Euclid.

# A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI online

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Online LibraryEuclidA text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 10 of 27)
Font size BD is divided into two parts at C,

Because >. ^. ^
(m iig. 2.

.\ in both cases,

the sum of the sqq. on CB, BD = twice the rect. CB, BD with

the sq. on CD. ii. 7.

To each add the sq. on DA.

Then the sum of the sqq. on CB, BD, DA =^ twice the rect.

CB, BD with the sum of the sqq. on CD, DA.

But the sum of the sqq. on BD, DA = the sq. on AB,

for the angle ADB is a rt. angle. i. 47.

Similarly tlie sum of the sqq. on CD, DA= the sq. on AC.

.-. the sum of the sqq. on AB, BC, =:^ twice the rect. CB, BD,

with the sq. on AC.

That is, the sq. on AC is less than the sqq. on AB, BC

by twice the rect. CB, ED. q.e.d.

142 F.rrT.TD's Kl.K.MKX'l'S.

Obs. If the perpendicular AD coijicides with AC, tlmt is, if ACB
ia a right angle, it may be shewn that the proposition merely repeat.s
the result of i. 47.

Note. The result of Prop. 12 may be written
AB2=BC-+CA2 + 2BC.CD.

Remembering the definition of the Projection of a straight lino
given on page 97, the student will see that this proposition may be
enunciated as follows :

In an obtuse-angled triangle the square on the aide opposite the
obtuse angle is greater than the sum of the squares on the sides contain-
ing the obtuse angle by twice the rectangle contained by either of those
sideSy and the projection of the other side upon it.

Prop. 13 may be written

AC2 = AB2+BC2-2CB.BD,

and it may also be enunciated as follows :

In every triangle the square on the side subtending an acute angle,
is less than the squares on the sides containing that angle, by twice the
rectangle contained by either of these sides, and the projection of the
other side upon it.

EXERCISES.

The following theorem should be noticed ; it is proved by the help
ofii. 1.

1. If four points A, B, C, D are taken in order on a straight line,
then will

ON II. 12 AND 13.

2. If from one of the base angles of an isosceles triangle a per-
pendicular is drawn to the opposite side, then twice the rectangle
contained by that side and the segment adjacent to the base is equal
to the square on the base.

3. If one angle of a triangle is one-third of two right angles,
shew that the square on the opposite side is less than the sum of the
squares on the sides forming that angle, by the rectangle contained by
these two sides. [See Ex. 10, p. 101.]

4. If one angle of a triangle is two-thirds of two right angles,
shew that the square on the opposite side is greater than the squares
on the sides fonuing that angle, by the rectangle contained by these
pides, [See Ex. 10, p. 101.]

book ii. prof. 14.
Propositiox 14. Problem.

143

Ih describe a square that shalLbe equal to a given recfi-
lineal figure.

Let A be the given rectilineal figure.

It is required to describe ca square equal to A.

Describe the par*" BCDE equal to the fig. A, and liaving

the angle CBE a right angle. i. 45.

Then if BC = BE, the fig. BD is a square; and what was

required is done.

But if not, produce BE to F, making EF equal to ED; i. 3.

and bisect BF at G. i. 10.

From centre G, with radius GF, describe the semicircle BHF :

produce DE to meet the semicircle at H.

Then shall the sq. on EH be equal to the given fig. A.

Join GH.

Then because BF is divided equally at G and unequally

at E,
.-. the rect. BE, EF with the sq. on GE = the sq. on GF ii. 5.

= the sq. on GH.
But the sq. on GH = the sum of the sqq. on GE, EH ;

for the angle HEG is a rt. angle. i. 47.

.-, the rect. BE, EF with the sq. on GE = the sum of the
sqq. on GE, EH.

From these take the sq. on G E :
then the rect. BE, EF = the sq. on HE.
But the rect. BE, EF = the fig. BD ; for EF = ED ; Constr.
BD = the given fig. A. Constr.

