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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Font size CA respectively, also from^ let fall A G perpendicular on the base BC,
Then ^ G^ is equal to the sum of BD, FE, FF.

Wf 1

Prom P draw PA, PB, PC to the angles A, B, C.
Then the triangle ABCis equal to the three triangles FAB, PBC,
PCA.

116 GEOMETRICAL EXERCISES

But since every rectangle is double of a triangle of the same base
and altitude, (I. 41.)

therefore the rectangle AG, BC, is equal to the three rectangles
AB, PD', AC, Pi^ and BC, PE.

Whence the line AG h equal to the sum of the lines PD, PE, PF.
If the point P fall on one side of the triangle, or coincide with E:
then the triangle ABC is equal to the two triangles A PC, BPA :
whence AG h equal to the sum of the two perpendiculars PP>, PE.
If the point P fall without the base BC of the triangle :

then the triangle ABCh equal to the difference between the sum
of the two triangles APC, BPA, and the triangle PCB.

Whence AG is equal to the difference between the sum of PE,
PF, and PE.

6. If the straight line AB be divided into two unequal parts in
E, and into two unequal parts in E, the rectangle contained by AE,
EB, wall be greater or less than the rectangle contained by AE, EB,
according as E is nearer to, or further from, the middle point of AB,
than E.

7. Produce a given straight line in such a manner that the square
on the whole line thus produced, shall be equal to twice the square on
the given line.

8. If AB be the line so divided in the points C and E, (fig. Euc.
II. 5.) shew that ^J?'- = 4 . CE' + 4..AE.EB.

9. Divide a straight line into two parts, such that the sum of their
squares may be the least possible.

10. Divide a line into two parts, such that the sum of their
squares shall be double the square on another line.

11. Shew that the difference between the squares on the two un-
equal parts (fig. Euc. il. 9.) is equal to twdce the rectangle contained
by the whole line, and the part between the points of section.

12. Shew how in all the possible cases, a straight line may be
geometricalhj divided into two such parts, that the sum of their squares
fehall be equal to a given square.

13. Divide a given straight line into two parts, such that the
squares on the whole line and on one of the parts shall be equal to twice
the square on the other part.

14. Any rectangle is the half of the rectangle contained by the
diameters of the squares on its two sides.

15. If a straight line be divided into tw^o equal and into two un-
equal parts, the squares on the two unequal parts are equal to twice
the rectangle contained by the tw^o unequal parts, together with four
times the square on the line between the points of section.

16. If the points C, Z) be equidistant from the extremities of the
straight line A B, shew that the squares constructed on AE and A C,
exceed twice the rectangle AC, AEhy the square constructed on CE.

17. If any point be taken in the plane of a parallelogram from
which perpendiculars are let fall on the diagonal, and on the sides
which include it, the rectangle of the diagonal and the perpendicular

ON BOOK II.

117

on it, is equal to the sum or difference of the rectangles of the sides
and the perpendiculars on them.

18. ABCD is a rectangular parallelogram, of which A, C are
opposite angles, E any point in BC, i^any point in CJJ. Prove that
twice the area of the triangle AEF together with the rectangle BE,
BE is equal to the parallelogram A C.

II.

19. Shew how to produce a given line, so that the rectangle con-
tained by the whole line thus produced, and the produced part, shall be
equal to'the square (1) on the given line (2) on the part produced.

20. If in the figure Euc. Ii. 11, we join BE and CH, and produce
CH to meet ^jFin L, CL is perpendicular to' BE.

21. If a line be divided, as in Euc. ii. 11, the squares on the whole
line and one of the parts are together three times the square on the
other part.

22. If in the fig. Euc. ll. 11, the points E, D be joined cutting
AHB, GIlKinf, d respectively; then shall Ef=Dd.

III.

23. If from the three angles of a triangle, lines be drawn to the
points of bisection of the opposite sides, the squares on the distances
between the angles and the common intersection, are together one-third
of the squares on the sides of the triangle.

24. ABCis a triangle of which the angle at Cis obtuse, and the
angle at B is half a right angle : D is the middle point ofAB, and CE
is drawn perpendicular to AB. Shew that the square on ^ Cis double
of the squares on AD and DE.

25. If an angle of a triangle be two-thirds of two right angles,
shew that the square on the side subtending that angle is equal to the
squares on the sides containing it, together with the rectangle con-
tained by those sides.

