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

. **(page 13 of 27)**

Online Library → Euclid → A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 13 of 27)

Font size

of the arcs cut off' by it, then the bisector of the angle APB cuts the

conjugate arc in the same point, ivhatever be the position of P.

8. Two circles intersect at A and B; and through these points

straight lines are drawn from any point P on the circumference of

one of the circles: shew that when produced they intercept on the

other circumference an arc which is constant for all positions of P.

9. A triangle ABC is inscribed in a circle, and the bisectors of

the angles meet the circumference at X, Y, Z. Find each angle of

the triangle XYZ in terms of those of the original triangle.

ON PROPOSITIONS 28, 29.

10. The straight lines which join the extremities of parallel chords

in a circle (i) towards the same parts, (ii) towards opposite parts, are

equal.

11. Through A, a point of intersection of two equal circles two

straight lines PAQ, XAY are drawn: shew that the chord PX is equal

to the chord QY.

12. Through the points of intersection of two circles two parallel

straight lines are drawn terminated by the circumferences : shew that

the straight lines which join their extremities towards the same parts

are equal.

13. Two equal circles intersect at A and B; and through A any

straight line PAQ is drawn terminated by the circumferences: shew

that BP:=BQ.

14. ABC is an isosceles triangle inscribed in a circle, and the

bisectors of the base angles meet the circumference at X and Y. Shew

that the figure BXAYC must have four of its sides equal.

What relation must subsist among the angles of the triangle ABC,

in order that the figure BXAYC may be equilateral?

1D8 EUCLlDb ELEMENTS.

Note. We have given Euclid's demonstrations of Propositions

26, 27, 28, 29 ; but it should be noticed that all these propositions

also admit of direct proof by the method of superposition.

To illustrate this method we will apply it to Proposition 26.

Proposition 26. [Alternative Proof.]

In equal circles, the arcs which subtend equal angles, whether at

the centres or circumferences, shall be equal.

Let ABC, DEF be equal circles, and let the Z ^ BGC, EHF at the

centres be equal, and consequently the z ' BAG, EDF at the o''*"

equal: iii. 20.

then shall the arc BKG = the arc ELF.

For if the Â© ABG be applied to the DEF, so that the centre G

may fall on the centre H,

then because the circles are equal. Hyp.

.'. their O*^' must coincide ;

hence by revolving the upper circle about its centre, the lower circle

remaining fixed,

B may be made to coincide with E,

and consequently GB with HE.

And because the Z BGC = the z EHF,

.â€¢. GC must coincide with HF:

and since G C = H F, Hyp.

.: C must fall on F.

Now B coinciding with E, and C with F,

and the o^^^ of the ABC with the O''^ of the Â© DEF,

.â™¦. the arc BKC must coincide with the arc ELF.

.-. the arc BKC = the arc ELF.

Q.E.D.

book iii. prop. 30. 199

Proposition 30. Problem.

To bisect a given arc.

Let ADB be the given arc:

it is required to bisect it.

Join AB; and bisect it at C. i. 10.

At C draw CD at rt. angles to AB, meeting the given

arc at D. i. 11.

Tlien shall the arc ADB be bisected at D.

Join AD, BD.

Then in the A^ ACD, BCD,

( AC = BC, Constr.

Because â€¢< and CD is common;

(and the l ACD = the l BCD, being rt. angles:

.'. AD = BD. I. 4.

And since in the 0ADB, the chords AD, BD are equal,

.". the arcs cut off by them are equal, the minor arc equal

to the minor, and the major arc to the major: iii. 28.

and the arcs AD, BD are both minor arcs,

for each is less than a semi-circumference, since DC, bisecting

the chord AB at rt. angles, must pass through the centre

of the circle. iii. 1. Cor.

.'. the arc AD = the arc BD :

that is, the arc ADB is bisected at D. Q. e. f.

EXERCISES.

1. If a tangent to a circle is parallel to a chord, the point of

contact will bisect the arc cut off by the chord.

2. Trisect a quadrant, or the fourth part of the circumference, of

a circle.

200 EUCLID.S j;i.K.MENTS.

Proposition 31. Theorem.

llie angle in a semicircle is a right anyle :

the anyle in a segvietit greater than a semicircle is less

than a right arigle :

and the angle in a segment less thato a semicircle is

greater than a right angle.

Let A BCD be a circle, of wliich BC is a diameter, and

E the centre ; and let AC be a chord dividing the circle into

the segments ABC, ADC, of which the segment ABC is

greater, and the segment is ADC less than a semicircle:

then (i) tlie angle in the semicircle BAC shall be a rt. angle ;

(ii) the angle in the segment ABC shall be less than a

rt. angle ;

(iii) the angle in the segment ADC shall be greater

than a rt. angle.

In the arc ADC take any point D;

Join BA, AD, DC, AE; and produce BA to F.

(i) Then because EA = EB, iii. J)ef. 1.

.'. the L EAB = the l EBA. 'l. 5.

And because EA = EC,

.'. the L EAC = the z. ECA.

.'. the whole l BAC = the sum of the l " EBA, ECA:

but tlie ext. l FAC = the sum of the two int. l ^ CBA, BCA;

.'. the z. BAC ^ the <l FAC;

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

.'. the L BAC, in the semicircle BAC, is a rt. angle.

BOOK III. PROP. 31. 201

(ii) 111 the A ABC, because the two l ^ ABC, BAC are

together less tlian two rt. angles; I. 17.

and of these, the l. BAC is a rt. angle ; Proved.

:. the L ABC, which is the angle in the segment ABC, is

less than a rt, angle.

(iii) Because A BCD is a quadrilateral inscribed in the

0ABC,

.'. the L ^ ABC, ADC together = two rt. angles; in. 22.

and of these, the z_ ABC is less than a rt. angle: Proved.

.'. the L. ADC, which is the angle in the segment ADC, is

greater than a rt. angle. q. E. D.

EXERCISES.

1. A circle described on the hypotenuse of a right-angled triangle

as diameter, passes through the opposite angular j)o int.

2. A system of right-angled triangles is described upon a given

straight line as hypotenuse : find the locus of the opposite angular

points.

