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

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.'. the I. DAC = the l DCA.

But the L DAC is a rt. angle; H^yp-

.*. the L DCA is a rt. angle;

that is, two angles of the A DAC are together equal to two

rt. angles; which is impossible. i. 17.

Hence AE meets the circle at A, but being produced,

does not cut it;

that is, AE is a tangent to the circle, iii. Bef 10.

H. E.

12

178

EUCLID'S ELEMENTS.

(ii) Also through A no other straight Hue but AE can

be drawn so as not to cut the circle.

For, if possible, let AF be another st. line drawn through

A so as not to cut the circle.

From D draw DG perp. to AF; I. 12.

and let DG meet the O^Â® at H.

Then in the A DAG, because the z_ DGA is a rt. angle,

.'. the L DAG is less than a rt. angle; i. 17.

.*. DA is greater than DG. I. 19.

But DA ^ DH, in. Def. 1.

.'. DH is greater than DG,

the part greater than the whole, which is impossible.

.". no st. line can be drawn from the point A, so as not to

cut the circle, except AE.

Corollaries, (i) A tangent touches a circle at one

point only.

(ii) There can he hut one tangent to a ci7'cle at a given

point.

book iii. prop. 17. 179

Proposition 17. Problem.

To draw a tangent to a circle from a givejt i}oint either

o7i, or without the circumference.

Let BCD be the given circle, and A the given point:

it is required to draw from A a tangent to the Â©CDB.

Case I. If the given point A is on the O *'Â®.

Find E, the centre of the circle. iii. 1.

Join EA.

At A draw AK at rt. angles to EA. i. 11.

Then AK being perp. to a diameter at one of its extremities,

is a tangent to the circle. iii. 16.

Case II. If the given point A is without the O ^^^

Find E, the centre of the circle; iii. 1.

and join AE, cutting the BCD at D.

From centre E, with radius EA, describe the Â©AFG.

At D, draw GDF at rt. angles to EA, cutting the 0AFG at

F and G. I. 11.

Join EF, EG, cutting the BCD at B and C,

Join AB, AC.

Then both AB and AC shall be tangents to the 0CDB.

For in the A^ AEB, FED,

AE = FE, being radii of the 0GAF;

Because ^and EB = ED, being radii of the BDC;

and the included angle AEF is common;

.'. the L ABE = the L FDE. i. 4.

12â€”2

180

EUCLID'S ELEMENTS.

But the z. FDE is a rt. angle, Goiistr.

.'. the L ABE is a rt. angle ;

hence AB, being drawn at rt. angles to a diameter at one

of its extremities, is a tangent to the O BCD. iii. 16.

Similarly it may be shewn that AC is a tangent, q. e. f.

Corollary. If two tangents are drawn to a circle from

an external point, then (i) they are equal ; (ii) they subtend

equal angles at the centre ; (iii) they Tnake equal angles with

tlw straight line tvhich joins the given point to the centre.

For, in the above figure,

Since ED is perp. to FG, a chord of the FAG,

.â€¢.DF=DG.

Then in the AÂ« DEF, DEG,

[DE is common to both,

Because ^ and EF = EG ;

[ and DF = DG ;

.*. the L DEF = the l DEG.

Again in the A* AEB, AEC,

I^AE is common to both,

Because < and EB = EC,

[and the l AEB = the l AEC

.-. AB = AC:

and the z_ EAB = the l EAC.

III. 3.

in. Be/. 1.

Proved.

I. 8.

Proved.

I. 4.

Q.E.D.

Note. If the given point A is within the circle, no solution is

possible.

Hence we see that this problem admits of tico solutions, one solu-

tion, or no solution, according as the given point A is without, on, or

within the circumference of a circle.

For a simpler method of drawing a tangent to a circle from a given

point, see page 202.

book iii. prop. 18. 181

Proposition 18. Theorem.

The straight line drawn from the centre of a circle to the

point of contact of a tangent is perpendicular to the tangent.

Let ABC be a circle, of which F is the centre;

and let the st. line DE touch the circle at C :

then shall FC be perp. to DE.

For, if not, suppose FG to be perp. to DE, i. 12.

and let it meet the O^^at B.

Then in the A FCG, because the L FGC is a rt. angle, Hyp.

.'. the L FCG is less than a rt. angle : i. 17.

.'. the L FGC is greater than the z. FCG ;

.*. FC is greater than FG : i. 19.

but FC = FB ;

.'. FB is greater than FG,

the part greater than the whole, which is impossible.

.'. FC cannot be otherwise than perp. to DE :

that is, FC is perp. to DE. q.e.d.

EXERCISES.

1. Draw a tangent to a circle (i) parallel to, (ii) at right angles to

a given straight line.

2. Tangents drawn to a circle from the extremities of a diameter

are parallel.

3. Circles which touch one another internally or externally have a

common tangent at their point of contact.

4. In two concentric circles any chord of the outer circle ivhich

touches the inner, is bisected at the point of contact.

5. In two concentric circles, all chords of the outer circle which

touch the inner, are equal.

182

EUCLID'S ELEMENTS.

Proposition 19. Theorem.

The straight line dratcn perpendicular to a tangent to a

circle from the point of contact joassea through the centre.

D c e

Let ABC be a circle, and DE a tangent to it at tlie point C ;

and let CA be drawn perp. to DE :

then shall CA pass through the centre.

For if not, suppose the centre to be outside CA, as at F.

Join CF.

Then because DE is a tangent to the circle, and FC

is drawn from the centre F to the point of contact,

.*. the L FCE is a rt. angle. iii. 18.

But the z. ACE is a rt. angle ; Hyp.

.'. the ^ FCE = the ^ACE;

the part equal to the whole, which is impossible.

.'. the centre cannot be otherwise than in CA;

that is, CA passes through the centre.

Q.E.D.

exercises on the tangent.

Propositions 16, 17, 18, 19.

1. The centre of any circle which touches two intersecting straight

lines must lie on the bisector of the angle between them.

2. AB and AC are two tangents to a circle whose centre is O;

shew that AC bisects the chord of contact BC at right angles.

BOOK III. PROP. 19. 183

3. If two circles are concentric all tangents drawn from points on

the circumference of the outer to the inner circle are equal.

4. The diameter of a circle bisects all chords which are parallel

to the tangent at either extremity.