I

and the fig.

the sq. on EH = the given fig. A, q.e.f.

144 EUCLID'S ELEMENTS.

THEOREMS AND EXAMPLES ON BOOK 11.

ON n. 4 AND 7.

1. Shew by ii. 4 that the square on a straight line is four times
the square on half the line.

[This result is constantly used in solving examples on Book ii,
especially those which follow from ii. 12 and 13.]

2. If a straight line is divided into any three parts, the square on
the whole line is equal to the sum of the squares on the three parts
together with twice the rectangles contained by each pair of these
parts.

Shew that the algebraical formula corresponding to this theorem is
{a + b + c)^=a^ + h- + c^ + 2hc + 2ca + 2ab.

3. In a right-angled triangle, if a perpendicular is drawn from the
right angle to the hypotenuse, the square on this perpendicular is equal
to the rectangle contained by tlie segments of tlie hypotenuse.

4. In an isosceles triangle, if a perpendicular be drawn from one
of the angles at the base to the opposite side, shew that the square on
the perpendicular is equal to twice the rectangle contained by the
segments of that side together with the square on the segment

5. Any rectangle is half the rectangle contained by the diagonals
of the squares described upon its two sides.

6. In any triangle if a perpendicular is drawn from the vertical
angle to the base, the sum of the squares on the sides forming that
angle, together with twice the rectangle contained by the segments of
the base, is equal to the square on the base together with twice the
square on the perpendicular.

ON II. 5 AND 6.

The student is reminded that these important propositions are
both included in the following enunciation.

The difference of the squares on two straight lines is equal to the
rectangle contained by their smn and difference.

7. In a right-angled triangle the square on one of the sides form-
ing the right angle is equal to the rectangle contained by the sum and
difference of the hypotenuse and the other side. [i. 47 and ii. 5.]

THEOREMfc* A:ND EXAMPLES ON BOOK II.

145

8. The difference of the squares on two sides of a triangle is equal
to ticice the rectangle contained by the base and the intercept between
the middle point of the base and the foot of the perpendicular dravm
from the vertical angle to the base.

Let ABC be a triangle, and let P be the middle point of the base
BC : let AQ be drawn perp. to BC.

Then shall AB=^- AC2 = 2BC . PQ.

First, let AQ fall within the triangle.

Now AB2=BQ2 + QA2, i. 47.

also AC2=QC2 + QA2,

.-. AB2-AC2=BQ2_QC^ Ax. 3.

= (BQ + QC) (BQ - QC) ii. 5.

= BC.2PQ Ex. 1, p. 129.

= 2BC.PQ. Q.E.D.

The case in which AQ falls outside the triangle presents no
difficulty.

9. The square on any straight line draicn from tJie vertex of an
isosceles triangle to the base is less than the square on one of the equal
sides by the rectangle contained by the segments of the base.

10. The square on any straight line drawn from the vertex of an
isosceles triangle to the base produced, is greater than the square on
one of the equal sides by the rectangle contained by the segments into
xoJdch the base is divided externally.

11. If a straight line is drawn through one of the angles of
an equilateral triangle to meet the opposite side produced, so that the
rectangle contained by the segments of the base is equal to the square
on the side of the triangle ; shew that the square on the line so drawn
is double of the square on a side of the triangle.

12. If XY be drawn parallel to the base BC of an isosceles
triangle ABC, then the difference of the squares on BY and CY ia
equal to the rectangle contained by BC, XY. [See above, Ex. 8.]

13. In a right-angled triangle, if a perpendicular be drawn from
the right angle to the hypotenuse, the square on either side forming
the right angle is equal to the rectangle contained by the hypotenuse
and the segment of it adjacent to that side.

H. K 10

kr

]40 EUCLID'S ki.i:mk.nts.

ON II. 9 AND 10.