26. The square described on a straight line drawn from one of
the angles at the base of a triangle to the middle point of the opposite
side, is equal to the sum or difi'erence of the square on half the side
bisected, and the rectangle contained between the base and that part of
it, or of it produced, which is intercepted between the same angle and
a perpendicular drawn from the vertex.

27. ABC is a. triangle of which the angle at Cis obtuse, and the
angle at B is half a right angle : D is the middle point of AB, and
CE is di'awn perpendicular to AB. Shew that the square on AC is
double of the squares on AD and DE.

28. Produce one side of a scalene triangle, so that the rectangle
under it and the produced part may be equal to the difi'erence of the
squares on the other two sides.

29. Given the base of any triangle, the area, and the line bisecting
the base, construct the triangle.

i

118 GEOMETRICAL EXERCISES

IV.

30. Shew that the square on the h\T)otenuse of a right-angled
triangle, is equal to four times the area of the triangle together with
the square on the difference of the sides.

31. In the triangle ABC, if AD be the perpendicular let fall
upon the side BC', then the square on -^ C together with the rectangle
contained by BC, BD is equal to the square on AB together with
the rectangle CB, CD.

32. ABCis a triangle, right angled at C, and CD is the perpen-
dicular let fall from C upon AB ; if HK is equal to the sum of the
sides AC, CB, ^nd LM to the sum of AB, CD, shew that the square
on UK together with the square on CD is equal to the square on LM,

3'j. ABCh a triangle having the angle at ^ a right angle : it is
required to find in AB a point P such that the square on ^ C may
exceed the squares on AP and PC by half the square on AB.

34. In a right-angled triangle, the square on that side which is the
greater of the two sides containing the right angle, is equal to the
rectangle by the sum and difference of the other sides.

35. The h)-potenuse AB of a right-angled triangle ABC is tri-
sected in the points D, D; prove that if CD, CD be joined, the sum
of the squares on the sides of the triangle CDE is equal to two-thirds
of the square on AB.

36. From the hypotenuse of a right-angled triangle portions are
cut off equal to the adjacent sides : shew that the square on the middle
segment is equivalent to twice the rectangle under the extreme
segments.

V.

37. Prove that the square on any straight line drawn from the
vertex of an isosceles triangle to the base, is less than the square on a
side of the triangle by the rectangle contained by the segments of the
base : and conversely.

38. If from one of the equal angles of an isosceles triangle a per-
pendicular be drawn to the opposite side, the rectangle contained by
that side and the segment of it intercepted between the perpendicular
and base, is equal to the half of the square described upon the base.

39. If in an isosceles triangle a perpendicular be let fall from one
of the equal angles to the opposite side, the square on the perpendicu-
lar is equal to the square on the line intercepted between the other
equal angle and the perpendicular, together with twice the rectangle
contained by the segments of that side.

40. The square on the base of an isosceles triangle whose vertical
angle is a right angle, is equal to four times the area of the triangle.

41. Describe an isosceles obtuse-angled triangle, such that the
square on the side subtending the obtus'e angle may be three times the
square on either of the sides containing the obtuse angle.

42. li AB^ one of the sides of an isosceles triangle AB Che pro-
duced beyond the base to D, so that BD = AB, shew that

CD^^AB' \2.BC\

ON BOOK n. 119

43. If ABC be an isosceles triangle, and DJ3 be drawn parallel
to the base BC, and UB be joined ; prove that BJS' = BCy DE^ CE\

44. If ^i?C be an isosceles triangle of which the angles at B and
C are each double of ^ ; then the square on ^ C is equal to the square
on ^C together with the rectangle contained by ^Cand BC.

VI.

45. Shew that in a parallelogram the squares on the diagonals are
equal to the sum of the squares on all the sides.

46. If ABCD be any rectangle, A and C being opposite angles,
and any point either within or without the rectangle :

OA^+OC'=OB'+OD\

47. In any quadrilateral j&gure, the sum of the squares on the
diagonals together with four times the square on the line joining their
middle points, is equal to the sum of the squares on all the sides.

48. In any trapezium, if the opposite sides be bisected, the sum
of the squares on the other two sides, together with the squares on the
diagonals, is equal to the sum of the squares on the bisected sides,
together with four times the square on the line joining the points of
bisection.

49. The squares on the diagonals of a trapezium are together
double the squares on the two lines joining the bisections of the
opposite sides.

50. In any trapezium two of whose sides are parallel, the squares
on the diagonals are together equal to the squares on its two sides which
are not parallel, and twice the rectangle contained by the sides which
are parallel.

