

Sil

he point where the inscribed circle touches the third side BC. If F be

lot the point of contact, let some other point G be the point of contact.

Through D draw DH parallel to AC, and DK parallel to BC. By the

imilar triangles, CG may be proved equal to CF, or G the point of con-

act coincides with F, the point where the line drawn from A through O

aeets BC.

q2

340 GEOMETRICAL EXERCISES, &C.

23. In the figure, Euc. iv. 5. Let AF bisect the angjle at A, and be

produced to meet the circumference in Gr. Join GB, GO and find the

center H of the circle inscribed in the triangle ABC. The lines GH, GB,

GC are equal to one another.

24. Let ABC be any triangle inscribed in a circle, and let the per-

pendiculars AD, BE, CF intersect in G. Produce AD to meet the cir-

cumference in H, and join BII, CH. Then the triangle BHC may be

shewn to be equal in all respects to the triangle BGC, and the circle

which circumscribes one of the triangles will also circumscribe the other.

Similarly may be shewn by producing BE and CF, &c.

25. First. Prove that the perpendiculars Aa, B6, Cc pass through

the same point O, as Theo. 112, p. 171. Secondly. That the triangles

Ac6, Bca, Cab are equiangular to ABC. Euc. iii. 21. Thirdly. That the

angles of the triangle abc are bisected by the perpendiculars ; and lastly,

by means of Prob. 4, p. 71, that ab + be + ca is a minimum.

26. The equilateral triangle can be proved to be the least triangle

which can be circumscribed about a circle.

27. Through C draw CH parallel to AB and join AH. Then HAC

the difference of the angles at the base is equal to the angle HFC. Euc.

III. 21, and HFC is bisected by FG.

28. LetF, G, (figure, Euc.iv. 5,) be the centers of the circumscribed

and inscribed circles ; join GF, GA, then the angle GAF which is equal to

the difference of the angles GAD, FAD, may be shewn to be equal to

half the difference of the angles ABC and ACB.

29. This Theorem may be stated more generally, as follows :

Let AB be the base of a triangle, AEB the locus of the vertex ; D the

bisection of the remaining arc AD B of the circumscribing circle ; then the

locus of the center of the inscribed circle is another circle whose center is

D and radius DB. For join CD : then P the center of the inscribed

circle is in CD. Join AP, PB ; then these lines bisect the angles CAB,

CBA, and DB, DP, DA may be proved to be equal to one another.

30. Let ABC be a triangle, having C a right angle, and upon AC, BC,

let semicircles be described : bisect the hypotenuse in D, and let fall DE,

DF perpendiculars on AC, BC respectively, and produce them to meet

the circumferences of the semicircles in P, Q ; then DP may be proved

to be equal to DQ.

31. Let the angle BAC be a right angle, fig. Euc. iv. 4. Join AD.

Then Euc. iii. 17, note p. ]o5.

32. Suppose the triangle constructed, then it may be shewn that the

difference between the hypotenuse and the sum of the two sides is equal

to the diameter of the inscribed circle.

33. Let P, Q be the middle points of the arcs AB, AC, and let PQ

be joined, cutting AB, AC in DE ; then AD is equal to AE. Find the

center O and join OP, QO.

34. With the given radius of the circumscribed circle, describe a

circle. Draw BC cutting ofi" the segment BAC containing an angle

equal to the given vertical angle. Bisect BC in D, and draw the diame-

ter EDF : join FB, and with center F and radius FB describe a circle:

this will be the locus of the centers of the inscribed circle (see Theorem

33, supra.) On DE take DG equal to the given radius of the inscribed

circle, and through G draw GH parallel to BC, and meeting the locus of

the centers in H. H is the center of the inscribed circle.

85. This may readily be effected in almost a similar way to the pre-

ceding Problem.

36. With the given radius describe a circle, then by Euc. iii. 34.

ON BOOK IV. 341

37. Let ABC be a triangle on the given base BC and having its ver-

tical angle A equal to the given angle. Then since the angle at A is

constant, A is a point in the arc of a segment of a circle described on EC.

Let D be the center of the circle inscribed in the triangle ABC. Join

DA, DB, DC: then the angles at B, C, A, are bisected. Euc. iv. 4.

Also since the angles of each of the triangles ABC, DBC are equal to two

right angles, it follows that the angle BDC is equal to the angle A and

half the sum of the angles B and C. But the sum of the angles B and C

can be found, because A is given. Hence the angle BDC is known, and

therefore D is the locus of the vertex of a triangle described on the base

BC and having its vertical angle at D double of the angle at A.