14. Deduce Prop. 9 from Props. 4 and 5, using also the theorem
tliat the square on a straight line is four times the square on half the
line.

15. Deduce Prop. 10 from Props. 7 and 6, using also the theorem
mentioned in the preceding Exercise.

16. If a straight line is divided equally and also unequally, the
squares on the two unequal segments are together equal to twice the
rectangle contained hy these segments together with four times the
square on the line between the points of section.

ON II. 11.

17. If a straight line is divided internally in medial section, and
from the greater segment apart he taken equal to the less; shew tliat
the greater segment is also divided in medial section.

18. If a straight line is divided in medial section, the rectangle
contained by the sum and difference of the segments is equal to
the rectangle contained by the segments.

19. If AB is divided at H in medial section, and if X is the
middle point of the greater segment AH, shew that a triangle whose
sides are equal to AH, XH, BX respectively must be right-angled.

20. If a straight line AB is divided internally in medial section at
H, prove that the sum of the squares on AB, BH is three times the
square on AH.

21. Divide a straight line externally in medial section.

[Proceed as in ii. 11, but instead of drawing EF, make EF' equal
to EB in the direction remote from A; and on AF' describe the square
AF'G'H' on the side remote from AB. Then AB will be divided exter-
nally at H as required.]

ON II. 12 AND 13.

22. In a triangle ABC the angles at B and C are acute: if E and
F are the feet of perpendiculars drawn from the opposite angles to the
sides AC, AB, shew that the square on BC is equal to the sum of the
rectangles AB, BF and AC, CE.

23. ABC is a triangle right-angled at C, and DE is drawn from
a point D in AC perpendicular to AB : shew that the rectangle
AB, AE is equal to the rectangle AC, AD.

THEOREMS ANi) EXAMPLES OS BOOK II.

14<

24. In any triangle the sum of the squares on two sides is equal to
twice the square on half the third side together loith tioice the square on
the median which bisects the third side.

Q C

Let ABC be a triangle, and AP the median bisecting the side BC.
Then shall AB2 + AC2 = 2 BP2+2AP2.

Draw AQ perp. to BC.

Consider the case in which AQ falls within the triangle, but does
not coincide with A P.

Then of the angles APB, APC, one must be obtuse, and the other
acute: let APB be obtuse.

Then in the A APB, AB2=BP2-!-AP2 + 2 BP . PQ. ii. 12.

Also in the A APC, AC^^CP^-l- AP2-2CP . PQ. ii. 13.
ButCP=BP,
.-. CP2=BP2; and the rect. BP, PQ = the rect. CP, PQ.

AB2 + AC2=:2.BP2+2.AP2. Q.E.D.

The student will have no difficulty in adapting this proof to the
cases in which AQ falls without the triangle, or coincides with A P.

25. The sum of the squares on the sides of a parallelogram is equal
to the sum of the squares on the diagonals.

26. In any quadrilateral the squares on the diagonals are toge-
ther equal to twice the sum of the squares on the straight lines join-
ing the middle points of opposite sides. [See Ex. 9, p. 97.]

27. If from any j)oint within a rectangle straight lines are drawn
to the angular points, the sum of the squares on one pair of the lines
drawn to opposite angles is equal to the sum of the squares on the
other pair.

28. The sum of the squares on the sides of a quadrilateral is
greater than the sum of the squares on its diagonals by four times
the square on the straight line which joins the middle points of the
diagonals.

29. O is the middle point of a given straight line AB, and from
O as centre, any circle is described : if P be any point on its cireum-^
ference, shew that the sum of the squares on AP, BP is constant.

10-2

148 Euclid's elements.

30. Given the base of a triangle, and the sum of the squares on
the sides forming the vertical angle ; find the locus of the vertex.

31. ABC is an isosceles triangle in which AB and AC are equal.
AB is produced beyond the base to D, so that BD is equal to AB.
Shew that the square on CD is equal to the square on AB together
with twice the square on BC.