51. If the two sides of a trapezium be parallel, shew that its
area is equal to that of a rectangle contained by its altitude and half
the sum of the parallel sides.

52. If a trapezium have two sides parallel, and the other two equal,
shew that the rectangle contained by the two parallel sides, together
with the square on one of the other sides, will be equal to the square
on the straight line joining two opposite angles of the trapezium.

53. If squares be described on the sides of any triangle and the
angular points of the squares be joined ; the sum of the squares on the
sides of the hexagonal figure thus formed is equal to four times the
sum of the squares on the sides of the triangle.

VII.

54. Find the side of a square equal to a given equilateral triangle.

55. Find a square which shall be equal to the sum of two given
rectilineal figures.

56. To divide a given straight line so that the rectangle under its
segments may be equal to a^iven rectangle.

57. Construct a rectangle equal to a given square and having the
difference of its sides equal to a given straight line.

58. Shew how to describe a rectangle equal to a giypn square, and
having one of its sides equal to a given straight line.

BOOK III.

DEFIlViTIONS.
I.

Equal circles are those of which the diameters are equal, or from
the centers of which the straight lines to the circumferences are
equal.

This is not a definition, but a theorem, the truth of which is evident; for,
if the circles be applied to one another, so that their centers coincide, the
circles must likewise coincide, since the straight lines from the centers are
equal.

n.

A straight line is said to touch a circle when it meets the circle,
and being produced does not cut it.

III.

Circles are said to touch one another, which meet, but do not cut
one another.

i

Straight lines are said to be equally distant from the center of a
circle, when the perpendiculars di-awn to them from the center are
equal.

And the sti-aight line on which the greater perpendicular falls, is
said to be further from the center.

VI.

A segment of a circle is the figure contained by a straight line, and
the arc or the part of the circumference which it cuts off.

BOOK III. PROP. 1.

121

VII.

The angle of a segment is that which is contained by a straight
line and a part of the circumference.

VIII.

An angle in a segment is any angle contained by two straight lines
drawn from any point in the arc of tlie segment, to the extremities of
the straight line which is the base of the segment.

IX.

An angle is said to insist or stand upon the part of the circum-
ference intercepted between the straight lines that contain the angle.

A sector of a circle is the figure contained by two straight lines
drawn from the center and the arc between them.

XI.

Similar segments of circles are those in which the angles are equal,
or which contain equal angles.

^^

PROPOSITION I. PROBLEM.

To find the center of a given circle.
Let AJBChe the given circle : it is required to find its center.

Draw within it any straight line AB to meet tiie circumference in
A, JB; and bisect AB in I); (i. 10.) from the point JD di-aw DC at
right angles to AB, (i. 11.) meeting the circumference in C, produce
CD to JEJ to meet the circumference again in D, and bisect CJE in F.

Then the point i^ shall be the center of the circle ABC.

For, if it be not, if possible, let G be the center, and join GA, GD, GB.

Then, because DA is equal to DB, (constr.)

and DG common to the two triangles ADG, BDG,

the two sides AD, DG, are equal to the two BD, DG, each to each;

and the base GA is equal to the base GB, (i. def. 15.)

because they are drawn from the center G :

therefore the angle ADG is equal to the angle GDB : (l. 8.)

but when a straight line standing upon another straight line makes

the adjacent angles equal to one another, each of the angles is a right

angle ; (I. def. 10.)

therefore the angle GDB is a right angle :

but FDB is likewise a right angle ; (constr.)

wherefore the angle FDB is equal to the angle GDB, (ax. 1.)

the greater angle equal to the less, which is impossible ;

therefore G is not the center of the circle ABC.

In the same manner it can be shewn that no other point out of the

line CE is the center ;

and since CE is bisected in F,
any other point in CE divides CE into unequal parts, and cannot
be the center.

Therefore no point but jPis the center of the circle ABC,
Which was to be found.
Cor. From this it is manifest, that if in a circle a straight line
bisects another at right angles, the center of the cii'cle is in the line
which bisects the other.

PROPOSITION II. THEOREM.

If any two points be taken in the circumferetice of a circle^ the straight
line which joins them shall fall xcithin the circle.

Let ABC he a circle, and A, B any two points in the circumference.
Then the straight line drawn from A to B shall fall within the circle.

BOOK I. PROP. II, III. 123

r

I^K A E B

I^P For if ^jB do not fall within the circle,

let it fall, if possible, without the circle as AEB ;

find D the center of the cii'cle ABC, (iii. 1.) and join DA, BB ;

in the circumference AB take any point F,

join DF, and produce it to meet AB in F.

