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

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42. Given the base, the vertical angle, and the length of the

bisector of the vertical angle : construct the triangle.

43. Given the perpendicular from the vertex on the base, the

bisector of the vertical angle, and the median which bisects the base :

construct the triangle.

44. Given the bisector of the vertical angle, the median bisect-

ing the base, and the difference of the angles at the base : construct the

triangle.

BOOK IV.

Book IV. consists entirely of problems, dealing with

various rectilineal figures in relation to the circles which

pass through their angular points, or are touched by their

sides.

Definitions.

1. A Polygon is a rectilineal figure bounded by more

than four sides.

A Polygon of Jive sides is ca^

â€ž six sides

,5 seven sides

,, eight sides

,, ten sides

â€ž twelve sides

3) fifteen sides

led a Pentagon,

Hexagon,

Heptagon,

Octagon,

Decagon,

Dodecagon,

Quindecagon.

2. A Polygon is Regular when all its sides are equal,

and all its angles are equal.

3. A rectilineal figure is said to be _

inscribed in a circle, when all its angular

points are on the circumference of the circle :

and a circle is said to be circumscribed

about a rectilineal figure, when the circum-

ference of the circle passes through all the

angular points of the figure.

4. A rectilineal figure is said to be

circumscribed about a circle, when each side

of the figure is a tangent to the circle :

and a circle is said to be inscribed in a recti-

lineal figure, when the circumference of the

circle is touched by each side of the figure.

5. A straight line is said to be placed in a circle, wlien

its extremities are on the circumference of the circle.

BOOK iV. PROP. 1. 251

Proposition 1. Problem.

In a given circle to i^lace a chord equal to a given

straight line, which is not greater than the diameter of the

circle.

Let ABC be the given circle, and D the given straight

line not greater tlian the diameter of the circle :

it is required to place in the 0ABC a chord equal to D.

Draw CB, a diameter of the 0ABC.

Then if CB = D, the thing required is done.

But if not, CB must be greater than D. Hyp-

From CB cut off CE equal to D : I. 3.

and from centre C, with radius CE, describe the 0AEF,

cutting the given circle at A.

Join CA.

Then CA shall be the chord required.

For CA = CE, being radii of the AEF :

and CE = D : Constr.

:. CA= D.

Q. E. F.

EXERCISES.

1. In a given circle place a chord of given length so as to pass

through a given point (i) without, (ii) within the circle.

When is this problem impossible ?

2. In a given circle place a chord of given length so that it may

be parallel to a given straight line.

262 kuclid'h elements.

Proposition 2. Problem.

In a given circle to inscribe a triangle- equiangvZar to a

given triangle.

Let ABC be the given circle, and DEF the given triangle:

it is required to inscribe in the ABC a triangle equiangular

to the A DEF.

At any point A, on the O*^^ of the 0ABC, draw the

tangent GAH. ill. 17.

At A make the z. GAB equal to the l DFE ; i. 23.

and make the l HAC equal to the L DEF. i. 23.

Join BC.

Then ABC shall be the triangle required.

Because GH is a tangent to the 0ABC, and from A its

point of contact the chord AB is drawn,

.'. the L GAB = the L ACB in tlie alt. segment: iii. 32.

.'. the L ACB = the l DFE. Constr.

Similarly the z. HAC = the l ABC, in the alt. segment:

.*. the z. ABC = the z. DEF. Constr.

Hence the third l BAG = the third z. EDF,

for the three angles in each triangle are together equal to

two rt. angles, i. 32.

.'. the A ABC is equiangular to the A DEF, and it is

inscribed in the 0ABC.

Q. E. F.

BOOK IV. PROP. 3. 253

Proposition 3. Problem.

About a given circle to circumscribe a triangle equi-

angular to a given triangle.

G E F H

M B

Let ABC be the given circle, and DEF the given A :

it is required to circumscribe about the Â©ABC a triangle

equiangular to the A DEF.

Produce EF both ways to G and H.

Find K the centre of the Â©ABC, iii. 1.

and draw any radius KB.

At K make the l BKA equal to the ^ DEG ; i. 23,

and make the z. BKG equal to the ^ DFH.