3. A straight rod of given length slides between two straight

rulers placed at right angles to one another : find the locus of its

middle point.

4. Two circles intersect at A and B ; and through A two diameters

AP, AQ are drawn, one in each circle : shew that the points P, B, Q.

are collinear. [See Def. p. 102.]

5. A circle is described on one of the equal sides of an isosceles

triangle as diameter. Shew that it passes through the middle point

of the base.

6. Of two circles which have internal contact, the diameter of the

inner is equal to the radius of the outer. Shew that any chord of

the outer circle, drawn from the point of contact, is bisected by the

circumference of the inner circle.

7. Circles described on any two sides of a triangle as diameters

intersect on the third side, or the third side produced.

8. Fi7id the locus of the middle points of chords of a circle draicn

through a fixed point.

Distinguish between the cases when the given point is within, on,

or without the circumference.

9. Describe a square equal to the difference of two given squares.

10. Through one of the points of intersection of two circles draAV

a chord of one circle which shall be bisected by the other.

11. On a given straight line as base a system of equilateral four-

sided figures is described : find the locus of the intersection of their

diagonals.

202

EUCLID S ELEMENTS.

Note 1. The extension of Proposition 20 to straight and rejlex

angles furnishes a simple alternative proof of

the first theorem contained in Proposition 31,

viz.

The angle in a semicircle is a right angle.

For, in the adjoining figure, the angle at

the centre, standing on the arc BHC, is

double the angle at the 0*=*, standing on the

same arc.

Now the angle at the centre is the straight angle BEC ;

.-. the z SAC is half of the straight angle BEC:

and a straight angle = two rt. angles;

.â€¢. the z BAG = one half of two rt. angles.

: one rt. angle.

Q.E.D.

Note 2. From Proposition 31 we may derive a simple practical

solution of Proposition 17, namely,

To draw a tangent to a circle from a given external 'point.

Let BCD be the given

circle, and A the given exter-

nal point:

it is required to draw from

A a tangent to the Â© BCD.

Find E, the centre of the

circle, and join AE.

On AE describe the semi-

circle ABE, to cut the given

circle at B.

Join AB.

Then AB shall be a tangent

to theÂ© BCD.

For the Z ABE, being in a semicircle, is a rt. angle. iii. 31.

.-. AB is drawn at rt. angles to the radius EB, from its ex-

tremity B;

.-. AB is a tangent to the circle. iii. 16.

Q.E.F.

Since the semicircle might be described on either side of AE, it is

clear that there will be a second solution of the problem, as shewn by

the dotted lines of the figure.

BOOK III. PRor. 32.

20S

Proposition 32. Theorem.

If a straight line touch a circle^ and from the point of

contact a chord he drawn, the angles which this chord makes

with the tangent shall he equal to the angles hi the alternate

segments of the circle.

Let EF touch the given 0ABC at B, and let BD be a

chord drawn from B, the point of contact:

then shall (i) the z. DBF = the angle in the alternate

segment BAD:

(ii) the z. DBE = the angle in the alternate

From B draw BA perp. to EF.

Take any point C in the arc BD;

and join AD, DC, CB.

I. 11,

(i) Then because BA is drawn perp. to the tangent EF,

at its point of contact B,

.*. BA passes through the centre of the circle: iii. 19.

.*. the L ADB, being in a semicircle, is a rt. angle: ill. 31.

.'. in the AABD, the other z. ^ ABD, BAD together = a rt.

angle; i. 32.

that is, the z. ^ ABD, BAD together = the z. ABF.

From these equals take the common z. ABD;

.'. the z_DBF = the z. BAD, which is in the alternate seg-

ment.

204 Euclid's kj.ement.s.

A

(ii) Because A BCD is a quadrilateral inscribed in a

circle,

.'.the L^ BCD, BAD together = two rt. angles: iii. 22,

but the L^ DBE, DBF together = two rt. angles; i. 13.

.'. the /.Â« DBE, DBF together = the z. Â» BCD, BAD:

and of these the l DBF = the /L BAD; Proved.

.'. the i. DBE = the z_ DCB, which is in the alternate seg-

ment. Q. E. D.

EXERCISES.

1. State and prove the converse of this proposition.

2. Use this Proposition to shew that the tangents drawn to a

circle from an external point are equal.

3. If two circles touch one another, any straight line drawn

through the point of contact cuts off similar segments.

Prove this for (i) internal, (ii) external contact.

4. If two circles touch one another, and from A, the point of con-

tact, two chords APQ, AXY are drawn: then PX and QY are parallel.

Prove this for (i) internal, (ii) external contact.

5. Two circles intersect at the points A, B: and one of them

passes through O, the centre of the other : prove that OA bisects the

angle between the common chord and the tangent to the first circle

at A.

6. Two circles intersect at A and B ; and through P, any point

on the circumference of one of them, straight lines PAC, PBD are

drawn to cut the other circle at C and D: shew that CD is parallel

to the tangent at P.

7. If from the point of contact of a tangent to a circle, a chord

be drawn, the perpendiculars dropped on the tangent and chord from

the middle point of either arc cut off by the chord are equal.

book iii. prop. 83. -205

Proposition 33. Problem.

On a given straight line to describe a segment of a circle

tohich shall contain an angle equal to a given angle.

H H

\

Let AB be the given st. line, and C the given angle:

it is required to describe on AB a segment of a circle which

shall contain an angle equal to C.

At A in BA, make the z. BAD equal to the l C. i. 23.

From A draw AE at rt. angles to AD. i. 11.

Bisect AB at F; I. 10.

and from F draw FG at rt. angles to AB, cutting AE at G.

Join GB.

Then in the A^ AFG, BFG.

i AF = BF, Cojistr.

Because -< and FG is common,

(and the l AFG =the L BFG, being rt, angles;

.'. GA-GB : I. 4.

,'. the circle described from centre G, with radius GA, will

pass through B.

Describe this circle, and call it ABH:

then the segment AHB shall contain an angle equal to C.

Because AD is drawn at rt. angles to the radius GA from

its extremity A,

.". AD is a tangent to the circle: in. 16.

and from A, its point of contact, a chord AB is drawn;

.*. the L BAD = the angle in the alt. segment AHB. iii. 32.