5. Find the locus of the centres of all circles which touch a given

straight line at a given point.

6. Find the locus of the centres of all circles which touch each

of two parallel straight lines.

7. Find the locus of the centres of all circles which touch each of

tiDO intersecting straight lines of unlimited length.

8. Describe a circle of given radius to touch two given straight

lines.

9. Through a given point, within or without a circle, draw a

chord equal to a given straight line.

In order that the problem may be possible, between what limits

must the given line lie, when the given point is (i) without the circle,

(ii) within it?

10. Two parallel tangents to a circle intercept on any third tan-

gent a segment which subtends a right angle at the centre.

11. In any quadrilateral circumscribed about a circle, the sum of

one pair of opposite sides is equal to the sum of the other pair.

12. Any parallelogram which can be circumscribed about a circle,

must be equilateral.

13. If a quadrilateral be described about a circle, the angles sub-

tended at the centre by any two opposite sides are together equal to

two right angles.

14. AB is any chord of a circle, AC the diameter through A, and

AD the perpendicular on the tangent at B: shew that AB bisects the

angle DAC.

15. Find the locus of the extremities of tangents of fixed length

drawn to a given circle.

16. In the diameter of a circle produced, determine a point such

that the tangent drawn from it shall be of given length.

17. In the diameter of a circle produced, determine a point such

that the two tangents drawn from it may contain a given angle.

18. Describe a circle that shall pass through a given point, and

touch a given straight line at a given point. [See page 183. Ex. 5.]

19. Describe a circle of given radius, having its centre on a given

straight line, and touching another given straight line.

20. Describe a circle that shall have a given radius, and touch a

given circle and a given straight line. How many such circles can

be drawn?

184 euclid's elements.

Proposition 20. Theorem.

The angle at the centre of a circle is double of an angle

at the circumference y standing on the same arc.

Fig. 2

Let ABC be a circle, of which E is the centre ; and let

BEC be an angle at the centre, and BAG an angle at the O^'',

standing on the same arc BC :

then shall the ^ BEC be double of the l BAC.

Join AE, and produce it to F.

Case I. When the centre E is within the angle BAC.

Then in the A EAB, because EA = EB,

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

.'. the sum of the l^ EAB, EBA = twice the l. EAB.

But the ext. z. BEF = the sum of the l ^ EAB, EBA; i. 32.

.'. the L BEF = twice the z. EAB.

Similarly the l FEC = twice the l EAC.

.'. the sum of the l ^ BEF, FEC == twice the sum of

the z. Â« EAB, EAC ;

that is, the l BEC = twice the l BAC.

Case II. When the centre E is without the z. BAC.

As before, it may be shewn that the z. FEB == twice the z. FAB ;

also the z. FEC = twice the l FAC;

.". the difference of the z. ^ FEC, FEB ^ twice the difference

of the 1. Â« FAC, FAB ;

that is, the z. BEC = twice the z. BAC.

Q.E.D.

BOOK III. TROP. 21.

185

Note. If the arc BFC, on which the angles

stand, is greater than a semi-circumference, it

is clear that the angle BEC at the centre will be

reflex: but it may still be shewn as, in Case I.,

that the reflex z BEC is double of the z BAC

at the o^, standing on the same arc BFC.

Proposition 21. Theorem.

Angles in the same segment of a circle are equal.

il

Let A BCD be a circle, and let BAD, BED be angles in

the same segment BAED:

then shall the ^ BAD = the /.BED.

Find F, the centre of the circle. iii. 1.

Case I. When the segment BAED is greater than a

semicircle.

Join BF, DF.

Then the z. BFD at the centre = twice the .1 BAD at the

O^^, standing on the same arc BD: ill. 20.

and similarly the z. BFD = twice the z. BED. ill. 20.

.'. the L BAD = the z. BED.

Case II. When the segment BAED is not greater than

a semicircle.

OF THE

IVEES

186 EUCLID'S ELEMENTS.

Join AF, and produce it to meet the O''^ at C.

Join EC.

Then since A EDO is a semicircle;

.'. the segment BAEC is greater than a semicircle:

.'. the L BAG = the l BEC, in this segment. Case 1.

Similarly the segment CAED is greater than a semicircle;

.'. the z. CAD = the ^ CED, in this segment.

'. the sum of the z. " BAC, CAD = the sum of the l"" BEC,

CED:

that is, the ^ BAD = the z. BED. Q. E. D.

EXERCISES.

1. P is any point on the arc of a segment of which AB is the

chord. Shew that the sum of the angles PAB, PBA is constant.

2. PQ and RS are two chords of a circle intersecting at X: prove

that the triangles PXS, RXQ are equiangular.

3. Two circles intersect at A and B ; and through A any straight

line PAQ is drawn terminated by the circumferences: shew that PQ

subtends a constant angle at B.

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

straight lines PAQ, XAY are drawn terminated by the circumferences :

shew that the arcs PX, QY subtend equal angles at B.

5. P is any point on the arc of a segment whose chord is AB : and

the angles PAB, PBA are bisected by straight lines which intersect at

O. Find the locus of the point O.

BOOK III. PROP. 21.

18'

Note. If the extension of Proposition 20, given in the note on

page 185, is adopted, a separate treatment of a

the second case of the present proposition is ^x-^^^^^E

unnecessary.

For, as in Case I.,

the reflex z BFD = twice the z BAD; iii.20.

also the reflex i BFD=: twice the z BED;

.-.the z BAD = the z BED.

The converse of Proposition 21 is very important. For the con-

struction used in its proof, viz. To describe a circle abotit a given

triangle, the student is referred to Book iv. Proposition 5. [Or see

Theorems and Examples on Book i. Page 103, No. 1.]

Converse of Proposition 21.

Equal angles standing on the same base, and on the same side of

it, have their vertices on an arc of a circle, of which the given base

is the cliord.

Let BAG, BDC be two equal angles standing

on the same base BC :

then shall the vertices A and D lie upon a

segment of a circle having BC as its chord.

Describe a circle about the a BAG : iv. 5.

then this circle shall pass through D.

For, if not, it must cut BD, or BD produced,

at some other point E.

Join EG.

Then the z BAG = the z BEG, in the same segment:

but the Z BAG = the Z BDG, by hypothesis;

.-. the Z BEG = the Z BDG;

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. 1. 16.