38. Suppose the parallelogram to be rectangular and inscribed in the

given triangle and to be equal in area to half the triangle : it may be

shewn that the parallelogram is equal to half the altitude of the triangle,

and that there is a restriction to the magnitude of the angle which two

adjacent sides of the parallelogram make with one another.

39. Let ABC be the given triangle, and A'B'C the other triangle, to

the sides of which the inscribed triangle is required to be parallel.

Through any point a in AB draw a^ parallel to A'B' one side of the given

triangle and through a, b draw ac, be respectively parallel to AC, BC.

Join Ac and produce it to meet BC in D ; through D draw DE, DF,

parallel to ca, c6, respectively, and join EF. Then DEF is the triangle

required.

40. This point will be found to be the intersection of the diagonals

of the given parallelogram.

41. The difference of the two squares is obviously the sum of the four

triangles at the corners of the exterior square.

42. (1) Let ABCD be the given square: join AC, at A in AC,

make the angles CAE, CAF, each equal to one-third of aright angle, and

join EF.

(2) Bisect AB any side in P, and draw PQ parallel to AD or BC,

then at P make the angles as in the former case.

43. Each of the interior angles of a regular octagon may be shewn to

be equal to three-fourths of two right angles, and the exterior angles

made by producing the sides, are each equal to one fourth of two right

angles, or one-half of a right angle.

44. Let the diagonals of the rhombus be drawn ; the center of the

inscribed circle may be shewn to be the point of their intersection.

45. Let ABCD be the required square. Join O, O' the centers of the

circles and draw the diagonal AEC cutting 00' in E. Then E is the

middle point of 00' and the angle AEO is half a right angle.

46. Let the squares be inscribed in, and circumscribed about a circle,

and let the diameters be drawn, the relation of the two squares is manifest.

47. Let one of the diagonals of the square be drawn, then the isos-

celes right-angled triangle which is half the square, may be proved to be

greater than any other right-angled triangle upon the same hypotenuse.

48. Take half of the side of the square inscribed in the given circle,

this will be equal to a side of the required octagon. At the extremities

on the same side of this line make two angles each equal to three-fourths

of two right angles, bisect these angles by two straight lines, the point

at which they meet will be the center of the circle which circumscribes

the octagon, and either of the bisecting lines is the radius of the circle.

49. First shew the possibility of a circle circumscribing such a figure,

and then determine the center of the circle.

50. By constructing the figures and drawing lines from the center of

84:2 GEOMETRICAL EXERCISES, &C.

the circle to the angles of the octagon, the areas of the eight triangles

may be easily shewn to be equal to eight times the rectangle contained

by the radius of the circle, and half the side of the inscribed square.

51. Let AB, AC, AD, be the sides of a square, a regular hexagon and

an octagon respectively inscribed in the circle whose center is O. Pro-

duce AC to E making AE equal to AB ; from E draw EF touching the

circle in F, and prove EF to be equal to AD.

52. Let the circle required touch the given circle in P, and the given

line in Q. Let C be the center of the given circle and C that of the re-

quired circle. Join CC, C'Q, QP ; and let QP produced meet the given

circle in R, join RC and produce it to meet the given line in V. Then

RCV is perpendicular to VQ. Hence the construction.

53. Let A, B be the centers of the given circles and CD the given

straight line. On the side of CD opposite to that on which the circles

are situated, draw a line EF parallel to CD at a distance equal to the

radius of the smaller circle. From A the center of the larger circle de-

scribe a concentric circle GH with radius equal to the difference of the

radii of the two circles. Then the center of the circle touching the

circle GH, the line EF, and passing through the center of the smaller

circle B, may be shewn to be the center of the circle w'hich touches the

circles whose centers are A, B, and the line CD.

51. Let AB, CD be the two lines given in position and E the center

of the given circle. Draw two lines FG, HI parallel to AB, CD respec-

tively and external to them. Describe a circle passing through E and

touching FG, HL Join the centers E, O, and with center O and radius

equal to the difference of the radii of these circles describe a circle ; this

will be the circle required.

55. Let the circle ACF having the center G, be the required circle

touching the given circle whose center is B, in the point A, and cutting

the other given circle in the point C. Join BG, and through A draw a

line perpendicular to BG ; then this line is a common tangent to the

circles whose centers are B, G. Join AC, GC. Hence the construction.

56. liCt C be the give^ point in the given straight line AB, and D

the center of the given circle. Through C draw a line CE perpendicular

to AB ; on the other side of AB, take CE equal to the radius of the given

circle. Draw ED, and at D make the angle EDF equal to the angle

DEC, and produce EC to meet DF. This gives the construction for one

case, when the given line does not cut or touch the other circle.