32. In a right-angled triangle the sum of the squares on the
straight lines drawn from the right angle to the points of tri-
section of the hypotenuse is equal to five times the square on the
line between the points of trisection.

33. Three times the sum of the squares on the sides of a tri-
angle is equal to four times the sum of the squares on the medians.

34. ABC is a triangle, and O the point of intersection of its
medians : shew that

AB2+BC2 + CA2 = 3(OA2 + OB2+OC2).

35. A BCD is a quadrilateral, and X the middle point of the
straight line joining the bisections of the diagonals ; with X as centre
any circle is described, and P is any point upon this circle : shew that
PA^-f PB2+ PC-+ PD2 is constant, being equal to

X A3 + X B2 + XC2 + X D2 + 4X P2.

36. The squares on the diagonals of a trapezium are together
equal to the sum of the squares on its two oblique sides, with twice
the rectangle contained by its parallel sides.

PH0BLEM8.

37. Construct a rectangle equal to the difference of two squares.

38. Divide a given straight line into two parts so that the rect-
angle contained by them may be equal to the square described on a
given straight line which is less than half the straight line to be
divided.

39. Given a square and one side of a rectangle which is equal
to the square, find the other side.

40. Produce a given straight line so that the rectangle contained
by the whole line thus produced, and the part produced, may be
equal to the square on half the line.

41. Produce a given straight line so that the rectangle con-
tained by the whole line thus produced and the given line shall be
equal to a given square.

42. Pivide a straight line AB into two parts at C, such that the
rectangle contained by BC and another line X may be equal to the
square on AC.

I

PAET 11.
BOOK III

Book III. deals with the properties of Circles.
Definitions.

1. A circle is a plane figure bounded
by one line, which is called the circum-
ference, and is such that all straight lines
drawn from a certain point within the
figure to the circumference are equal to
one another : this point is called the centre
of the circle.

2. A radius of a circle is a straight line drawn from
the centre to the circumference.

3. A diameter of a circle is a straight line drawn
through the centre, and terminated both ways, by the
circumference.

4. A semicircle is the figure bounded by a diameter
of a circle and the part of the circumference cut oft' by the
diameter.

From these definitions we draw the following inferences:

(i) The distance of a point from the centre of a circle is less than
the radius, if the point is within the circumference : and the distance
of a point from the centre is greater than the radius, if the point is
without the "circumference.

(ii) A point is within a circle if its distance from the centre is
less than the radius : and a point is without a circle if its distance
from the centre is greater than the radius.

(iii) Circles of equal radius are equal in all respects ; that is to
say, their areas and circumferences are equal.

(iv) A circle is divided by any diameter into two parts which are
equal in all respects.

160 EUCLID'S ELEMENTS.

o. Circles which liave tlie same centre are snifl to })e
concentric.

6. An arc of a circle is any part of the circumferene^r

7. A chord of a circle is the straight line which joins
any two points on the circumference.

From these definitions it may be seen that a
chord of a circle, which does not pass through
the centre, divides the circumference into two
unequal arcs ; of these, the greater is called the
major arc, and the less the rainor arc. Thus
the major arc is greater, and the minor arc let>s
than the semicircumference.

The major and minor arcs, into which a cir-
cumference is divided by a chord, are said to be
conjugate to one another.

8. Chords of a circle are said to be
equidistant from the centre, when the
perpendiculars drawn to them from the
centre are equal :

and one chord is said to be further from
the centre than another, when the per-
pendicular drawn to it from the centre is
greater than the perpendicular drawn to
the other.

9. A secant of a circle is a straight
line of indefinite length, which cuts the
circumference in two points.

10. A tangent to a circle is a straight
line which meets the circumference, but
being produced, does not cut it. Such a
line is said to touch the circle at a point;
and the point is called the point of
contact.

DEFINlTIpKS.