.-. the circle- which passes through B, A, G, cannot pass otherwise

thaTi through D.

That is, the vertices A and D are on an arc of a circle of which

the chord is BG. q. k.d.

The following corollary is important.

All triangles drawn on the same base, and with equal vertical angles,

have their vertices on an arc of a circle, of which the given base is the

cliord.

Ob, The locus of the vertices of triangles drawn on the same base

loith equal vertical angles is an arc of a circle.

III. 21.

188 euclid's elements.

Proposition 22. Theorem.

TJie opposite angles of amy quadrilateral inscribed in a

circle are together equal to two right angles.

Let A BCD be a quadrilateral inscribed in the 0ABC;

then shall, (i) the l ^ ADC, ABC together = two rt. angles ;

(ii) the L^ BAD, BCD together = two rt. angles.

Join AC, BD.

Then the z. ADB = the l ACB, in the segment ADCB; iii. 21.

also the ^ CDB ^ the L. CAB, in the segment CDAB.

.'. the L ADC = the sum of the l. ^ ACB, CAB.

To each of these equals add the z_ ABC:

then the two z. ^ ADC, ABC together = the three z. " ACB,

CAB, ABC.

But the L^ ACB, CAB, ABC, being the angles of a

triangle, together = two rt. angles. I. 32.

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

Similarly it may be shewn that

the L^ BAD, BCD together = two rt. angles.

Q. K. D.

EXERCISES.

1. If a circle can be described about a parallelogram, the

parallelogram must be rectangular.

2. ABC is an isosceles triangle, and XY is drawn parallel to the

base BC: shew that the four points B, C, X, Y lie on a circle.

3. If one side of a quadrilateral iiiscribed in a circle is produced,

the exterior angle is equal to the opposite interior angle of the quadri-

lateral.

BOOK III. PROP. 22. 189

Proposition 22. [Alternative Proof.]

Let ABCD be a quadrilateral inscribed in the Â© ABC :

then shall the L ^ ADC, ABC together = two rt. angles.

Join FA, FC.

Then the Z AFC at the centre = twice the

Z ADC at the C"^, standing on the same arc

ABC. in. 20.

Also the reflex angle AFC at the centre

= twice the Z ABC at the O"^, standing on the

same arc ADC. in. 20.

Hence the Z^ ADC, ABC are together half

the sum of the z AFC and the reflex angle AFC ;

but these make up four rt. angles : 1. 15. Gor. 2.

.-. the z Â» ADC, ABC together = two rt. angles. q.e.d.

Definition. Four or more points through which a circle

may be described are said to be concyclic.

Converse of Proposition 22.

If a 'pair of opposite angles of a quadrilateral are together equal to

tivo right angles, its vertices are concyclic.

Let ABCD be a quadrilateral, in which the opposite angles at

B and D together = two rt. angles;

then shall the four points A, B, C, D be

concyclic.

Through the three points A, B, C describe

a circle : iv. 5.

then this circle must pass through D.

For, if not, it will cut AD, or AD produced,

at some other point E.

Join EC.

Then since the quadrilateral ABCE is inscribed in a circle,

.-. the Z s ABC, AEC together = two rt. angles. m. 22.

But the Z s ABC, ADC together=two rt. angles ; Hyp.

hence the Z Â» ABC, AEC = the Z ^ ABC, ADC.

Take from these equals the Z ABC:

then the Z AEC = the z ADC;

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. 1. 16.

.-. the circle which passes through A, B, C cannot pass otherwise

than through D :

that is the four vertices A, B, C, D are concyclic. q.e.d.

UK) Euclid's elements.

Definition. ISimilar segments of circles arc tliosc which

contain equal angles.

Proposition 23. Theorem.

O71 the same cliord and on the same side of ity there

cannot he two similar segments of circles, not coinciding with

one another.

If possible, on the same chord AB, and on the same

side of it, let there be two similar segments of circles ACB,

ADO, not coinciding with one another.

Then since the arcs ADB, ACB intersect at A and B,

.'. they cannot cut one another again; ill. 10.

.'. one segment falls within the other.

In the outer arc take any point D ;

join AD, cutting the inner arc at C :

join CB, DB.

Then because the segments are similar,

.*. the /L ACB = the ^ADB; iii. Def.

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. i. 16.

Hence the two similar segments ACB, ADB, on the same

chord AB and on the same side of it, must coincide.

Q.E.D.

exercises on proposition 22.

1. The straight lines which bisect any angle of a quadrilateral

figure inscribed in a circle and the opposite exterior angle, meet on

the circumference.

2. A triangle is inscribed in a circle : shew that the sum of the

angles in the three segments exterior to the triangle is equal to four

right angles.

3. Divide a circle into two segment?, so that the angle contained

by the one shall be double of the angle contained by the other.

book iii. prop. 24. 191

Proposition 24. Theorem.

Similar segments of circles on equal chords are equal to

one another.

Let AEB and CFD be similar segments on equal chords

AB, CD:

then shall the segment ABE = the segment CDF.

For if the segment ABE be applied to the segment CDF,

so that A falls on C, and AB falls along CD;

then since AB = CD,

.'. B must coincide with D.

.". the segment AEB must coincide with the segment CFD ;

for if not, on the same chord and on the same side of it

there would be two similar segments of circles, not co-

inciding with one another : which is impossible. iii. 23.

.'. the segment AEB = the segment CFD. q. e.d.

EXERCISES.

1. Of two segments standing on the same chord, the greater

segment contains the smaller angle.

2. A segment of a circle stands on a chord AB, and P is any

point on the same side of AB as the segment: shew that the angle

APB is greater or less than the angle in the segment, according as P

is within or without the segment.

3. P, Q, R are the middle "points of the sides of a triangle,

and X is the foot of the perpendicular let fall from one vertex on the

opposite side : shew that the four points P, Q, R, X are concyclic.

[See page 96, Ex. 2 : also page 100, Ex. 2.]

4. Use the preceding exercise to shew that the middle points of the

sides of a triangle and the feet of the perpendiculars let fall from the

vertices on the opposite sides, are concyclic.

192

Euclid's elements.

Proposition 25. Problem*.

An arc of a circle being given, to describe the whole cir-

cumference of which the given arc is a part.