But the ^ BAD = the l C : Constr.

.'. tlie angle in the segment AHB = the z. C.

.*; AHB is the segment required; Q. E;FÂ»

206 EUCLID'S ELEMENTS.

NoTK. In the particular case when the given angle C is a rt, angle,

the segment required will be the

Beraicircle described on the given ^^H

St. line AB; for the angle in a

semicircle is a rt. angle. iii. 31.

EXERCISES.

[The following exercises depend on the corollary to Proposition 21

given on page 187, namely

The locus of the vertices of triangles which stand on the same base

and have a given vertical angle, is the arc of tlie segment standing on

this base, and containing an angle equal to the given angle.

Exercises 1 and 2 afford good illustrations of the method of find-

ing required points by the Intersection of Loci. See page 117.]

1. Describe a triangle on a given base, having a given vertical

angle, and liaving its vertex on a given straight line.

2. Construct a triangle, having given the base, the vertical angle

and (i) one other side.

(ii) the altitude.

(iii| the length of the median which bisects the base.

(iv) the point at which the perpendicular from tlie vertex

meets the base.

3. Construct a triangle having given the base, the vertical angle,

and the point at tvhich the base is cut by the bisector of the vertical

angle.

[Let AB be the base, X the given point in it, and K the given

angle. On AB describe a segment of a circle containing an angle

equal to K ; complete the o''^ by drawing the arc APB. Bisect the arc

APB at P: join PX, and produce it to meet the C"" at C. Then ABC

shall be the required triangle.]

4. Construct a triangle having given the base, the vertical angle,

and the sum of the remaining sides.

[Let AB be the given base, K the given angle, and H the given line

equal to the sum of the sides. On AB describe a segment containing

an angle equal to K, also another segment containing an angle equal

to half the z K. From centre A, with radius H, describe a circle

cutting the last drawn segment at X and Y. Join AX (or AY) cutting

the first segment at C. Then ABC shall be the required triangle.]

5. Construct a triangle having given the base, the vertical angle,

and the difference of the remaining sides.

book iii. prop. 34. 207

Proposition 34. Problem.

From a given circle to cut off a segment which shall

contain an angle equal to a given angle.

Let ABC be the given circle, and D the given angle:

it is required to cut off from the ABC a segment which

shall contain an angle equal to D.

Take any point B on the O '^'',

and at B draw the tangent EBF. iii. 17.

At B, in FB, make the L FBC equal to the l D. i. 23.

Then the segment BAC shall contain an angle equal to D.

Because EF is a tangent to the circle, and from B, its

point of contact, a chord BC is drawn,

.'. the L FBC = the angle in the alternate segment BAC.

III. 32.

But the L FBC = the ^ D;- Constr.

.". the angle in the segment BAC = the l D.

Hence from the given 0ABC a segment BAC has been

cut off, containing an angle equal to D. q, e. f.

EXERCISES.

1. The chord of a given segment of a circle is produced to a fixed

point : on this straight line so produced draw a segment of a circle

similar to the given segment.

2, _ Through a given point without a circle draw a straight line

that will cut off a segment capable of containing an angle equal to a

given angle.

20H kucj.id'k ki.kmknts.

Proposition 35. Theorem.

If two chords of a circle cut one anotlieVy tits rectangle

contained hy the segments of one shall he eqtud to the rect-

angle contained hy tlie segments of the other.

Let AB, CD, two chords of the 0ACBD, cut one another

at E:

then shall the rect. AE, EB-the rect. CE, ED.

Find F the centre of the Â© ACB : iii. 1.

From F draw FG, FH perp. respectively to A B, CD. i. 12.

Join FA, FE, FD.

Then because FG is drawn from the centre F perp. to AB,

.'. AB is bisected at G. iii. 3.

For a similar reason CD is bisected at H.

Again, because AB is divided equally at G, and unequally at E,

.*. the rect. AE, EB with the sq. on EG ^ the sq. on AG. ll. 5.

To each of these equals add the sq. on G F ;

then the rect. AE, EB with the sqq. on EG, GF:=the sum of

the sqq. on AG, GF.

But the sqq. on EG, GF = the sq. on FE; I. 47.

and the sqq. on AG, GF = the sq. on AF;

for the angles at G are rt. angles.

.'. the rect. AE, EB with the sq. on FE^^the sq. on AF.

Similarly it may be shewn that

the rect. CE, ED with the sq. on FE-the sq. on FD.

But the sq. on AF - the sq. on FD; for AF = FD.

.'. the rect. AE, EB with the sq. on FE = the rect. CE, ED

with the sq. on FE.

From these equals take the sq. on FE:

then the rect. AE, EB = the rect. CE, ED, <j.e.d.

BOOK III. TRor. 35. 209

CoKOLLARY. If through a fixed 'point vnthin a circle

any number of chords are drawn, the rectangles contained

hy their segments are all equal.

Note. The following special cases of this proposition deserve

notice.

(i) when the given chords both pass through the centre :

(ii) when one chord passes through the centre, and cuts the

other at right angles :

(iii) when one chord passes through the centre, and cuts the

other obliquely.

In each of these cases the general proof requires some modifica-

tion, which may be left as an exercise to the student.

EXERCISUiS.

1. Tico straiglit ILnen AB, CD intersect at E, so that the rectangle

AE, EB is equal to the rectangle CE, ED: sheio that the four points

A, B, C, D are concyclic.

2. The rectangle contained by the segments of any chord drawn

through a given point within a circle is equal to the square on half

the shortest chord which may be drawn through that point.

3. ABC is a triangle right-angled at C ; and from C a perpen-

dicular CD is drawn to the hypotenuse : shew that the square on CD

is equal to the rectangle AD, DB.

4. ABC is a triangle; and AP, BQ the perpendiculars dropped

from A and B on the opposite sides, intersect at O : shew that the

rectangle AG, OP is equal to the rectangle BO, OQ.

5. Two circles intersect at A and B, and through any point in AB

their common chord two chords are drawn, one in each circle ; shew

that their four extremities are concyclic.