57. This is a particular case of the general problem ; To describe a

circle passing through a given point and touching two straight lines

given in position.

Let A be the given point between the two given lines which when

produced meet in the point B. Bisect the angle at B by BD and through

A draw AD perpendicular to BD and produce it to meet the two given

lines in C, E. Take DF equal to DA, and on CB take CG such that the

rectangle contained by CF, CA is equal to the square on CG. The circle

described through the points F, A, G, will be the circle required. De-

duce the particular case when the given lines are at right angles to one

another, and the given point in the line which bisects the angle at B. If

the lines are parallel, when is the solution possible ?

58. Let A, B, be the centers of the given circles, which touch

externally in E ; and let C be the given point in that whose center is B.

Make CD equal to AE and draw AD ; make the angle DAG equal to

the angle ADG : then G is the center of the circle required, and GC

its radius.

ON BOOK IV. S4.0

59. If the three points be such as when joined by straight lines a

triangle is formed ; the points at which the inscribed circle touches the

sides of the triangle, are the points at which the three circles touch one

another. Euc. iv. 4. Different cases arise from the relative position

of the three points.

60. Bisect the angle contained by the two lines at the point where

the bisecting line meets the circumference, draw a tangent to the circle

and produce the two straight lines to meet it. In this triangle inscribe

a circle.

61. From the given angle draw a line through the center of the circle,

and at the point where the line intersects the circumference, draw a

tangent to the circle, meeting two sides of the triangle. The circle

inscribed within this triangle will be the circle required.

62. Let the diagonal AD cut the arc in P, and let O be the center of

the inscribed circle. Draw OQ perpendicular to AB, Draw PE a

tangent at P meeting AB produced in E : then BE is equal to PD. Join

PQ, PB. Then AB may be proved equal to QE. Hence AQ is equal

to BE or DP.

63. Suppose the center of the required circle to be found, let fall

two perpendiculars from this point upon the radii of the quadrant,

and join the center of the circle with the center of the quadrant and

produce the line to meet the arc of the quadrant. If three tangents be

drawn at the three points thus determined in the two semicircles and

the arc of the quadrant, they form a right-angled triangle which

circumscribes the required circle.

64. Let AB be the base of the given segment, C its middle point.

Let DCE be the required triangle having the sum of the base DE and

perpendicular CF equal to the given line. Produce CF to H making

FH equal to DE. Join HD and produce it, if necessary, to meet AB

produced in K. Then CK is double ofDF. Draw DL perpendicular

toCK.

65. From the vertex of the isosceles triangle let fall a perpendicular

on the base. Then, in each of the triangles so formed, inscribe a circle,

Euc. IV. 4 ; next inscribe a circle so as to touch the two circles and the

two equal sides of the triangle. This gives one solution : the problem

is indeterminate.

C)Q. If BD be shewn to subtend an arc of the larger circle equal to

one-tenth of the whole circumference : â€” then BD is a side of the decagon

in the larger circle. And if the triangle ABD can be shewn to be

inscriptible in the smaller circle, BD will be the side of the inscribed

pentagon.

67. It may be shewn that the angles ABF, BFD stand on two arcs,

one of which is three times as large as the other.

68. It may be proved that the diagonals bisect the angles of the

pentagon , and the five-sided figure formed by their intersection, may be

shewn to be both equiangular and equilateral.

69. The figure ABODE is an irregular pentagon inscribed in a circle ;

it may be shewn that the five angles at the circumference stand upon

arcs whose sum is equal to the whole circumference of the circle ; Euc.

III. 20.

70. If a side CD (figure, Euc. iv. 11) of a regular pentagon be

produced to K, the exterior angle ADK of the inscribed quadrilateral

figure ABCD is equal to the angle ABC, one of the interior angles of the

pentagon. From this a construction may be made for the method of

folding the ribbon.

344 GEOMETRICAL EXERCISES ON BOOK IV.

71. In the figure, Euc. iv. 10, let DC be produced to meet the circum-

ference in F, and join FB. Then FB is the side of a regular pentagon

inscribed in the larger circle, D is the middle of the arc subtended by

the adjacent side of the pentagon. Then the diflference of F.D and Bl)

is equal to the radius AB. Next, it may be shewn, that FD is divided

in the same manner in C as AB, and by Euc. ii. 4, 11, the squares on

FD and DB are three times the square on AB, and the rectangle of FD

and DB is equal to the square on AB.