Let ABC be an arc of a circle :

it is required to describe the whole O^^ of which the arc

ABC is a part.

In the given arc take any three points A, B, C.

Join AB, BC.

Draw DF bisecting AB at rt. angles, i. 10. 11.

and draw EF bisecting BC at rt.

angles.

Then because DF bisects the chord AB at rt. angles,

.'. the centre of the circle lies in DF. iii. 1. Cor

Again, because EF bisects the chord BC at rt. angles,

.'. the centre of the circle lies in EF. iii. 1. Cor.

:. the centre of the circle is F, the only point common to

DF, EF.

Hence the O ^^ of a circle described from centre F, with

radius FA, is that of which the given arc is a part. Q. E. P.

* Note. Euclid gave this proposition a somewhat different form,

as follows :

A segment of a circle being given, to describe the circle of which

it is a segment.

Let ABC be the given segment standing on the chord AC.

Draw DB, bisecting AC at rt. angles. 1. 10. b

Join AB.

At A, in BA, make the i BAE equal to the

ZABD. 1.23.

Let AE meet BD, or BD produced, at E.

Then E shall be the centre of the required circle.

[Join EC ; and prove (i) EA= EC ; i. 4.

(ii)EA=EB. 1. 6.]

BOOK ITT. ]'ROP. 26. 193

Proposition 26. Theorem.

In equal circles the arcs which subtend equal angles^

v:)hetlier at the centres or at the circumferences^ shall he equal.

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

at the centres be equal, and consequently the l ^ BAG, EDF

at the O *'*" equal : iii. 20.

then shall the arc BKG â€” the arc ELF.

Join BG, EF.

Then because the 0^ ABG, DEF are equal,

.'. their radii are equal.

Hence in the AÂ« BGG, EHF,

( BG = EH,

Because â™¦-< andGG = HF,

(and the z. BGG = the ^ EHF ; Hyp.

.-. BG = EF. I. 4.

Again, because the z. BAG = the z. EDF, Hyp.

.'. the segment BAG is similar to the segment EDF;

III. Def. 15.

and they are on equal chords BG, EF;

.'. the segment BAG = the segment EDF. ill. 24.

But the whole ABG = the whole DEF;

.'. the remaining segment BKG = the remaining segment ELF,

.'. the arc BKG =^ the arc ELF.

Q.E, D.

[For an Alternative Prdof and Exercises see pp. 197, 198.]

H. E. 1 3

194 EUCLID'S ET.EMKXTS

Proposttton 27. Tftkorkm.

In equal circles the angles, v^hefher at the centres or the

circumferences, which stand on equal arcs, shall he equal.

Let ABC, DEF be equal circles,

and let the arc BC = the arc EF:

then shall the z. BGC = the l EHF, at the centres;

and also the l BAG = the l EDF, at the O *'*'".

If the L^ BGG, EHF are not equal, one must be the

greater.

If possible, let the z. BGG be the greater.

At G, in BG, make the z. BGK equal to the L EHF. I. 23,

Then because in the equal 0Â« ABC, DEF,

the /L BGK = the l EHF, at the centres; Constr.

.'. the arc BK = the arc EF. in. 26.

But tlie arc BC = tlie arc EF, Ifi/p.

.*. the arc BK =- the arc BC,

a part equal to tlie whole, which is impossible.

.'. the L BGG is not unequal to the z. EHF; â€¢

that is, the z. BGG = the l EHF.

And since the l BAG at the O*"" is half tlie l. BGG at tlie

centre, in. 20.

and likewise the l EDF is half the l EHF,

.'. the ^ BAG = the ^i EDF. Q.E.D.

[For Exercises see pp. 197, 198.]

BOOK in. PROP. 28. 195

Proposition 28. Theorem.

In equal circles the arcs, which are cut of hy equal

chords, shall be equal, the major arc equal to the major arc,

and the rninor to the minor.

Let ABC, DEF be two equal circles,

and let the chord BC â€” the chord EF :

then shall the major arc BAG = the major arc EDF;

and the minor arc BGC = the minor arc EHF.

Find K and L the centres of the 0'' ABG, DEF: iii. 1.

and join BK, KG, EL, LF.

Then because the 0* ABG, DEF are equal,

.'. their radii are equal.

Hence in the AÂ« BKG, ELF,

( BK = EL,

Because < KG = LF,

(and BG = EF; J^l/P-

.'. the ^ BKG = the ^ELF; 1.8.

.". the arc BGG = the arc EHF; iii. 26.

and these are the minor arcs.

But the whole O ''^ ABGG ^ the whole 0^Â« DEHF; Hyp.

.*. the remaining arc BAG = the remaining arc EDF:

and these are the major arcs. Q. e.d.

[For Exercises see pp. 197, 198.]

13-2

196 Kl'Cl.lhV IMKMKNTS.

Proposition 29. Theorem.

In equal circles tlie chordsy which cut off equal arcs, shall

he equal.

A D

Let ABC, DEF be equal circles,

and let the arc BGC = the arc EHF:

then shall the chord BC = the chord EF.

Find K, L the centres of the circles. in. 1.

Join BK, KC, EL, LF.

Then in the equal 0Â« ABC, DEF,

because the arc BGC = the arc EHF,

.'. the L BKC - the l ELF. in. 27.

Hence in the AÂ« BKC, ELF,

I' BK = EL, being radii of equal circles;

Because - KC â€” LF, for the same reason,

[and the ^ BKC = the L ELF; Proved.

:. BC = EF. I. 4.

Q. E. D.

EXERCISES

ON PROPOSITIONS 26, 27.

1. If two chords of a circle are parallel, they intercept equal arcs.

2. The straight lines, which join the extremities of two equal

arcs of a circle towards the same parts, are parallel.

3. In a circle, or in e-vinl circles, sectors are equal if their angles

at the centres (in- cqiud.

EXERCISES ON TROPS. 28, 29. 197

4. If two chords of a circle intersect at right angles, the opposite

arcs are together equal to a semicircumference.

5. If two chords intersect loithin a circle, they form an angle

equal to that subtended at the circuviference by the sum of the arcs they

cut off.

6. Jf tivo chords intersect xoithout a circle, they form an anyle

equal to that sribtended at the circumference by the difference of the arcs

they cut off.