6. A and B are two points within a circle such that the rectangle

contained by the segments of any chord drawn through A is equal to

the rectangle contained by the segments of any chord through B :

shew that A and B are equidistant from the centre.

7. If through E, a point ivithout a circle, two secants EAB, ECD

are drawn; shew that the rectangle EA, EB is equal to the rectangle

EC, ED.

[Proceed as in ni. 35, using ii. 6.]

8. Through A, a point of intersection of two circles, two straight

lines CAE, DAF are drawn, each passing through a centre and termi-

nated by the circumferences: shew that the rectahgle CA, AE is equal

to the rectangle DA, AFÂ»

II. E. 14

210 euclid's elements.

Proposition 36. Theorem.

If from any point without a circle a tangent and a

secant be di'awn, then the rectangle contained by the whole

secant and the part of it loithout the circle shall be equal to

the square on the tangent.

Let ABC be a circle; and from D a point without it, let

there be drawn the secant DC A, and the tangent DB:

then the rect. DA, DC shall be equal to the sq. on DB.

Find E, the centre of the 0ABC: iii. 1.

and from E, draw EF perp. to AD. i. 12.

Join EB, EC, ED.

Then because EF, passing through the centre, is perp.

to the chord AC,

.'. AC is bisected at F. iii. 3.

And since AC is bisected at F and produced to D,

.'.the rect. DA, DC with the sq. on FC-^the sq. on FD. H. 6.

To each of these equals add the sq. on EF :

then the rect. DA, DC witli the sqq. on EF, FC = the sqq. on

EF, FD.

But the sqq. on EF, FC = the sq. on EC ; for EFC is a rt. angle;

= the sq. on EB.

And the sqq. on EF, FD = the sq. on ED ; for EFD is a rt. angle;

= the sqq. on EB, BD; for EBD is a

rt. angle. in. 18.

/. the rect. DA, DC with the sq. on EB=:the sqq. on EB, BD.

From these equals take the sq. on EB:

then the rect. DA, DC = the sq. on DB. Q. e.d.

Note. This proof may easily be adapted to the case when the

Becant passes through the centre of the circle.

]iooK 111. riiop. 3ti.

i>il

Corollary. If from a given point zvithout a circle

any number of secants are drawn, the o'ectanffles contained

hy the whole secants and the ^^ar^s of them without the circle

are all equal ; for each of these rectangles is equal to the

square on the tangent drawn from the given point to the

circle. p

For instance, in the adjoining figure,

each of the rectangles PB, PA and PD, PC

and PF_, PE is equal to the square on the

tangent PQ:

.*. the rect. PB, PA

= the rect. PD, PC

^^ the rect. PF, PE.

Note, llemembcring that the segments into which the chord AB

is divided at P, are the Hnes PA, PB, (see Part I. page 131) we are

enabled to include the corollaries of Propositions 35 and 36 in a

single enunciation.

If any numher of chords of a circle arc draicn through a [jiven

point icithin or without a circle, the rectangles contained hy the

segments of the chords are equal.

EXERCISES.

1. Use this proposition to shew that tangents drawn to a circle

from an external point are equal.

2s If two circles intersect, tangents drawn to them from any

point in their common chord produced are equal.

3. If two circles intersect at A and B, and PQ is a tangent to

both circles; shew that AB produced bisects PQ.

4. If P is any i^oint on the straight line AB produced, shew that

the tangents drawn from P to all circles which pass through A and B

are equal.

5. ABC is a triangle right-angled at C, and from any point P in

AC, a perpendicular PQ is drawn to the hypotenuse: shew that the

rectangle AC, AP is equal to the rectangle AB, AQ.

0. ABC is a triangle right-angled at C, and from C a perpen-

dicular CD is drawn to the hypotenuse; shew that the rect. AB, AD

is equal to the square on AC.

14-2

212 EUCLID'S I : I, K. Mi; NTS.

PkOPOSTTION o7. 'J'llKUUEM.

//* from a jtoi/d /rifhuut a circle there h<' <h-<in-)h two

straight lines, one of tvhich cuts the circle, and the other

tneets it, and if the rectangle contained hy the whole line

lohich cuts the circle and the pa,rt of it vnthout the circle he

equal to the square on the line which meets tlte circle, then

the line which meets the circle shall he a tangcfU to it.

Let ABC be a circle; and from D, a point without it,

let there be drawn two st. lines DCA and DB, of which

DC A cuts the circle at C and A, and DB meets it; and let

the rect. DA, DC =^^ the sq. on DB:

then shall DB be a tangent to the circle.

From D draw DE to touch the Â©ABC: in. 7.

let E be the point of contact.

Find the centre F, and join FB, FD, FE. in. 1.

Then since DCA is a secant, and DE a tangent to the circle,

.'. the rect. DA, DC = the sq. on DE, III. 36.

But, by liypotliesis, the rect. DA, DC = the sq. on DB;

.*. the sq. on DE = the sq. on DB,

.-. DE= DB.

Hence in the A^ DBF, DEF.

[ DB = DE,

Proved.

Because - and BF = EF;

[and DF is common;

III. Def. 1.

.-. the L DBF - the l DEF.

I. 8.

But DEF is a rt. angle ;

III. 18.

.'. DBF is also a it. angle;

and since BF is a radius,

.*. DB touches the Â©ABC at the point

B.

Q. K. U.

NOTE ON thp: method of limits as applied to tangency.

Euclid defines a tangent to a circle as a straight line which meets

the circumference, hut being produced, does not cut it: and from this

definition he deduces the fundamental theorem that a tangent is j)er-

pendicular to the radius drawn to the point of contact. Prop. IG.

But this result may also be established by the Method of Limits,

which regards the tangent as the idtimate position of a secant tohen its

two points of intersection tvith the circumference are brought into coin-

cidence [See Note on page 151]: and it may be shewn that every

theorem relating to the tangent may be derived from some more

general proposition relating to the secant, by considering the ultimate

case when the two points of intersection coincide.

1. To prove by the Method of Limits tliat a tangent to a circle

is at right angles to the radius drawn to the point of contact.