Through A, B, C draw LM, MN, NL perp. to KA, KB, KG.

Then LMN shall be the triangle required.

Because LM, MN, NL are drawn perp. to radii at their

extremities,

.'. LM, MN, NL are tangents to the circle. lii. 16.

And because the four angles of the quadrilateral AKBM

together = four rt. angles ; i. 32. Cor.

and of these, the l^ KAM, KBM, are rt. angles; Constr.

.'. the L^ AKB, AMB, together = two rt. angles.

But the L^ DEG, DEF together = two rt. angles; i. 13.

.*. the L^ AKB, AMB -the L^ DEG, DEF;

and of these, the l AKB = the ^ DEG ; Constr.

:. the ^AMB=.the L DEF.

Similarly it may be shewn that the ^ LNM = the lDFE.

.â€¢. the third z. MLN =the third l EDF. i. 32.

.'. the A LMN is equiangular to the A DEF, and it is

circumscribed about the Â©ABC. (J.B. f.

â€¢io4

KtiCl.lDS Kl.hi.MKN

Proposition 4. Problem.

To inscribe a circle in a give^i tricvagle.

Let ABC be the given triangle :

it is required to inscribe a circle in the A ABC.

Bisect the l ^ ABC, ACB by the st. lines Bl, CI, which

intersect at I. i. 9.

From I draw IE, IF, IG perp. to AB, BC, CA. i. 12.

Then in the AÂ« EIB, FIB,

i the z. EBI = the z. FBI ; Constr.

Because -land the z. BEI = the z. BFI, being rt. angles ;

( and Bl is common ;

.-. IE=:|F. I. 26.

Similarly it may be shewn that IF = IG.

.'. IE, IF, IG are all equal.

From centre I, with radius IE, describe a circle:

this circle must pass through the points E, F, G ;

and it will be inscribed in the A ABC.

For since IE, IF, IG are radii of the EFG ;

and since the z. ^ at E, F, G are rt. angles ;

.'. the 0EFG is touched at these points by AB, BC, CA:

III. 16.

.'. the 0EFQ is inscribed in the A ABC.

Q. E. F.

NotÂ«. From page 103 it is seen tliat if A I be joined, then A I

bisects the angle BAC.

BOOK IV. PROP. 4.

255

Hence it follows that the bisectors of the angles of a triangle are

concurrent, the point of intersection being the centre of the inscribed

circle.

The centre of the circle inscribed in a triangle is sometimes called

its in-centre.

Definitiox.

From Ij draw I^G

Because

A circle whicli touches one side of a triangle and the

other two sides produced is said to be an escribed circle of

* the triangle.

To draio an escribed circle of a given triangle.

Let ABC be the given triangle, of which

the two sides AB, AC are produced to E

and F:

it is required to describe a circle touching

BC, and AB, AC produced.

Bisect the z ** CBE, BCF by the st. lines

which intersect at I^, i. 9.

I^H, I^K perp. to AE,

BC, AF. I. 12.

Then in the a' I^BG, I^BH,

I- the z ljBG = the Z liBH, Coiistr.

and the z I^GB^the Z I^HB,

being rt. angles;

{ also IjB is common ;

.-. liG=:liH.

Similarly it may be shewn that I^H = Ij^K ;

.-. IjG, liH, IjK are all equal.

From centre 1^ with radius I^G, describe a circle:

this circle must pass through the points G, H, K :

and it will be an escribed circle of the a ABC.

For since I^H, I^G, I^K are radii of the HGK,

and since the angles at H, G, K are rt. angles,

.-, the GHK is touched at these points by BC, and by AB, AC

jjroduced :

.'. the GHK is an escribed circle of the a ABC. q.e.f.

It is clear that every triangle has three escribed circles.

Note. From page 104 it is seen that if Alj be joined, then Alj

bisects the angle BAG : hence it follows that

The bisectors of tico exterior angles of a triangle and the bisector of

the third angle are concurrent, the point of intersection being the centre

of an escribed circle.

25G euclid'8 elements.

Proposition 5. Problem.

To circumscribe a circle about a given tTricmgle.

'^>, /^?^^ bA^o

Let ABC be the given triangle :

it is required to circumscribe a circle about the A ABC.