7. If AB is a fixed chord of a circle, and P any point on one

But the L DAC is a rt. angle; H^yp-

.*. the L DCA is a rt. angle;

that is, two angles of the A DAC are together equal to two

rt. angles; which is impossible. i. 17.

Hence AE meets the circle at A, but being produced,

does not cut it;

that is, AE is a tangent to the circle, iii. Bef 10.

H. E.

12

178

EUCLID'S ELEMENTS.

(ii) Also through A no other straight Hue but AE can

be drawn so as not to cut the circle.

For, if possible, let AF be another st. line drawn through

A so as not to cut the circle.

From D draw DG perp. to AF; I. 12.

and let DG meet the O^Â® at H.

Then in the A DAG, because the z_ DGA is a rt. angle,

.'. the L DAG is less than a rt. angle; i. 17.

.*. DA is greater than DG. I. 19.

But DA ^ DH, in. Def. 1.

.'. DH is greater than DG,

the part greater than the whole, which is impossible.

.". no st. line can be drawn from the point A, so as not to

cut the circle, except AE.

Corollaries, (i) A tangent touches a circle at one

point only.

(ii) There can he hut one tangent to a ci7'cle at a given

point.

book iii. prop. 17. 179

Proposition 17. Problem.

To draw a tangent to a circle from a givejt i}oint either

o7i, or without the circumference.

Let BCD be the given circle, and A the given point:

it is required to draw from A a tangent to the Â©CDB.

Case I. If the given point A is on the O *'Â®.

Find E, the centre of the circle. iii. 1.

Join EA.

At A draw AK at rt. angles to EA. i. 11.

Then AK being perp. to a diameter at one of its extremities,

is a tangent to the circle. iii. 16.

Case II. If the given point A is without the O ^^^

Find E, the centre of the circle; iii. 1.

and join AE, cutting the BCD at D.

From centre E, with radius EA, describe the Â©AFG.

At D, draw GDF at rt. angles to EA, cutting the 0AFG at

F and G. I. 11.

Join EF, EG, cutting the BCD at B and C,

Join AB, AC.

Then both AB and AC shall be tangents to the 0CDB.

For in the A^ AEB, FED,

AE = FE, being radii of the 0GAF;

Because ^and EB = ED, being radii of the BDC;

and the included angle AEF is common;

.'. the L ABE = the L FDE. i. 4.

12â€”2

180

EUCLID'S ELEMENTS.

But the z. FDE is a rt. angle, Goiistr.

.'. the L ABE is a rt. angle ;

hence AB, being drawn at rt. angles to a diameter at one

of its extremities, is a tangent to the O BCD. iii. 16.

Similarly it may be shewn that AC is a tangent, q. e. f.

Corollary. If two tangents are drawn to a circle from

an external point, then (i) they are equal ; (ii) they subtend

equal angles at the centre ; (iii) they Tnake equal angles with

tlw straight line tvhich joins the given point to the centre.

For, in the above figure,

Since ED is perp. to FG, a chord of the FAG,

.â€¢.DF=DG.

Then in the AÂ« DEF, DEG,

[DE is common to both,

Because ^ and EF = EG ;

[ and DF = DG ;

.*. the L DEF = the l DEG.

Again in the A* AEB, AEC,

I^AE is common to both,

Because < and EB = EC,

[and the l AEB = the l AEC

.-. AB = AC:

and the z_ EAB = the l EAC.

III. 3.

in. Be/. 1.

Proved.

I. 8.

Proved.

I. 4.

Q.E.D.

Note. If the given point A is within the circle, no solution is

possible.

Hence we see that this problem admits of tico solutions, one solu-

tion, or no solution, according as the given point A is without, on, or

within the circumference of a circle.

For a simpler method of drawing a tangent to a circle from a given

point, see page 202.

book iii. prop. 18. 181

Proposition 18. Theorem.

The straight line drawn from the centre of a circle to the

point of contact of a tangent is perpendicular to the tangent.

Let ABC be a circle, of which F is the centre;

and let the st. line DE touch the circle at C :

then shall FC be perp. to DE.

For, if not, suppose FG to be perp. to DE, i. 12.

and let it meet the O^^at B.

Then in the A FCG, because the L FGC is a rt. angle, Hyp.

.'. the L FCG is less than a rt. angle : i. 17.

.'. the L FGC is greater than the z. FCG ;

.*. FC is greater than FG : i. 19.

but FC = FB ;

.'. FB is greater than FG,

the part greater than the whole, which is impossible.

.'. FC cannot be otherwise than perp. to DE :

that is, FC is perp. to DE. q.e.d.

EXERCISES.

1. Draw a tangent to a circle (i) parallel to, (ii) at right angles to

a given straight line.

2. Tangents drawn to a circle from the extremities of a diameter

are parallel.

3. Circles which touch one another internally or externally have a

common tangent at their point of contact.

4. In two concentric circles any chord of the outer circle ivhich

touches the inner, is bisected at the point of contact.

5. In two concentric circles, all chords of the outer circle which

touch the inner, are equal.

182

EUCLID'S ELEMENTS.

Proposition 19. Theorem.

The straight line dratcn perpendicular to a tangent to a

circle from the point of contact joassea through the centre.

D c e

Let ABC be a circle, and DE a tangent to it at tlie point C ;

and let CA be drawn perp. to DE :

then shall CA pass through the centre.

For if not, suppose the centre to be outside CA, as at F.

Join CF.

Then because DE is a tangent to the circle, and FC

is drawn from the centre F to the point of contact,

.*. the L FCE is a rt. angle. iii. 18.

But the z. ACE is a rt. angle ; Hyp.

.'. the ^ FCE = the ^ACE;

the part equal to the whole, which is impossible.

.'. the centre cannot be otherwise than in CA;

that is, CA passes through the centre.

Q.E.D.

exercises on the tangent.

Propositions 16, 17, 18, 19.

1. The centre of any circle which touches two intersecting straight

lines must lie on the bisector of the angle between them.

2. AB and AC are two tangents to a circle whose centre is O;

shew that AC bisects the chord of contact BC at right angles.

BOOK III. PROP. 19. 183

3. If two circles are concentric all tangents drawn from points on

the circumference of the outer to the inner circle are equal.

4. The diameter of a circle bisects all chords which are parallel

to the tangent at either extremity.

