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

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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

AB.CD+BC.AD = AC.BD.

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

adjacent to the base.

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.

Hence adding the above results

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. Â£. 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,

and have a given radius.

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.

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

AB.CD+BC.AD = AC.BD.

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

adjacent to the base.

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.

Hence adding the above results

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. Â£. 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,

and have a given radius.

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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