Then, because DA is equal to DB, (l. def. 15.)

therefore the angle DBA is equal to the angle DAB ; (l. 5.)

and because AF, a side of the triangle DAF, is produced to B,

the exterior angle DEB is greater than the interior and opposite

angle D^jE; (l. 16.)

but DAF was proved to be equal to the angle DBF',

therefore the angle DFB is greater than the angle DBF ;

but to the greater angle the greater side is opposite, (l. 19.)

therefore DB is greater than DF :

but DB is equal to DF-, (l. def. 15.)

wherefore DF is greater than DF,

the less than the greater, which is impossible ;

therefore the sti-aight line di*awn from A to B does not fall without

the circle.
In the same manner, it may be demonstrated that it does not fall
upon the circumference ;

therefore it falls within it.
Wherefore, if any two points, &c. Q. E. D.

PROPOSITION III. THEOREM.

If a straight line drawn through the center of a circle bisect a straight
line in it which does not pass through the center, it shall cut it at right
angles : and conversely, if it cut it at right angles, it shall bisect it.

Let ABC he a circle ; and let CD, a straight line drawn through
the center, bisect any straight line AB, which does not pass through
the center, in the point F.

Then CD shall cut ^i? at right angles.

Take J^ the center of the circle, (ill. 1.) and join FA, FB.

Then, because ^jPis equal to FB, (hyp.)

and FE common to the two triangles AFE, BFF,

g2

124 Euclid's elements.

there are ^wo sides in the one equal to two sides in the other, each
to each ;
and the base ^A is equal to the base JSB; (l. def. 15.)
therefore the angle AFJE is equal to the angle BFE', (l. 8.)
but when a straight line standing upon another straight line makes
the adjacent angles equal to one another,

each of them is a right angle; (l. def. 10.)
therefore each of the angles AFE, BFE, is a right angle :
wherefore the straight line CD, drawn through the center, bisecting
another AB that does not pass through the center, cuts the same at
right angles.

Conversely, let CD cut AB dX right angles.
Then CD shall also bisect AB, that is, ^i^ shall be equal to FB.

because, EB, EA, from the center are equal to one another,
(I. def. 15.)

therefore the angle EAF is equal to the angle EBF', (I. 5.)
and the right angle AFE is equal to the right angle BFE ; (I. def. 10.)

therefore, in the two triangles, EAF, EBF,
there are two angles in the one equal to two angles in the other, each

to each ;
and the side EF, which is opposite to one of the equal angles in each,
is common to both ;

therefore the other sides are equal ; (l. 26.)

therefore ^i^ is equal to FB.
Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION IV. THEOREM.

If in a circle tico straight lines cut one another, which do not both pass
through the center, they do not bisect each other.

Let ABCD be a circle, and A C, BD two straight lines in it which
cut one another in the point E, and do not both pass through the center.
Then AC, BD, shall not bisect one another.

For, if it be possible, let ^^ be equal to EC, and ^^to ED.

If one of the lines pass through the center,
it is plain that it cannot be bisected by the other which does not
pass through the center :

but if neither of them pass through the center,

find i^the center of the circle, (ill. 1.) and join EF.

Then because FE, a straight line drawn through the center, bisects

another ^(7 which does not pass through the center, (hyp.)

therefore FE cuts AC oX right angles : (ill. 3.)

wherefore FEA is a right angle.

BOOK 111. PROP. V, VI.

125

Again, because the straight line FE bisects the straight line BD
which does not pass tlirough the center, (hyp.)

therefore FE cuts BD at right angles : (ill. 3.)

wherefore FEB is a right angle :

but FEA was shewn to be a right angle ;

therefore the angle FEA is equal to the angle FEB, (ax. 1.)

the less equal to the greater, which is impossible ;

therefore A C, BD do not bisect one another.

Wherefore, if in a circle, &c. Q. e.d.

PROPOSITION V. THEOREM.

If two circles cut one another ^ they shall not have the same center.

Let the two circles ABC, CDG, cut one another in the points B, C.
They shall not have the same center.

If possible, let E be the center of the two circles ; join EC,

and draw any straight line EFG meeting the circumferences in i^ and G.

And because E is the center of the circle ABC,

therefore ^i^is equal to EC: (l. def. 15.)

again, because E is the center of the circle CDO^

therefore EG is equal to EC-, (i. def. 15.)

but EFwsiS shewn to be equal to EC;

therefore ^i^is equal to EG, (ax. 1.)

the less line equal to the greater, which is impossible.