151

If a secant, which cuts a circle at the
points P and Q, gradually changes its position
in such a way that P remains lixed, the point
Q will ultimately approach the fixed point P,
until at length these points may be made to
coincide. When the straight line PQ reaches
this limiting position, it becomes the tangent
to the circle at the point P.

Hence a tangent may be defined as a
straight line which passes through two coinci-
dent points on the circumference.

11. Circles are said to touch one another when they
meet, but do not cut one another.

When each of the circles which meet is outside the other, they are
said to touch one another externally, or to have external contact:
when one of the circles is within the other, they are said to touch one
another internally, or to have internal contact.

12. A segment of a circle is the figure bounded by a
chord and one of the two arcs into which tlie chord divides
the circumference.

The chord of a segment is sometimes called its base.

[ir2

EUClilD.S Kljh^MKN'I'.S.

1 3. An angle' in a segment is one
formed by two straiglit lines drawn from
any point in the arc of tlie segment to
the extremities of its chord.

[It will be shewn in Proposition 21, that all angles in the same
segment of a circle are equal.]

14. An angle at the circumference
of a circle is one formed by straight lines
drawn from a point on the circumference
to the extremities of an arc : sucli an
angle is said to stand upon the arc, wliicli
it subtends.

15. Similar segments
of circles are those which
contain equal angles.

16. A sector of a circle is a figure
bounded by two radii and the arc inter-
cepted between them.

\

Symbols and Abbreviations.

In addition to the symbols and abbreviations given on
page 10, we shall use the following.

for circle, O^'' for circumference.

book iii. prop. 1. 153

Proposition 1. Problem.

To find the centre of a given circle.

E^ ^B

Let ABC be a given circle:

it is required to find its centre.

In the given circle draw any chord AB,

and bisect AB at D. i. 10.

From D draw DC at right angles to AB; i. 11.

and produce DC to meet the O ''^ at E and C.

Bisect EC at F. i. 10.

Then shall F be the centre of the ABC.
First, the centre of the circle must be in EC :
for if not, let the centre be at a point G without EC.
Join AG, DG, BG.
Then in the AÂ« GDA, GDB,
{ DA=DB, Co7istr.

Because -! and G D is common ;

[ and GA= GB, for by supposition they are radii;
.â€˘. the ^ G DA -= the ^ G DB ; i. 8.

.". these angles, being adjacent, are rt. angles.

But the ^ CDB is a rt. angle ; Constr.

.'. the z. GDB = the ^ CDB, Ax.W.

the part equal to the whole, which is impossible.
.â– . G is not the centre.
So it may be shewn that no point outside EC is the centre ;

.'. the centre lies in EC.
.'. F, the middle point of the diameter EC, must be the
centre of the ABC. Q.E.F.

Corollary. The straight line which bisects a chord of
a circle at right angles passes through the centre.
[For Exercises, see page 156.]

154 Euclid's klemknt^

Proposition 2. Thkorem.

If any two points are taken in the circumference of a
circle, the chord lohich joins them falls within the circle.

Let ABC be a circle, and A and B any two points on
itsO''^:

then shall the chord AB fall within the circle.

Find D, the centre of the 0ABC; iii. 1.

and in AB take any point E.
Join DA, DE, DB.
In the A DAB, because DA = DB, iii. Def 1.

.'. the L DAB = the L DBA. I. 5.

But the ext. l DEB is greater than the int. opp. ÂŁ. DAE;

I. 16.

.'. also the L DEB is greater than the z_ DBE;

.'. in the A DEB, the side DB, which is opposite the greater

angle, is greater than DE which is opposite the less: i. 19.

that is to say, DE is less than a radius of the circle ;

.". E falls within the circle.
So also any other point between A and B may be shewn
to fall within the circle.

.*. AB falls within the circle. Q. e. d.