72. If one of the diagonals be drawn, this line with three sides of the

pentagon forms a quadrilateral figure of which three consecutive sides

are equal. The problem is reduced to the inscription of a quadrilateral

in a square.

73. This may be deduced from Euc. iv. 11.

74. The angle at A the center of the circle (fig. Euc. iv. 10.) is one-

tenth of four right angles, the arc BD is therefore one-tenth of the

circumference, and the chord BD is the side of a regular decagon

inscribed in the larger circle. Produce DC to meet the circumference

in F and join BF, then BF is the side of the inscribed pentagon, and AB

is the side of the mscribed hexagon. Join FA. Then FCA may be

proved to be an isosceles triangle and FB is a line drawn from the

vertex meeting the base produced. If a perpendicular be drawn from

F on BC, the difference of the squares on FB, FC may be shewn to be

equal to the rectangle AB, BC, (Euc. i. 47 ; ii. o. Cor.) ; or the square

on AC.

75. Divide the circle into three equal sectors, and draw tangents to

the middle points of the arcs, the problem is then reduced to the

inscription of a circle in a triangle.

76. Let the inscribed circles whose centers are A, B touch each

other in G, and the circle whose center is C, in the points D, E ; join

A, D ; A, E ; at D, draw DF perpendicular to DA, and EF to EB,

meeting in F. Let F, G be joined, and FG be proved to touch the two

circles in G whose centers are A and B.

77. The problem is the same as to find how many equal circles may

be placed round a circle of the same radius, touching this circle and

each other. The number is six.

78. This is obvious from Euc. iv. 7, the side of a square circum-

scribing a circle being equal to the diameter of the circle.

79. Each of the vertical angles of the triangles so formed, may be

proved to be equal to the difi'erence between the exterior and interior

angle of the heptagon.

80. Every regular polygon can be divided into equal isosceles tri-

angles by drawing lines from the center of the inscribed or circumscribed

circle to the angular points of the figure, and the number of triangles

will be equal to the number of sides of the polygon. If a perpendicular

FG be let fall from F (figure, Euc. iv. 14) the center on the base CD of

FCD, one of these triangles, and if GF be produced to H till FH be

equal to FG, and HC, HD be joined, an isosceles triangle is formed,

such that the angle at H is half "the angle at F. Bisect HC, HD in K,

L, and join KL; then the triangle HKL may be placed round the

vertex H, twice as many times as the triangle CFD round the vertex F.

81. The sum of the arcs on which stand the 1st, 3rd, oth, &c. angles,

is equal to the sum of the arcs on which stand the 2nd, 4th, 6th, &c.

angles.

82. The proof of this property depends on the fact, that an isosceles

triangle has a greater area than any scalene triangle of the same perimeter.

I

GEOMETRICAL EXERCISES ON BOOK VI.

HINTS. &c.

6. In the figure Euc. vi. 23, let the parallelograms be supposed to be

rectangular.

Then the rectangle AC : the rectangle DG :: BC : CG, Euc. vi. 1.

and the rectangle DG : the rectangle CF :: CD : EC,

whence the rectangle AC : the rectangle CF : : BC . CD : CG . EC.

In a similar way it may be shewn that the ratio of any two parallelo-

grams is as the ratio compounded of the ratios of their bases and altitudes

7. Let two sides intersect in O, through O draw POQ parallel to

the base AB. Then by similar triangles, PO may be proved equal to

OQ : and POFA, QOEB, are parallelograms : whence AE is equal

toFB.

8. Apply Euc. VI. 4, v. 7.

9. Let ABC be a scalene triangle, having the vertical angle A, and

suppose ADE an equivalent isosceles triangle, of which the side AD is

equal to AE. Then Euc. vi. 15, 16, AC.AB=AD.AE, or ADÂ«,

Hence AD is a mean proportional between AC, AB. Euc. vi. 8.

10. The lines drawn making equal angles with homologous sides,

divide the triangles into two corresponding pairs of equiangular triangles ;

by Euc. VI. 4, the proportions are evident.

11. By constructing the figure, the angles of the two triangles may

easily be shewn to be respectively equal.

12. A circle may be described about the four-sided figure ABDC.

By Euc. I. 13 ; Euc. iii. 21, 22. The triangles ABC, ACE may be

shewn to be equiangular.

13. Apply Euc. I. 48 ; ii. 5. Cor., vi. 16.

14. This property follows as a corollary to Euc. vi. 23 : for the two

triangles are respectively the halves of the parallelograms, and are

therefore in the ratio compounded of the ratios of the sides which contain

the same or equal angles : and this ratio is the same as the ratio of the

rectangles by the sides.