Therefore E is not the center of the circles ABC, CDG.

Wherefore, if two circles, &c. Q. E. D.

PROPOSITION VI. THEOREM.

If one circle touch another internally, they shall not have the same center.

Let the circle CDE touch the circle ^ J? C internally in the point C.
They shall not have the same center,
c

If possible, let i^be the center of the two circles: join FC,

and draw any straight line FEB, meeting the circumferences in E and B.

And because F is the center of the circle ABC,

FB is equal to FC-, (l. def. 15.)

126

also, because F is the center of the circle CDU,

FE is equal to FC: (l. def. 15.)

but FB was shewn to be equal to FC',

therefore FE is equal to FB, (ax. 1.)

the less line equal to the greater, which is impossible :

therefore i^is not the center of the circles ABC, CDE.

Therefore, if two circles, &c. q.e.d.

PROPOSITION VII. THEOREM.

If any point be taken in the diameter of a circle ichich is not the center,
of all the straight lines which can be drawn fi^om it to the circumference,
the greatest is that in which the center is, and the other part of that
diameter is the least ; and, of the rest, that which is nearer to the
line which passes through the center is always greater than one more remote :
and from the same point there can be drawn only two equal straight lines
to the circumference one upon each side of the diameter.

Let ABCD be a circle, and^Z) its diameter, in which let any point
F be taken which is not tlie center :

let the center be E.
Then, of all the straight lines FB, FC, FG Â«fec. that can be drawn
from F to the circumference,

FA, that in which the center is, shall be the greatest,
and FD, the other part of the diameter AD, shall be the least:
and of the rest, FB, the nearer to FA, shall be greater than FC
the more remote, and FC greater than FG.

Join BE, CE, GE.

Because two sides of a triangle are greater than the third side, (l. 20.)

therefore BE, EFare greater than BE:

but AE is equal to BE; (l. def. 15.)

therefore AE, EF, that is, ^i^is greater than BF,

Again, because BE is equal to CE,

and FE common to the triangles BEE, CEF,

the two sides BE, EF are equal to the two CE, EF, each to each ;

but the angle BEE is greater than the angle CEF; (ax. 9.)

therefore the base BFh greater than the base CF. (l. 24.)

For the same reason Ci^is greater than GF.

Again, because GF, FE are greater than EG, (l. 20.)

and EG is equal to ED ;

therefore GF, FE are greater than ED :

take away the common part FE,

and the remainder GF is greater than the remainder FD. (ax. 5.)

BOOK 111. PROP. VIII. 127

_ Therefore, FA is the greatest,

and FD the least of all the straight lines from F to the circumference ;
and J3F is greater than CF, and CF than GF.
Also, there can be di'awn only two equal straight li;ies from the
point F to the circumference, one upon each side of the diameter.

At the point F, in the straight line FF, make the angle FFII
equal to the angle FFG, (l. 23.) and join FH.

Then, because GF is equal to FH, (l. def. 15.)

and FF common to the two triangles GFF, JIFF;

the two sides GF, FF are equal to the two HF, FF, each to each ;

and the angle GFF is equal to the angle JIFF', (constr.)

therefore the base FG is equal to the base FH: (i. 4.)

but, besides FH, no other straight line can be drawn from F to the

circumference equal to FG :

for, if possible, let it be FK:

and because FK is equal to FG, and FG to FH,

therefore FK is equal to FH; (ax. 1.)

that is, a line nearer to that which passes through the center, is equal

to one which is more remote ;

which has been proved to be impossible.
Therefore, if any point be taken, &c. Q. E. D.

PROPOSITION VIII. THEOREM.

If any point he taJcen without a circle, and straight lines be drawn from
it to the circumference, whereof one passes through the center ; of those
which fall upon the concave part of the circumference, the greatest is that
tchich passes through the center ; and of the rest, that which is nearer to the
one passing through the ce7iter is ahcays greater than one more remote: hut
of those which fall upon the convex piart of the circumference, the least is
that hetioeen the point loithout the circle and the diameter; and of the rast,
that which is nearer to the least is always less than one more remote ; and
only two equal straight lines can he drawn from the same point to the circum-
ference, one upon each side of the line which passes through the center.

Let ABC he a circle, and D any point without it, from which let
the straight lines DA, DF, DF, DC he drawn to the circumference,
whereof DA passes through the center.