Definition. A part of a curved line is said to be concave to a
point when, any chord being taken in it, all straight lines drawn
from the given point to the intercepted arc are cut by the chord : if,
when any chord is taken, no straight line drawn from the given point
to the intercepted arc is cut by the chord, the curve is said to be
convex to that point.

Proposition 2 proves that the whole circumference of a circle
is concave to its centre.

BOOK III. PRor. 3. 155

Proposition 3. Theoreji.
If a straight line drawn through the centre of a circle
bisects a chord which does not pass through the centre, it shall
cut it at right angles :

and, conversely, if it cut it at right angles, it shcdl bisect it.

B

Let ABC be a circle ; and let CD be a st. line drawn
through the centre, and AB a chord which does not pass
through the centre.

First. Let CD bisect AB at F :

then shall CD cut AB at rt. angles.
Find E, the centre of the circle; ill. L

and join EA, EB.
Then in the A^ AFE, BFE,
( AF = BF, Hyp.

Because \ and FE is common ;

. ( and AE = BE, being radii of the circle ;

.-. the L AFE -the L BFE; i. 8.

,*. these angles, being adjacent, are rt. angles,

that is, DC cuts AB at rt. angles. q.e.d.

Conversely. Let CD cut AB at rt. angles :
then shall CD bisect AB at F.
As before, find E the centre ; and join EA, EB.
In the A EAB, because EA= EB, iii. Def. 1.

.â€˘.thez.EAB=the/. EBA. 1.5.

Then in the AÂ« EFA, EFB,
i the L EAF = the z. EBF, Proved.

Because -| and the l EFA = the z_ EFB, being rt. angles ; Hyp.
I and EF is common;

.'. AF=BF. I. 26.

Q.E.D.
[For Exercises, see page 156.]

156 Euclid's elements.

exercises.
ON Proposition 1.

1. If two circles intersect at the points A, B, shew that the line
which joins their centres bisects their common chord AB at right
angles.

2. AB, AC are two equal chords of a circle; shew that the
straight line which bisects the angle BAG passes through the centre.

3. Two chords of a circle are given in position and magnitude:
find the centre of the circle.

4. Describe a circle that shall pass through three given points,
which are not in the same straight line.

5. Find the locus of the centres of circles ichich pass through two
given points.

6. Describe a circle that shall pass through two given points,

ON Proposition 2.

7. A straight line cannot cut a circle in more titan two points.

on Pboposition 3.

8. Through a given point within a circle draw a chord which
shall be bisected at that point.

9. The parts of a straight line intercepted between the circum-
ferences of two concentric circles are equal.

10. The line joining the middle points of two parallel chords of a
circle passes through the centre.

11. Find the locus of the middle points of a system of parallel
chords drawn in a circle.

12. If two circles cut one another, any two parallel straight lines
drawn through the points of intersection to cut the circles, are equal.

13. PQ and XY are two parallel chords in a circle : shew that
the points of intersection of PX, QY, and of PY, QX, lie on the
straight line which passes through the middle points of the given
chords.

BOOK III. PEOP. 4. 157

. / Proposition 4. Theorem.

If in a circle two chords cut one anotJter, which do not
hoth jmss' through the centre, they can7iot both he bisected at
their point of intersection.

Let A BCD be a circle, and AC, BD two chords which
intersect at E, but do not both pass through the centre :
then AC and BD shall not be hoth bisected at E.

Case I. If one chord passes through the centre, it is
a diameter, and the centre is its middle point;
.'.it cannot be bisected by the other chord, which by hypo-
thesis does not pass through the centre.

Case II. If neither chord passes through the centre;
then, if possible, let E be the middle point of hoth;
that is, let AE = EC; and BE = ED.
Find F, the centre of the circle: iii. 1.

Join EF.

Then, because FE, which passes through the centre,
bisects the chord AC, Hyp.

.'. the L FEC is a rt. angle. ill. 3.

And because FE, which passes through the centre, bi-
sects the chord BD, Hyp.

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