Let ABD be a circle, whose centre

is C; and PABQ a secant cutting the

conjugate arc in the same point, ivhatever be the position of P.

8. Two circles intersect at A and B; and through these points

straight lines are drawn from any point P on the circumference of

one of the circles: shew that when produced they intercept on the

other circumference an arc which is constant for all positions of P.

9. A triangle ABC is inscribed in a circle, and the bisectors of

the angles meet the circumference at X, Y, Z. Find each angle of

the triangle XYZ in terms of those of the original triangle.

ON PROPOSITIONS 28, 29.

10. The straight lines which join the extremities of parallel chords

in a circle (i) towards the same parts, (ii) towards opposite parts, are

equal.

11. Through A, a point of intersection of two equal circles two

straight lines PAQ, XAY are drawn: shew that the chord PX is equal

to the chord QY.

12. Through the points of intersection of two circles two parallel

straight lines are drawn terminated by the circumferences : shew that

the straight lines which join their extremities towards the same parts

are equal.

13. Two equal circles intersect at A and B; and through A any

straight line PAQ is drawn terminated by the circumferences: shew

that BP:=BQ.

14. ABC is an isosceles triangle inscribed in a circle, and the

bisectors of the base angles meet the circumference at X and Y. Shew

that the figure BXAYC must have four of its sides equal.

What relation must subsist among the angles of the triangle ABC,

in order that the figure BXAYC may be equilateral?

1D8 EUCLlDb ELEMENTS.

Note. We have given Euclid's demonstrations of Propositions

26, 27, 28, 29 ; but it should be noticed that all these propositions

also admit of direct proof by the method of superposition.

To illustrate this method we will apply it to Proposition 26.

Proposition 26. [Alternative Proof.]

In equal circles, the arcs which subtend equal angles, whether at

the centres or circumferences, shall be equal.

Let ABC, DEF be equal circles, and let the Z ^ BGC, EHF at the

centres be equal, and consequently the z ' BAG, EDF at the o''*"

equal: iii. 20.

then shall the arc BKG = the arc ELF.

For if the Â© ABG be applied to the DEF, so that the centre G

may fall on the centre H,

then because the circles are equal. Hyp.

.'. their O*^' must coincide ;

hence by revolving the upper circle about its centre, the lower circle

remaining fixed,

B may be made to coincide with E,

and consequently GB with HE.

And because the Z BGC = the z EHF,

.â€¢. GC must coincide with HF:

and since G C = H F, Hyp.

.: C must fall on F.

Now B coinciding with E, and C with F,

and the o^^^ of the ABC with the O''^ of the Â© DEF,

.â™¦. the arc BKC must coincide with the arc ELF.

.-. the arc BKC = the arc ELF.

Q.E.D.

book iii. prop. 30. 199

Proposition 30. Problem.

To bisect a given arc.

Let ADB be the given arc:

it is required to bisect it.

Join AB; and bisect it at C. i. 10.

At C draw CD at rt. angles to AB, meeting the given

arc at D. i. 11.

Tlien shall the arc ADB be bisected at D.

Join AD, BD.

Then in the A^ ACD, BCD,

( AC = BC, Constr.

Because â€¢< and CD is common;

(and the l ACD = the l BCD, being rt. angles:

.'. AD = BD. I. 4.

And since in the 0ADB, the chords AD, BD are equal,

.". the arcs cut off by them are equal, the minor arc equal

to the minor, and the major arc to the major: iii. 28.

and the arcs AD, BD are both minor arcs,

for each is less than a semi-circumference, since DC, bisecting

the chord AB at rt. angles, must pass through the centre

of the circle. iii. 1. Cor.

.'. the arc AD = the arc BD :

that is, the arc ADB is bisected at D. Q. e. f.

EXERCISES.

1. If a tangent to a circle is parallel to a chord, the point of

contact will bisect the arc cut off by the chord.

2. Trisect a quadrant, or the fourth part of the circumference, of

a circle.

200 EUCLID.S j;i.K.MENTS.

Proposition 31. Theorem.

llie angle in a semicircle is a right anyle :

the anyle in a segvietit greater than a semicircle is less

than a right arigle :

and the angle in a segment less thato a semicircle is

greater than a right angle.

Let A BCD be a circle, of wliich BC is a diameter, and

E the centre ; and let AC be a chord dividing the circle into

the segments ABC, ADC, of which the segment ABC is

greater, and the segment is ADC less than a semicircle:

then (i) tlie angle in the semicircle BAC shall be a rt. angle ;

(ii) the angle in the segment ABC shall be less than a

rt. angle ;

(iii) the angle in the segment ADC shall be greater

than a rt. angle.

In the arc ADC take any point D;

Join BA, AD, DC, AE; and produce BA to F.

(i) Then because EA = EB, iii. J)ef. 1.

.'. the L EAB = the l EBA. 'l. 5.

And because EA = EC,

.'. the L EAC = the z. ECA.

.'. the whole l BAC = the sum of the l " EBA, ECA:

but tlie ext. l FAC = the sum of the two int. l ^ CBA, BCA;

.'. the z. BAC ^ the <l FAC;

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

.'. the L BAC, in the semicircle BAC, is a rt. angle.

BOOK III. PROP. 31. 201

(ii) 111 the A ABC, because the two l ^ ABC, BAC are

together less tlian two rt. angles; I. 17.

and of these, the l. BAC is a rt. angle ; Proved.

:. the L ABC, which is the angle in the segment ABC, is

less than a rt, angle.

(iii) Because A BCD is a quadrilateral inscribed in the

0ABC,

.'. the L ^ ABC, ADC together = two rt. angles; in. 22.

and of these, the z_ ABC is less than a rt. angle: Proved.

.'. the L. ADC, which is the angle in the segment ADC, is

greater than a rt. angle. q. E. D.

EXERCISES.

1. A circle described on the hypotenuse of a right-angled triangle

as diameter, passes through the opposite angular j)o int.

2. A system of right-angled triangles is described upon a given

straight line as hypotenuse : find the locus of the opposite angular

points.

3. A straight rod of given length slides between two straight

rulers placed at right angles to one another : find the locus of its

middle point.