Draw DS bisecting AB at rt. angles; i. 11.

and draw ES bisecting AC at rt. angles; .

llien since AB, AC are neither par', nor in the same st. line,

.'. DS and ES must meet at some point S.

Join SA ;

and if S be not in BC, join SB, SC.

Then in the AÂ« ADS, BDS,

I AD=BD

Because <and DS is common to both ;

(and the z. ADS = the ^ BDS, being rt. angles ;

.'. SA = SB.

Similarly it may be shewn that SC = SA.

.". SA, SB, SC are all equal.

From centre S, with radius SA, describe a circle :

this circle must pass through the points A, B, C, and is

therefore circumscribed about the A ABC. q.e.f.

It follows that

(i) when the centre of the circumscribed circle falls

within the triangle, each of its angles must be acute, for

each angle is then in a segment greater than a semicircle :

(ii) when the centre falls on one of the sides of the

triangle, the angle opposite to this side must be a right

angle, for it is the angle in a semicircle ;

BOOK IV. PROP. 5, 257

(iii) when the centre falls without the triangle, the

angle opposite to the side beyond which the centre falls,

must be obtuse, for it is the angle in a segment less than a

semicircle.

Therefore, conversely, if the given triangle be acute-angled,

the centre of the circumscribed circle falls within it : if it be

a right-angled triangle, the centre falls on the hypotenuse :

if it be an obtuse-atigled triangle, the centre falls without the

triangle.

Note. From page 103 it is seen that if S be joined to the middle

point of BC, then the joining Hne is perpendicular to BC.

Hence the perpendiculars draivn to the sides of a triangle from their

middle points are concurrent^ the point of intersection being the centre

of the circle circumscribed about the triangle.

The centre of the circle circumscribed about a triangle is some-

times called its circum-centrd.

EXERCISES.

On thk Inscribed, Circumscribed, and Escribed CiRCiiES of a

Triangle.

1. An equilateral triangle is inscribed in a circle, and tangents

are drawn at its vertices, prove that

(i) the resulting figure is an equilateral triangle:

(ii) its area is four times that of the given triangle.

2. Describe a circle to touch two parallel straight lines and a

third straight line which meets them. Shew that two such circles

can be drawn, and that they are equal.

3. Triangles lohich have equal bases and equal vertical angles

have equal circumscribed circles.

4. I is the centre of the circle inscribed in the triangle ABC, and

Ij is the centre of the circle which touches BC and AB, AC produced:

shew that A, i, Ij are collinear.

5. If the inscribed and circumscribed circles of a triangle are con-

centric, shew that the triangle is equilateral; and that the diameter of

the circumscribed circle is double that of the inscribed circle.

6. ABC is a triangle; and I, S are the centres of the inscribed

and circumscribed circles; if A, I, S are collinear, shew that AB = AC.

H. E. 17

2f)S EUCLID S ELEMENTS.

7. The sum of the diameters of the inscribed and circumscribed

circles of a right-angled triangle is equal to the sum of the sides

containing the right angle.

8. If the circle inscribed in a triangle ABC touches the sides at

D, E, F, shew that the triangle DEF is acute-angled; and express its

angles in terms of the angles at A, B, C.

9. If I is the centre of the circle inscribed in the triangle ABC,

and Ij the centre of the escribed circle which touches BC ; shew that

I, B, Ij, C are concyclic.

10. In any triangle the difference of two sides is equal to the dif-

ference of the segments into which the third side is divided at the

point of contact of the inscribed circle.

11. In the triangle ABC the bisector of the angle BAC meets the

base at D, and from I the centre of the inscribed circle a perpendicular

I E is drawn to BC : shew that the angle Bl D is equal to the angle CI E.

12. In the triangle ABC, I and S are the centres of the inscribed

and circumscribed circles : shew that I S subtends at A an angle equal

to half the difference of the angles at the base of the triangle.

13. In a triangle ABC, I and S are the centres of the inscribed

and circumscribed circles, and AD is drawn perpendicular to BC:

shew that A I is the bisector of the angle DAS.

14. Shew that the area of a triangle is equal to the rectangle

contained by its semi -perimeter and the radius of the inscribed circle.