15. Let ABC be the given triangle, and let the line EGF cut the

base BC in G. Join AG. Then by Euc. vi. 1, and the preceding

theorem (14,) it may be proved that AC is to AB as GE is to GF.

16. The two means and the two extremes form an arithmetic series

of four lines whose successive differences are equal ; the difference therefore

between the first and the fourth, or the extremes, is treble the difference

between the first and the second.

17. This may be effected in different ways, one of which is the

following. At one extremity A of the given line AB draw AC making

any acute angle with AB and join BC ; at any point D in BC draw DEF

paialiel to AC cutting AB in E and such that EF is equal to ED, draw

FC cutting AB in G. Then AB is harmonically divided in E, G.

18. In the figure Euc. vi. 13. If E be the middle point of AC ; then

AE or EC is the arithmetic mean, and DB is the geometric mean, between

AB and BC. If DE be joined and BF be drawn perpendicular on DE ;

then DF may be proved to be the harmonic mean between AB and BC.

19. In the fig. Euc. vi. 13. DB is the geometric mean between AB

and BC, and if AC be bisected in E, AE or EC is the Arithmetic mean.

The next is the same as â€” To find the segments of the hypotenuse of a

right-angled triangle made by a perpendicular from the right angle,

q5

346 GEOMETRICAL EXERCISES, &C.

having given the diflFerence between half the hypotenuse and the

perpendicular.

20. Let the line DF drawn from D the bisection of the base of the

triangle ABC, meet AB in E, and CA produced in F. Also let AG

drawn parallel to BC from the vertex A, meet DF in G. Then by means

of the similar triangles ; DF, FE, FG, may be shewn to be in harmonic

progression.

21. If a triangle be constructed on AB so that the vertical angle is

bisected by the line drawn to the point C. By Euc. vi. A, the point

required may be determined.

22. Let DB, DE, DCA be the three straight lines, fig. Euc. iii. 37 ;

let the points of contact B, E be joined by the straight line BC cutting

DA in G. Then BDE is an isosceles triangle, and DG is a line from the

vertex to a point G in the base. And two values of the square of BD

may be found, one from Theo. 37, p. 118: Euc. iii. 35; ii. 2; and

another from Euc. iii. 36; ii. 1. From these may be deduced, that

the rectangle DC, GA, is equal to the rectangle AD, CG. Whence

the, &c.

23. Let ABCD be a square and AC its diagonal. On AC take AE

equal to the side BC or AB : join BE and at E draw EF perpendicular

to AC and meeting BC in F. Then EC, the difference between the

diagonal AC and the side AB of the square, is less than AB ; and CE,

EF, FB may be proved to be equal to one another : also CE, EF are the

adjacent sides of a square whose diagonal is FC. On FC take FG equal

to CE and join EG. Then, as in the first square, the difference CG

between the diagonal FC and the side EC or EF, is less than the side EC.

Hence EC, the difference between the cUagonal and the side of the given

square, is contained twice in the side BC with a remainder CG : and CG

is the difference between the side CE and the diagonal CF of another

square. By proceeding in a similar way, CG, the difference between the

diagonal CF and the side CE, is contained twice m the side CE with a

remainder : and the same relations may be shewn to exist between the

difference of the diagonal and the side of every square of the series which

is so constructed. Hence, therefore, as the difference of the side and

diagonal of every square of the series is contained twice in the side with

a remainder, it follows that there is no line which exactly measures the

side and the diagonal of a square.

24. Let the given line AB be divided in C, D. On AD describe a

semicircle, and on CB describe another semicircle intersecting the former

in P ; draw PE perpendicular to AB ; then E is the point required.

25. Let AB be equal to a side of the given square. On AB describe

a semicircle ; at A draw AC perpendicular to AB and equal to a fourth

proportional to AB and the two sides of the given rectangle. Draw CD

parallel to AB meeting the circumference in D. Join AD, BD, which

are the required lines.

26. Let the two given lines meet when produced in A. At A draw

AD perpendicular to AB, and AE to AC, and such that AD is to AE in

the given ratio. Through D, E, draw DF, EF, respectively parallel to

AB, AC and meeting each other in F. Join AF and produce it, and

the perpendiculars drawn from any point of this line on the two given

lines will always be in the given ratio.

27. The angles made by the four lines at the point of their divergence,

remain constant. See Note on Euc. vi. A, p. 295.

28. Let AB be the given line from which it is required to cut off a

part BC such that BC shall be a mean proportional between the

remainder AC and another given line. Produce AB to D, making BD

ON BOOK VI.

Sil

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