5. Find the locus of the centres of all circles which touch a given

straight line at a given point.

6. Find the locus of the centres of all circles which touch each

of two parallel straight lines.

7. Find the locus of the centres of all circles which touch each of

tiDO intersecting straight lines of unlimited length.

8. Describe a circle of given radius to touch two given straight

lines.

9. Through a given point, within or without a circle, draw a

chord equal to a given straight line.

In order that the problem may be possible, between what limits

must the given line lie, when the given point is (i) without the circle,

(ii) within it?

10. Two parallel tangents to a circle intercept on any third tan-

gent a segment which subtends a right angle at the centre.

11. In any quadrilateral circumscribed about a circle, the sum of

one pair of opposite sides is equal to the sum of the other pair.

12. Any parallelogram which can be circumscribed about a circle,

must be equilateral.

13. If a quadrilateral be described about a circle, the angles sub-

tended at the centre by any two opposite sides are together equal to

two right angles.

14. AB is any chord of a circle, AC the diameter through A, and

AD the perpendicular on the tangent at B: shew that AB bisects the

angle DAC.

15. Find the locus of the extremities of tangents of fixed length

drawn to a given circle.

16. In the diameter of a circle produced, determine a point such

that the tangent drawn from it shall be of given length.

17. In the diameter of a circle produced, determine a point such

that the two tangents drawn from it may contain a given angle.

18. Describe a circle that shall pass through a given point, and

touch a given straight line at a given point. [See page 183. Ex. 5.]

19. Describe a circle of given radius, having its centre on a given

straight line, and touching another given straight line.

20. Describe a circle that shall have a given radius, and touch a

given circle and a given straight line. How many such circles can

be drawn?

184 euclid's elements.

Proposition 20. Theorem.

The angle at the centre of a circle is double of an angle

at the circumference y standing on the same arc.

Fig. 2

Let ABC be a circle, of which E is the centre ; and let

BEC be an angle at the centre, and BAG an angle at the O^'',

standing on the same arc BC :

then shall the ^ BEC be double of the l BAC.

Join AE, and produce it to F.

Case I. When the centre E is within the angle BAC.

Then in the A EAB, because EA = EB,

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

.'. the sum of the l^ EAB, EBA = twice the l. EAB.

But the ext. z. BEF = the sum of the l ^ EAB, EBA; i. 32.

.'. the L BEF = twice the z. EAB.

Similarly the l FEC = twice the l EAC.

.'. the sum of the l ^ BEF, FEC == twice the sum of

the z. Â« EAB, EAC ;

that is, the l BEC = twice the l BAC.

Case II. When the centre E is without the z. BAC.

As before, it may be shewn that the z. FEB == twice the z. FAB ;

also the z. FEC = twice the l FAC;

.". the difference of the z. ^ FEC, FEB ^ twice the difference

of the 1. Â« FAC, FAB ;

that is, the z. BEC = twice the z. BAC.

Q.E.D.

BOOK III. TROP. 21.

185

Note. If the arc BFC, on which the angles

stand, is greater than a semi-circumference, it

is clear that the angle BEC at the centre will be

reflex: but it may still be shewn as, in Case I.,

that the reflex z BEC is double of the z BAC

at the o^, standing on the same arc BFC.

Proposition 21. Theorem.

Angles in the same segment of a circle are equal.

il

Let A BCD be a circle, and let BAD, BED be angles in

the same segment BAED:

then shall the ^ BAD = the /.BED.

Find F, the centre of the circle. iii. 1.

Case I. When the segment BAED is greater than a

semicircle.

Join BF, DF.

Then the z. BFD at the centre = twice the .1 BAD at the

O^^, standing on the same arc BD: ill. 20.

and similarly the z. BFD = twice the z. BED. ill. 20.

.'. the L BAD = the z. BED.

Case II. When the segment BAED is not greater than

a semicircle.

OF THE

IVEES

186 EUCLID'S ELEMENTS.

Join AF, and produce it to meet the O''^ at C.

Join EC.

Then since A EDO is a semicircle;

.'. the segment BAEC is greater than a semicircle:

.'. the L BAG = the l BEC, in this segment. Case 1.

Similarly the segment CAED is greater than a semicircle;

.'. the z. CAD = the ^ CED, in this segment.

'. the sum of the z. " BAC, CAD = the sum of the l"" BEC,

CED:

that is, the ^ BAD = the z. BED. Q. E. D.

EXERCISES.

1. P is any point on the arc of a segment of which AB is the

chord. Shew that the sum of the angles PAB, PBA is constant.

2. PQ and RS are two chords of a circle intersecting at X: prove

that the triangles PXS, RXQ are equiangular.

3. Two circles intersect at A and B ; and through A any straight

line PAQ is drawn terminated by the circumferences: shew that PQ

subtends a constant angle at B.

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

straight lines PAQ, XAY are drawn terminated by the circumferences :

shew that the arcs PX, QY subtend equal angles at B.

5. P is any point on the arc of a segment whose chord is AB : and

the angles PAB, PBA are bisected by straight lines which intersect at

O. Find the locus of the point O.

BOOK III. PROP. 21.

18'

Note. If the extension of Proposition 20, given in the note on

page 185, is adopted, a separate treatment of a

the second case of the present proposition is ^x-^^^^^E

unnecessary.

For, as in Case I.,

the reflex z BFD = twice the z BAD; iii.20.

also the reflex i BFD=: twice the z BED;

.-.the z BAD = the z BED.

The converse of Proposition 21 is very important. For the con-

struction used in its proof, viz. To describe a circle abotit a given

triangle, the student is referred to Book iv. Proposition 5. [Or see

Theorems and Examples on Book i. Page 103, No. 1.]

Converse of Proposition 21.

Equal angles standing on the same base, and on the same side of

it, have their vertices on an arc of a circle, of which the given base

is the cliord.

Let BAG, BDC be two equal angles standing

on the same base BC :

then shall the vertices A and D lie upon a

segment of a circle having BC as its chord.

Describe a circle about the a BAG : iv. 5.

then this circle shall pass through D.

For, if not, it must cut BD, or BD produced,

at some other point E.

Join EG.

Then the z BAG = the z BEG, in the same segment:

but the Z BAG = the Z BDG, by hypothesis;

.-. the Z BEG = the Z BDG;

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. 1. 16.