15. The diagonals of a quadrilateral ABCD intersect at O: shew

that the centres of the circles circumscribed about the four triangles

AOB, BOC, COD, DOA are at the angular points of a parallelogram.

16. In any triangle ABC, if I is the centre of the inscribed circle,

and if A I is produced to meet the circumscribed circle at O ; shew that

O is the centre of the circle circumscribed about the triangle BIC.

17. Given the base, altitude, and the radius of the circumscribed

circle ; construct the triangle.

18. Describe a circle to intercept equal chords of given length on

three given straight lines.

19. In an equilateral triangle the radii of the circumscribed and

escribed circles are respectively double and treble of the radius of the

inscribed circle.

20. Two circles whose centres are A, B, C touch one another

externally two by two a^ D, E, F : shcwthat the inscribed circle of

the triangle ABC is the circumscribed circle of the triangle DEF.

book iv. prof. 6. 259

Proposition 6. Problem.

To inscribe a square in a given cli'cle.

A

Let A BCD be the given circle :

it is required to inscribe a square in the Â©A BCD.

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

and draw two diameters AC, BD perp. to one another, i. 11.

Join AB, BC, CD, DA.

Then the fig. ABCD shall be the square required.

For in the AÂ« BE A, DEA,

f BE^DE,

Because I and EA is common ;

[and the z. BEA = the L DEA, being rt. angles;

.â€¢. BA ^ DA. I. 4.

Similarly it may be shewn that CD ^^ DA, and that BC â€” CD.

IHb â€¢*â€¢ ^^^Â® â– ^o" ^^^'^ ^'^ equilateral.

^Kl And since BD is a diameter of the 0ABCD,

^^K .'. BAD is a semicircle;

j^^r .". the i_ BAD is a rt. angle. ill. 31.

Similarly the other angles of the hg. ABCD are rt. angles.

.'. the fig. ABCD is a square,

and it is inscribed in the given circle.

Q. E. F.

[Fox Exercibes see page 203.]

17-2

260

EUCLID'S ELEMUNIb.

Proposition 7. Problem.

To circumscribe a square about a given circle.

G A F

Let A BCD be the given circle :

it is required to circumscribe a square about it.

Find E the centre of the 0ABCD : iii. 1.

and draw two diameters AC, BD perp. to one another, i. 11.

Through A, B, C, D draw FG, GH, HK, KF perp. to EA, EB,

EC, ED.

Then the fig. G K shall be the square required.

Because FG, GH, HK, KF are drawn perp. to radii at their

extremities,

.â€¢. FG, GH, HK, KF are tangents to the circle, ill. 16.

And because the i. ^ AEB, EBG are both rt. angles, Constr.

.'. GH is par' to AC. i. 28.

Similarly FK is par' to AC :

and in like manner GF, BD, HK are par'.

Hence the figs. GK, GC, AK, GD, BK, GE are par"'^

.'. GF and HK each = BD ;

also GH and FK each = AC :

but AC - BD ;

.â€¢. GF, FK, KH, HG are all equal :

tliat is, the fig. GK is equilateral.

And since the tig. G E is a par",

.*. the L BGA = the l BEA ;

but the z. BEA is a rt. angle ;

.'. the z. at G is a rt. angle.

Similarly the i. ^ at F, K, H are rt. angles.

.'. the fig. GK is a square, and it has been circumscribed

about the 0ABCD. Q.E.F.

I. 34.

Constr.

BOOK IV. PROP. 8.

261

Proposition 8, Problem.

To inscribe a circle in a given square.

A E D

K

OA

Ey

H

Let A BCD bo the given square :

it is required to inscribe a circle in the sq. A BCD.

Bisect the sides AB, AD at F and E. I. 10.

Through E draw EH par' to A B or DC : i. 31,

and tlirough F draw FK par' to AD or BC, meeting EH at G.

Now AB = AD, being the sides of a square ;

and their halves are equal ; Constr

:. AF = AE. Ax. 7,

But the iig. AG is a par"â„¢; Constr.

:. AF=^GE, and AE = GF;

.-. GE-GF.

Similarly it may be shewn that G E = G K, and G K = G H :

.'. GF, GE, GK, GH are all equal.