4. Two circles intersect at A and B ; and through A two diameters

AP, AQ are drawn, one in each circle : shew that the points P, B, Q.

are collinear. [See Def. p. 102.]

5. A circle is described on one of the equal sides of an isosceles

triangle as diameter. Shew that it passes through the middle point

of the base.

6. Of two circles which have internal contact, the diameter of the

inner is equal to the radius of the outer. Shew that any chord of

the outer circle, drawn from the point of contact, is bisected by the

circumference of the inner circle.

7. Circles described on any two sides of a triangle as diameters

intersect on the third side, or the third side produced.

8. Fi7id the locus of the middle points of chords of a circle draicn

through a fixed point.

Distinguish between the cases when the given point is within, on,

or without the circumference.

9. Describe a square equal to the difference of two given squares.

10. Through one of the points of intersection of two circles draAV

a chord of one circle which shall be bisected by the other.

11. On a given straight line as base a system of equilateral four-

sided figures is described : find the locus of the intersection of their

diagonals.

202

EUCLID S ELEMENTS.

Note 1. The extension of Proposition 20 to straight and rejlex

angles furnishes a simple alternative proof of

the first theorem contained in Proposition 31,

viz.

The angle in a semicircle is a right angle.

For, in the adjoining figure, the angle at

the centre, standing on the arc BHC, is

double the angle at the 0*=*, standing on the

same arc.

Now the angle at the centre is the straight angle BEC ;

.-. the z SAC is half of the straight angle BEC:

and a straight angle = two rt. angles;

.â€¢. the z BAG = one half of two rt. angles.

: one rt. angle.

Q.E.D.

Note 2. From Proposition 31 we may derive a simple practical

solution of Proposition 17, namely,

To draw a tangent to a circle from a given external 'point.

Let BCD be the given

circle, and A the given exter-

nal point:

it is required to draw from

A a tangent to the Â© BCD.

Find E, the centre of the

circle, and join AE.

On AE describe the semi-

circle ABE, to cut the given

circle at B.

Join AB.

Then AB shall be a tangent

to theÂ© BCD.

For the Z ABE, being in a semicircle, is a rt. angle. iii. 31.

.-. AB is drawn at rt. angles to the radius EB, from its ex-

tremity B;

.-. AB is a tangent to the circle. iii. 16.

Q.E.F.

Since the semicircle might be described on either side of AE, it is

clear that there will be a second solution of the problem, as shewn by

the dotted lines of the figure.

BOOK III. PRor. 32.

20S

Proposition 32. Theorem.

If a straight line touch a circle^ and from the point of

contact a chord he drawn, the angles which this chord makes

with the tangent shall he equal to the angles hi the alternate

segments of the circle.

Let EF touch the given 0ABC at B, and let BD be a

chord drawn from B, the point of contact:

then shall (i) the z. DBF = the angle in the alternate

segment BAD:

(ii) the z. DBE = the angle in the alternate

From B draw BA perp. to EF.

Take any point C in the arc BD;

and join AD, DC, CB.

I. 11,

(i) Then because BA is drawn perp. to the tangent EF,

at its point of contact B,

.*. BA passes through the centre of the circle: iii. 19.

.*. the L ADB, being in a semicircle, is a rt. angle: ill. 31.

.'. in the AABD, the other z. ^ ABD, BAD together = a rt.

angle; i. 32.

that is, the z. ^ ABD, BAD together = the z. ABF.

From these equals take the common z. ABD;

.'. the z_DBF = the z. BAD, which is in the alternate seg-

ment.

204 Euclid's kj.ement.s.

A

(ii) Because A BCD is a quadrilateral inscribed in a

circle,

.'.the L^ BCD, BAD together = two rt. angles: iii. 22,

but the L^ DBE, DBF together = two rt. angles; i. 13.

.'. the /.Â« DBE, DBF together = the z. Â» BCD, BAD:

and of these the l DBF = the /L BAD; Proved.

.'. the i. DBE = the z_ DCB, which is in the alternate seg-

ment. Q. E. D.

EXERCISES.

1. State and prove the converse of this proposition.

2. Use this Proposition to shew that the tangents drawn to a

circle from an external point are equal.

3. If two circles touch one another, any straight line drawn

through the point of contact cuts off similar segments.

Prove this for (i) internal, (ii) external contact.

4. If two circles touch one another, and from A, the point of con-

tact, two chords APQ, AXY are drawn: then PX and QY are parallel.

Prove this for (i) internal, (ii) external contact.

5. Two circles intersect at the points A, B: and one of them

passes through O, the centre of the other : prove that OA bisects the

angle between the common chord and the tangent to the first circle

at A.

6. Two circles intersect at A and B ; and through P, any point

on the circumference of one of them, straight lines PAC, PBD are

drawn to cut the other circle at C and D: shew that CD is parallel

to the tangent at P.

7. If from the point of contact of a tangent to a circle, a chord

be drawn, the perpendiculars dropped on the tangent and chord from

the middle point of either arc cut off by the chord are equal.

book iii. prop. 83. -205

Proposition 33. Problem.

On a given straight line to describe a segment of a circle

tohich shall contain an angle equal to a given angle.

H H

\

Let AB be the given st. line, and C the given angle:

it is required to describe on AB a segment of a circle which

shall contain an angle equal to C.

At A in BA, make the z. BAD equal to the l C. i. 23.

From A draw AE at rt. angles to AD. i. 11.

Bisect AB at F; I. 10.

and from F draw FG at rt. angles to AB, cutting AE at G.

Join GB.

Then in the A^ AFG, BFG.

i AF = BF, Cojistr.

Because -< and FG is common,

(and the l AFG =the L BFG, being rt, angles;

.'. GA-GB : I. 4.

,'. the circle described from centre G, with radius GA, will

pass through B.

Describe this circle, and call it ABH:

then the segment AHB shall contain an angle equal to C.

Because AD is drawn at rt. angles to the radius GA from

its extremity A,

.". AD is a tangent to the circle: in. 16.

and from A, its point of contact, a chord AB is drawn;

.*. the L BAD = the angle in the alt. segment AHB. iii. 32.