.-. the circle- which passes through B, A, G, cannot pass otherwise

thaTi through D.

That is, the vertices A and D are on an arc of a circle of which

the chord is BG. q. k.d.

The following corollary is important.

All triangles drawn on the same base, and with equal vertical angles,

have their vertices on an arc of a circle, of which the given base is the

cliord.

Ob, The locus of the vertices of triangles drawn on the same base

loith equal vertical angles is an arc of a circle.

III. 21.

188 euclid's elements.

Proposition 22. Theorem.

TJie opposite angles of amy quadrilateral inscribed in a

circle are together equal to two right angles.

Let A BCD be a quadrilateral inscribed in the 0ABC;

then shall, (i) the l ^ ADC, ABC together = two rt. angles ;

(ii) the L^ BAD, BCD together = two rt. angles.

Join AC, BD.

Then the z. ADB = the l ACB, in the segment ADCB; iii. 21.

also the ^ CDB ^ the L. CAB, in the segment CDAB.

.'. the L ADC = the sum of the l. ^ ACB, CAB.

To each of these equals add the z_ ABC:

then the two z. ^ ADC, ABC together = the three z. " ACB,

CAB, ABC.

But the L^ ACB, CAB, ABC, being the angles of a

triangle, together = two rt. angles. I. 32.

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

Similarly it may be shewn that

the L^ BAD, BCD together = two rt. angles.

Q. K. D.

EXERCISES.

1. If a circle can be described about a parallelogram, the

parallelogram must be rectangular.

2. ABC is an isosceles triangle, and XY is drawn parallel to the

base BC: shew that the four points B, C, X, Y lie on a circle.

3. If one side of a quadrilateral iiiscribed in a circle is produced,

the exterior angle is equal to the opposite interior angle of the quadri-

lateral.

BOOK III. PROP. 22. 189

Proposition 22. [Alternative Proof.]

Let ABCD be a quadrilateral inscribed in the Â© ABC :

then shall the L ^ ADC, ABC together = two rt. angles.

Join FA, FC.

Then the Z AFC at the centre = twice the

Z ADC at the C"^, standing on the same arc

ABC. in. 20.

Also the reflex angle AFC at the centre

= twice the Z ABC at the O"^, standing on the

same arc ADC. in. 20.

Hence the Z^ ADC, ABC are together half

the sum of the z AFC and the reflex angle AFC ;

but these make up four rt. angles : 1. 15. Gor. 2.

.-. the z Â» ADC, ABC together = two rt. angles. q.e.d.

Definition. Four or more points through which a circle

may be described are said to be concyclic.

Converse of Proposition 22.

If a 'pair of opposite angles of a quadrilateral are together equal to

tivo right angles, its vertices are concyclic.

Let ABCD be a quadrilateral, in which the opposite angles at

B and D together = two rt. angles;

then shall the four points A, B, C, D be

concyclic.

Through the three points A, B, C describe

a circle : iv. 5.

then this circle must pass through D.

For, if not, it will cut AD, or AD produced,

at some other point E.

Join EC.

Then since the quadrilateral ABCE is inscribed in a circle,

.-. the Z s ABC, AEC together = two rt. angles. m. 22.

But the Z s ABC, ADC together=two rt. angles ; Hyp.

hence the Z Â» ABC, AEC = the Z ^ ABC, ADC.

Take from these equals the Z ABC:

then the Z AEC = the z ADC;

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. 1. 16.

.-. the circle which passes through A, B, C cannot pass otherwise

than through D :

that is the four vertices A, B, C, D are concyclic. q.e.d.

UK) Euclid's elements.

Definition. ISimilar segments of circles arc tliosc which

contain equal angles.

Proposition 23. Theorem.

O71 the same cliord and on the same side of ity there

cannot he two similar segments of circles, not coinciding with

one another.

If possible, on the same chord AB, and on the same

side of it, let there be two similar segments of circles ACB,

ADO, not coinciding with one another.

Then since the arcs ADB, ACB intersect at A and B,

.'. they cannot cut one another again; ill. 10.

.'. one segment falls within the other.

In the outer arc take any point D ;

join AD, cutting the inner arc at C :

join CB, DB.

Then because the segments are similar,

.*. the /L ACB = the ^ADB; iii. Def.

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. i. 16.

Hence the two similar segments ACB, ADB, on the same

chord AB and on the same side of it, must coincide.

Q.E.D.

exercises on proposition 22.

1. The straight lines which bisect any angle of a quadrilateral

figure inscribed in a circle and the opposite exterior angle, meet on

the circumference.

2. A triangle is inscribed in a circle : shew that the sum of the

angles in the three segments exterior to the triangle is equal to four

right angles.

3. Divide a circle into two segment?, so that the angle contained

by the one shall be double of the angle contained by the other.

book iii. prop. 24. 191

Proposition 24. Theorem.

Similar segments of circles on equal chords are equal to

one another.

Let AEB and CFD be similar segments on equal chords

AB, CD:

then shall the segment ABE = the segment CDF.

For if the segment ABE be applied to the segment CDF,

so that A falls on C, and AB falls along CD;

then since AB = CD,

.'. B must coincide with D.

.". the segment AEB must coincide with the segment CFD ;

for if not, on the same chord and on the same side of it

there would be two similar segments of circles, not co-

inciding with one another : which is impossible. iii. 23.

.'. the segment AEB = the segment CFD. q. e.d.

EXERCISES.

1. Of two segments standing on the same chord, the greater

segment contains the smaller angle.

2. A segment of a circle stands on a chord AB, and P is any

point on the same side of AB as the segment: shew that the angle

APB is greater or less than the angle in the segment, according as P

is within or without the segment.

3. P, Q, R are the middle "points of the sides of a triangle,

and X is the foot of the perpendicular let fall from one vertex on the

opposite side : shew that the four points P, Q, R, X are concyclic.

[See page 96, Ex. 2 : also page 100, Ex. 2.]

4. Use the preceding exercise to shew that the middle points of the

sides of a triangle and the feet of the perpendiculars let fall from the

vertices on the opposite sides, are concyclic.

192

Euclid's elements.

Proposition 25. Problem*.

An arc of a circle being given, to describe the whole cir-

cumference of which the given arc is a part.