From centre G, with radius GE, describe a circle;

this circle must pass through the points F, E, K, H :

and it will be touched by BA, AD, DC, CB; ill

for GF, GE, GK, GH are radii;

and the angles at F, E, K, H are rt. angles. i

Hence the FEKH is inscribed in the sq. A BCD.

Q. E. p.

16.

29.

[For Exercises see p. 263.]

262

EUCLID'S ELKMENTS.

Proposition 9. Problem,

To circum8C7'ibe a circle about a give)i square.

A

I. J)ef. 28,

I. Def. 28.

I. 8.

Let ABCD be the given square :

it is required to circumscribe a circle about the sq. ABCD.

Join AC, BD, intersecting at E.

Then in the AÂ« BAC, DAC,

I BA = DA,

Because â€¢< and AC is common ;

( and BC = DC ;

.-. the L BAC = the z. DAC :

that is, the diagonal AC bisects the l BAD.

Similarly the remaining angles of the square are bisected

])y the diagonals AC or BD.

Hence each of the z. ^ EAD, EDA is half a rt. angle ;

.-. the ^ EAD = the ^EDA:

.-. EA = ED. I. 6.

Similarly it may be shewn that ED : - EC, and EC - EB.

.*. EA, EB, EC, ED are all equal.

From centre E, with radius EA, describe a circle :

this circle must pass through the points A, B, C, D, and is

therefore circumscribed about the sq. ABCD. Q. E.P.

BOOK IT. PROP. 9. 263

Definition. A rectilineal figure about which a circle

may be described is said to be Cyclic.

EXERCISES ON PROPOSITIONS 6 â€” 9.

1. If a circle can he inscribed in a quadrilatei'al, slieio that the

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

2. If the sum of one pair of opposite sides of a quadrilateral is

equal to the sxim of the other pair, sheio that a circle may he inscribed

in the figure.

[Bisect two adjacent angles of the figure, and so describe a circle to

touch three of its sides. Then prove indirectly by means of the

last exercise that this circle must also touch the fourth side.]

3. Prove that a rhomhus and a square are the only parallelograms

in which a circle can he inscrihed.

4. All cyclic parallelograms are rectangular.

5. The greatest rectangle which can he inscrihed in a given circle

is a square,

6. Circumscribe a rhombus about a given circle.

7. All squares circumscribed about a given circle are equal.

8. The area of a square circumscribed about a circle is double of

the area of the inscribed square.

9. A BCD is a square inscribed in a circle, and P is any point on

the arc AD : shew that the side AD subtends at P an angle three times

as great as that subtended at P by any one of the other sides.

10. Inscribe a square in a given square A BCD so that one of its

angular points should be at a given point X in AB.

11. In a given square inscribe the square of minimum area.

12. Describe (i) a circle, (ii) a square about a given rectangle.

13. Inscribe (i) a circle, (ii) a square in a given quadrant.

14. In a given circle inscribe a rectangle equal to a given recti-

lineal figure.

15. ABCD is a square inscribed in a circle, and P is any point on

the circumference; shew that the sum of the squares on PA, PB, PC,

PD is double the square on the diameter. [See Ex. 24, p. 147.]

2G4 EUCLID'S ELKMKNTS.

Proposition 10. Problem.

To describe an isosceles triangle having each of ike angles

at the base double of the third angle.

Take any straight line AB.

Divide AB at C, so that the rect. BA, BC -the sq. on AC.

II. 11.

From centre A, with radius AB, describe the BDE ;

and in it place the chord BD equal to AC. i/. I.

Join DA.

Tlien ABD shall be the triangle required.

Join CD ;

and about the AACD circumscribe a circle. iv. 5.

Then the rect. BA, BC = the sq. on AC Constr.

= the sq. on BD. Constr.

Hence BD is a t^angent to the Â©ACD : in. 37.

and from the point of contact D a chord DC is drawn ;

.'. the ^ BDC =. the z_ CAD in the alt. segment, in. 32.

To each of these equals add the lCDA:

then the whole z. BDA =^ the sum of the z. * CAD, CDA.

But the ext. z. BCD = the sum of the z. Â» CAD, CDA ; i. 32.