But the ^ BAD = the l C : Constr.

.'. tlie angle in the segment AHB = the z. C.

.*; AHB is the segment required; Q. E;FÂ»

206 EUCLID'S ELEMENTS.

NoTK. In the particular case when the given angle C is a rt, angle,

the segment required will be the

Beraicircle described on the given ^^H

St. line AB; for the angle in a

semicircle is a rt. angle. iii. 31.

EXERCISES.

[The following exercises depend on the corollary to Proposition 21

given on page 187, namely

The locus of the vertices of triangles which stand on the same base

and have a given vertical angle, is the arc of tlie segment standing on

this base, and containing an angle equal to the given angle.

Exercises 1 and 2 afford good illustrations of the method of find-

ing required points by the Intersection of Loci. See page 117.]

1. Describe a triangle on a given base, having a given vertical

angle, and liaving its vertex on a given straight line.

2. Construct a triangle, having given the base, the vertical angle

and (i) one other side.

(ii) the altitude.

(iii| the length of the median which bisects the base.

(iv) the point at which the perpendicular from tlie vertex

meets the base.

3. Construct a triangle having given the base, the vertical angle,

and the point at tvhich the base is cut by the bisector of the vertical

angle.

[Let AB be the base, X the given point in it, and K the given

angle. On AB describe a segment of a circle containing an angle

equal to K ; complete the o''^ by drawing the arc APB. Bisect the arc

APB at P: join PX, and produce it to meet the C"" at C. Then ABC

shall be the required triangle.]

4. Construct a triangle having given the base, the vertical angle,

and the sum of the remaining sides.

[Let AB be the given base, K the given angle, and H the given line

equal to the sum of the sides. On AB describe a segment containing

an angle equal to K, also another segment containing an angle equal

to half the z K. From centre A, with radius H, describe a circle

cutting the last drawn segment at X and Y. Join AX (or AY) cutting

the first segment at C. Then ABC shall be the required triangle.]

5. Construct a triangle having given the base, the vertical angle,

and the difference of the remaining sides.

book iii. prop. 34. 207

Proposition 34. Problem.

From a given circle to cut off a segment which shall

contain an angle equal to a given angle.

Let ABC be the given circle, and D the given angle:

it is required to cut off from the ABC a segment which

shall contain an angle equal to D.

Take any point B on the O '^'',

and at B draw the tangent EBF. iii. 17.

At B, in FB, make the L FBC equal to the l D. i. 23.

Then the segment BAC shall contain an angle equal to D.

Because EF is a tangent to the circle, and from B, its

point of contact, a chord BC is drawn,

.'. the L FBC = the angle in the alternate segment BAC.

III. 32.

But the L FBC = the ^ D;- Constr.

.". the angle in the segment BAC = the l D.

Hence from the given 0ABC a segment BAC has been

cut off, containing an angle equal to D. q, e. f.

EXERCISES.

1. The chord of a given segment of a circle is produced to a fixed

point : on this straight line so produced draw a segment of a circle

similar to the given segment.

2, _ Through a given point without a circle draw a straight line

that will cut off a segment capable of containing an angle equal to a

given angle.

20H kucj.id'k ki.kmknts.

Proposition 35. Theorem.

If two chords of a circle cut one anotlieVy tits rectangle

contained hy the segments of one shall he eqtud to the rect-

angle contained hy tlie segments of the other.

Let AB, CD, two chords of the 0ACBD, cut one another

at E:

then shall the rect. AE, EB-the rect. CE, ED.

Find F the centre of the Â© ACB : iii. 1.

From F draw FG, FH perp. respectively to A B, CD. i. 12.

Join FA, FE, FD.

Then because FG is drawn from the centre F perp. to AB,

.'. AB is bisected at G. iii. 3.

For a similar reason CD is bisected at H.

Again, because AB is divided equally at G, and unequally at E,

.*. the rect. AE, EB with the sq. on EG ^ the sq. on AG. ll. 5.

To each of these equals add the sq. on G F ;

then the rect. AE, EB with the sqq. on EG, GF:=the sum of

the sqq. on AG, GF.

But the sqq. on EG, GF = the sq. on FE; I. 47.

and the sqq. on AG, GF = the sq. on AF;

for the angles at G are rt. angles.

.'. the rect. AE, EB with the sq. on FE^^the sq. on AF.

Similarly it may be shewn that

the rect. CE, ED with the sq. on FE-the sq. on FD.

But the sq. on AF - the sq. on FD; for AF = FD.

.'. the rect. AE, EB with the sq. on FE = the rect. CE, ED

with the sq. on FE.

From these equals take the sq. on FE:

then the rect. AE, EB = the rect. CE, ED, <j.e.d.

BOOK III. TRor. 35. 209

CoKOLLARY. If through a fixed 'point vnthin a circle

any number of chords are drawn, the rectangles contained

hy their segments are all equal.

Note. The following special cases of this proposition deserve

notice.

(i) when the given chords both pass through the centre :

(ii) when one chord passes through the centre, and cuts the

other at right angles :

(iii) when one chord passes through the centre, and cuts the

other obliquely.

In each of these cases the general proof requires some modifica-

tion, which may be left as an exercise to the student.

EXERCISUiS.

1. Tico straiglit ILnen AB, CD intersect at E, so that the rectangle

AE, EB is equal to the rectangle CE, ED: sheio that the four points

A, B, C, D are concyclic.

2. The rectangle contained by the segments of any chord drawn

through a given point within a circle is equal to the square on half

the shortest chord which may be drawn through that point.

3. ABC is a triangle right-angled at C ; and from C a perpen-

dicular CD is drawn to the hypotenuse : shew that the square on CD

is equal to the rectangle AD, DB.

4. ABC is a triangle; and AP, BQ the perpendiculars dropped

from A and B on the opposite sides, intersect at O : shew that the

rectangle AG, OP is equal to the rectangle BO, OQ.

5. Two circles intersect at A and B, and through any point in AB

their common chord two chords are drawn, one in each circle ; shew

that their four extremities are concyclic.