Let ABC be an arc of a circle :

it is required to describe the whole O^^ of which the arc

ABC is a part.

In the given arc take any three points A, B, C.

Join AB, BC.

Draw DF bisecting AB at rt. angles, i. 10. 11.

and draw EF bisecting BC at rt.

angles.

Then because DF bisects the chord AB at rt. angles,

.'. the centre of the circle lies in DF. iii. 1. Cor

Again, because EF bisects the chord BC at rt. angles,

.'. the centre of the circle lies in EF. iii. 1. Cor.

:. the centre of the circle is F, the only point common to

DF, EF.

Hence the O ^^ of a circle described from centre F, with

radius FA, is that of which the given arc is a part. Q. E. P.

* Note. Euclid gave this proposition a somewhat different form,

as follows :

A segment of a circle being given, to describe the circle of which

it is a segment.

Let ABC be the given segment standing on the chord AC.

Draw DB, bisecting AC at rt. angles. 1. 10. b

Join AB.

At A, in BA, make the i BAE equal to the

ZABD. 1.23.

Let AE meet BD, or BD produced, at E.

Then E shall be the centre of the required circle.

[Join EC ; and prove (i) EA= EC ; i. 4.

(ii)EA=EB. 1. 6.]

BOOK ITT. ]'ROP. 26. 193

Proposition 26. Theorem.

In equal circles the arcs which subtend equal angles^

v:)hetlier at the centres or at the circumferences^ shall he equal.

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

at the centres be equal, and consequently the l ^ BAG, EDF

at the O *'*" equal : iii. 20.

then shall the arc BKG â€” the arc ELF.

Join BG, EF.

Then because the 0^ ABG, DEF are equal,

.'. their radii are equal.

Hence in the AÂ« BGG, EHF,

( BG = EH,

Because â™¦-< andGG = HF,

(and the z. BGG = the ^ EHF ; Hyp.

.-. BG = EF. I. 4.

Again, because the z. BAG = the z. EDF, Hyp.

.'. the segment BAG is similar to the segment EDF;

III. Def. 15.

and they are on equal chords BG, EF;

.'. the segment BAG = the segment EDF. ill. 24.

But the whole ABG = the whole DEF;

.'. the remaining segment BKG = the remaining segment ELF,

.'. the arc BKG =^ the arc ELF.

Q.E, D.

[For an Alternative Prdof and Exercises see pp. 197, 198.]

H. E. 1 3

194 EUCLID'S ET.EMKXTS

Proposttton 27. Tftkorkm.

In equal circles the angles, v^hefher at the centres or the

circumferences, which stand on equal arcs, shall he equal.

Let ABC, DEF be equal circles,

and let the arc BC = the arc EF:

then shall the z. BGC = the l EHF, at the centres;

and also the l BAG = the l EDF, at the O *'*'".

If the L^ BGG, EHF are not equal, one must be the

greater.

If possible, let the z. BGG be the greater.

At G, in BG, make the z. BGK equal to the L EHF. I. 23,

Then because in the equal 0Â« ABC, DEF,

the /L BGK = the l EHF, at the centres; Constr.

.'. the arc BK = the arc EF. in. 26.

But tlie arc BC = tlie arc EF, Ifi/p.

.*. the arc BK =- the arc BC,

a part equal to tlie whole, which is impossible.

.'. the L BGG is not unequal to the z. EHF; â€¢

that is, the z. BGG = the l EHF.

And since the l BAG at the O*"" is half tlie l. BGG at tlie

centre, in. 20.

and likewise the l EDF is half the l EHF,

.'. the ^ BAG = the ^i EDF. Q.E.D.

[For Exercises see pp. 197, 198.]

BOOK in. PROP. 28. 195

Proposition 28. Theorem.

In equal circles the arcs, which are cut of hy equal

chords, shall be equal, the major arc equal to the major arc,

and the rninor to the minor.

Let ABC, DEF be two equal circles,

and let the chord BC â€” the chord EF :

then shall the major arc BAG = the major arc EDF;

and the minor arc BGC = the minor arc EHF.

Find K and L the centres of the 0'' ABG, DEF: iii. 1.

and join BK, KG, EL, LF.

Then because the 0* ABG, DEF are equal,

.'. their radii are equal.

Hence in the AÂ« BKG, ELF,

( BK = EL,

Because < KG = LF,

(and BG = EF; J^l/P-

.'. the ^ BKG = the ^ELF; 1.8.

.". the arc BGG = the arc EHF; iii. 26.

and these are the minor arcs.

But the whole O ''^ ABGG ^ the whole 0^Â« DEHF; Hyp.

.*. the remaining arc BAG = the remaining arc EDF:

and these are the major arcs. Q. e.d.

[For Exercises see pp. 197, 198.]

13-2

196 Kl'Cl.lhV IMKMKNTS.

Proposition 29. Theorem.

In equal circles tlie chordsy which cut off equal arcs, shall

he equal.

A D

Let ABC, DEF be equal circles,

and let the arc BGC = the arc EHF:

then shall the chord BC = the chord EF.

Find K, L the centres of the circles. in. 1.

Join BK, KC, EL, LF.

Then in the equal 0Â« ABC, DEF,

because the arc BGC = the arc EHF,

.'. the L BKC - the l ELF. in. 27.

Hence in the AÂ« BKC, ELF,

I' BK = EL, being radii of equal circles;

Because - KC â€” LF, for the same reason,

[and the ^ BKC = the L ELF; Proved.

:. BC = EF. I. 4.

Q. E. D.

EXERCISES

ON PROPOSITIONS 26, 27.

1. If two chords of a circle are parallel, they intercept equal arcs.

2. The straight lines, which join the extremities of two equal

arcs of a circle towards the same parts, are parallel.

3. In a circle, or in e-vinl circles, sectors are equal if their angles

at the centres (in- cqiud.

EXERCISES ON TROPS. 28, 29. 197

4. If two chords of a circle intersect at right angles, the opposite

arcs are together equal to a semicircumference.

5. If two chords intersect loithin a circle, they form an angle

equal to that subtended at the circuviference by the sum of the arcs they

cut off.

6. Jf tivo chords intersect xoithout a circle, they form an anyle

equal to that sribtended at the circumference by the difference of the arcs

they cut off.

7. If AB is a fixed chord of a circle, and P any point on one

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