.'. the L BCD =the z. BDA.

And since AB = AD, being radii of the Â©BDE,

.*. the z. DBA = the z. BDA : l. 5.

.'. the z. DB'C = the L DCB ;

BOOK IV. PROP. 10. 265

.'. DC=: DB; I. 6.

that is, DC = CA : Constr.

:. the L CAD = the l CDA ; i. 5.

.*. the sum of the l ^ CAD, CDA -= twice the angle at A.

But the ^ ADB = the sum of the z. Â« CAD, CDA ; Proved.

.'. each of the l^ ABD, ADB = twice the angle at A.

Q. E. F.

EXERCISES ON PROPOSITION 10.

1. In an isosceles triangle in which each of the angles at the

base is double of the vertical angle, shew that the vertical angle is

one-fifth of two right angles.

2. Divide a right angle into Jive equal parts.

3. Describe an isosceles triangle whose vertical angle shall be

three times either angle at the base. Point out a triangle of this kind

in the figure of Proposition 10.

4. In the figure of Proposition 10, if the two circles intersect at F,

sheiD that BD = DF.

5. In the figure of Proposition 10, shew that the circle ACD is

equal to the circle circumscribed about the triangle ABD.

6. In the figure of Proposition 10, if the two circles intersect at F,

shew that

(i) BD, DF are sides of a regular decagon inscribed in the

circle EBD.

(ii) AC, CD, DF are sides of a regular pentagon inscribed

in the circle ACD.

7. In the figure of Proposition 10, shew that the centre of the

circle circumscribed about the triangle DBC is the middle point of

the arc CD.

8. In the figure of Proposition 10, if I is the centre of the circle

inscribed in the triangle ABD, and I', S' the centres of the inscribed

and circumscribed circles of the triangle DBC, shew that S'l = 8'!'.

266 euctjd's elements.

Proposition 11. Problem.

To inscribe a regular pentagon in a given circle.

Let ABC be a given circle :

it is required to inscribe a regular pentagon in the Â©ABC.

Describe an isosceles AFGH, having each of the angles

at G and H double of the angle at F. I v. 10.

In the 0ABG inscribe the AACD equiangular to the

AFGH, â€¢ IV. 2.

so that each of the l^ ACD, ADC is double of the l CAD.

Bisect the l ^ ACD, ADC by CE and DB, which meet the

O '^'* at E and B. I. 9.

Join AB, BC, AE, ED.

Then ABCDE shall be the required regular pentagon.

Because each of the l^ ACD, ADC = twice the ^ CAD ;

and because the z. ^ ACD, ADC are bisected by CE, DB,

.-. the five L^ ADB, BDC, CAD, DCE, EGA are all equal.

.*. the five arcs AB, BC, CD, DE, EA are all equal, iii. 26.

.*. the five chords AB, BC, CD, DE, EA are all equal. III. 29.

.'. the pentagon ABCDE is equilateral.

Again the arc AB = the arc DE ; Proved.

to each of these equals add the arc BCD ;

.*. the whole arc ABCD = the whole arc BCDE :

hence the angles at the O^^ which stand upon these

equal arcs are equal ; iii. 27.

that is, the ^ AED = the z_ BAE.

In like manner the remaining angles of the pentagon

may be shewn to be equal ;

.'. the pentagon is equiangular.

Hence the pentagon, being both equilateral and equi-

angular, is regular ; and it is inscribed in the ABC. q.e.f.

BOOK IV. PROP. 12.

267

Proposition 12. Problem.

To circumscribe a regular pentagon about a given circle.

Lot A BCD be the given circle :

it is required to circumscribe a regular pentagon about it.

Inscribe a regular pentagon in the 0ABCD, iv. 11.

and let A, B, C, D, E be its angular points.

At the points A, B, C, D, E draw GH, HK, KL, LM, MG,

tangents to the circle. in. 17.

Then shall GHKLM be the required regular pentagon.

Find F the centre of the A BCD ; iii. 1.

and join FB, FK, FC, FL, FD.

Then in the two AÂ« BFK, CFK,

( BF = CF, being radii of the circle,

J and FK is common :

I and KB == KC, being tangents to tlie circle from

\ the same point K. in. 17. Cor.