6. A and B are two points within a circle such that the rectangle

contained by the segments of any chord drawn through A is equal to

the rectangle contained by the segments of any chord through B :

shew that A and B are equidistant from the centre.

7. If through E, a point ivithout a circle, two secants EAB, ECD

are drawn; shew that the rectangle EA, EB is equal to the rectangle

EC, ED.

[Proceed as in ni. 35, using ii. 6.]

8. Through A, a point of intersection of two circles, two straight

lines CAE, DAF are drawn, each passing through a centre and termi-

nated by the circumferences: shew that the rectahgle CA, AE is equal

to the rectangle DA, AFÂ»

II. E. 14

210 euclid's elements.

Proposition 36. Theorem.

If from any point without a circle a tangent and a

secant be di'awn, then the rectangle contained by the whole

secant and the part of it loithout the circle shall be equal to

the square on the tangent.

Let ABC be a circle; and from D a point without it, let

there be drawn the secant DC A, and the tangent DB:

then the rect. DA, DC shall be equal to the sq. on DB.

Find E, the centre of the 0ABC: iii. 1.

and from E, draw EF perp. to AD. i. 12.

Join EB, EC, ED.

Then because EF, passing through the centre, is perp.

to the chord AC,

.'. AC is bisected at F. iii. 3.

And since AC is bisected at F and produced to D,

.'.the rect. DA, DC with the sq. on FC-^the sq. on FD. H. 6.

To each of these equals add the sq. on EF :

then the rect. DA, DC witli the sqq. on EF, FC = the sqq. on

EF, FD.

But the sqq. on EF, FC = the sq. on EC ; for EFC is a rt. angle;

= the sq. on EB.

And the sqq. on EF, FD = the sq. on ED ; for EFD is a rt. angle;

= the sqq. on EB, BD; for EBD is a

rt. angle. in. 18.

/. the rect. DA, DC with the sq. on EB=:the sqq. on EB, BD.

From these equals take the sq. on EB:

then the rect. DA, DC = the sq. on DB. Q. e.d.

Note. This proof may easily be adapted to the case when the

Becant passes through the centre of the circle.

]iooK 111. riiop. 3ti.

i>il

Corollary. If from a given point zvithout a circle

any number of secants are drawn, the o'ectanffles contained

hy the whole secants and the ^^ar^s of them without the circle

are all equal ; for each of these rectangles is equal to the

square on the tangent drawn from the given point to the

circle. p

For instance, in the adjoining figure,

each of the rectangles PB, PA and PD, PC

and PF_, PE is equal to the square on the

tangent PQ:

.*. the rect. PB, PA

= the rect. PD, PC

^^ the rect. PF, PE.

Note, llemembcring that the segments into which the chord AB

is divided at P, are the Hnes PA, PB, (see Part I. page 131) we are

enabled to include the corollaries of Propositions 35 and 36 in a

single enunciation.

If any numher of chords of a circle arc draicn through a [jiven

point icithin or without a circle, the rectangles contained hy the

segments of the chords are equal.

EXERCISES.

1. Use this proposition to shew that tangents drawn to a circle

from an external point are equal.

2s If two circles intersect, tangents drawn to them from any

point in their common chord produced are equal.

3. If two circles intersect at A and B, and PQ is a tangent to

both circles; shew that AB produced bisects PQ.

4. If P is any i^oint on the straight line AB produced, shew that

the tangents drawn from P to all circles which pass through A and B

are equal.

5. ABC is a triangle right-angled at C, and from any point P in

AC, a perpendicular PQ is drawn to the hypotenuse: shew that the

rectangle AC, AP is equal to the rectangle AB, AQ.

0. ABC is a triangle right-angled at C, and from C a perpen-

dicular CD is drawn to the hypotenuse; shew that the rect. AB, AD

is equal to the square on AC.

14-2

212 EUCLID'S I : I, K. Mi; NTS.

PkOPOSTTION o7. 'J'llKUUEM.

//* from a jtoi/d /rifhuut a circle there h<' <h-<in-)h two

straight lines, one of tvhich cuts the circle, and the other

tneets it, and if the rectangle contained hy the whole line

lohich cuts the circle and the pa,rt of it vnthout the circle he

equal to the square on the line which meets tlte circle, then

the line which meets the circle shall he a tangcfU to it.

Let ABC be a circle; and from D, a point without it,

let there be drawn two st. lines DCA and DB, of which

DC A cuts the circle at C and A, and DB meets it; and let

the rect. DA, DC =^^ the sq. on DB:

then shall DB be a tangent to the circle.

From D draw DE to touch the Â©ABC: in. 7.

let E be the point of contact.

Find the centre F, and join FB, FD, FE. in. 1.

Then since DCA is a secant, and DE a tangent to the circle,

.'. the rect. DA, DC = the sq. on DE, III. 36.

But, by liypotliesis, the rect. DA, DC = the sq. on DB;

.*. the sq. on DE = the sq. on DB,

.-. DE= DB.

Hence in the A^ DBF, DEF.

[ DB = DE,

Proved.

Because - and BF = EF;

[and DF is common;

III. Def. 1.

.-. the L DBF - the l DEF.

I. 8.

But DEF is a rt. angle ;

III. 18.

.'. DBF is also a it. angle;

and since BF is a radius,

.*. DB touches the Â©ABC at the point

B.

Q. K. U.

NOTE ON thp: method of limits as applied to tangency.

Euclid defines a tangent to a circle as a straight line which meets

the circumference, hut being produced, does not cut it: and from this

definition he deduces the fundamental theorem that a tangent is j)er-

pendicular to the radius drawn to the point of contact. Prop. IG.

But this result may also be established by the Method of Limits,

which regards the tangent as the idtimate position of a secant tohen its

two points of intersection tvith the circumference are brought into coin-

cidence [See Note on page 151]: and it may be shewn that every

theorem relating to the tangent may be derived from some more

general proposition relating to the secant, by considering the ultimate

case when the two points of intersection coincide.

1. To prove by the Method of Limits tliat a tangent to a circle

is at right angles to the radius drawn to the point of contact.

Let ABD be a circle, whose centre

is C; and PABQ a secant cutting the

Online Library → Euclid → A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 13 of 27)