Because

.-. the

^BFK =

= the Z.CFK,

I. 8.

also the

Â£.BKF =

= the Z.CKF.

I. 8.

Cor.

Hence the

/.BFC^

^ twice the l

CFK,

and the

^BKC =

- twice the l

CKF.

Similarly it may be shewn

that the l CFD = twice the l CFL,

and that the l CLD = twice the l CLF.

But since the arc BC = the arc CD,

.-. the L BFC - the l CFD ;

and the halves of these angles are equal,

that is, the l CFK = the 4 CFL,

IV. 11.

III. 27,

EUCLID S EI.E.MKNTS.

Then in the AÂ« CFK, CFL,

( the z. CFK = the z. CFL, Proved.

Because -j and the z. FCK = the l FCL,beingrt. angles, hi. 18.

( and FC is common ;

.-. CK = CL, I. 26.

and the z. FKC -the L FLC.

Hence KL is double of KC; similarly HK is double of KB.

And since KC = KB, iii. 17. Cor.

.'. KL = HK.

In the same way it may be shewn that every two con-

secutive sides are equal ; ,

bisector of the vertical angle : construct the triangle.

43. Given the perpendicular from the vertex on the base, the

bisector of the vertical angle, and the median which bisects the base :

construct the triangle.

44. Given the bisector of the vertical angle, the median bisect-

ing the base, and the difference of the angles at the base : construct the

triangle.

BOOK IV.

Book IV. consists entirely of problems, dealing with

various rectilineal figures in relation to the circles which

pass through their angular points, or are touched by their

sides.

Definitions.

1. A Polygon is a rectilineal figure bounded by more

than four sides.

A Polygon of Jive sides is ca^

â€ž six sides

,5 seven sides

,, eight sides

,, ten sides

â€ž twelve sides

3) fifteen sides

led a Pentagon,

Hexagon,

Heptagon,

Octagon,

Decagon,

Dodecagon,

Quindecagon.

2. A Polygon is Regular when all its sides are equal,

and all its angles are equal.

3. A rectilineal figure is said to be _

inscribed in a circle, when all its angular

points are on the circumference of the circle :

and a circle is said to be circumscribed

about a rectilineal figure, when the circum-

ference of the circle passes through all the

angular points of the figure.

4. A rectilineal figure is said to be

circumscribed about a circle, when each side

of the figure is a tangent to the circle :

and a circle is said to be inscribed in a recti-

lineal figure, when the circumference of the

circle is touched by each side of the figure.

5. A straight line is said to be placed in a circle, wlien

its extremities are on the circumference of the circle.

BOOK iV. PROP. 1. 251

Proposition 1. Problem.

In a given circle to i^lace a chord equal to a given

straight line, which is not greater than the diameter of the

circle.

Let ABC be the given circle, and D the given straight

line not greater tlian the diameter of the circle :

it is required to place in the 0ABC a chord equal to D.

Draw CB, a diameter of the 0ABC.

Then if CB = D, the thing required is done.

But if not, CB must be greater than D. Hyp-

From CB cut off CE equal to D : I. 3.

and from centre C, with radius CE, describe the 0AEF,

cutting the given circle at A.

Join CA.

Then CA shall be the chord required.

For CA = CE, being radii of the AEF :

and CE = D : Constr.

:. CA= D.

Q. E. F.

EXERCISES.

1. In a given circle place a chord of given length so as to pass

through a given point (i) without, (ii) within the circle.

When is this problem impossible ?

2. In a given circle place a chord of given length so that it may

be parallel to a given straight line.

262 kuclid'h elements.

Proposition 2. Problem.

In a given circle to inscribe a triangle- equiangvZar to a

given triangle.

Let ABC be the given circle, and DEF the given triangle:

it is required to inscribe in the ABC a triangle equiangular

to the A DEF.

At any point A, on the O*^^ of the 0ABC, draw the

tangent GAH. ill. 17.

At A make the z. GAB equal to the l DFE ; i. 23.

and make the l HAC equal to the L DEF. i. 23.

Join BC.

Then ABC shall be the triangle required.

Because GH is a tangent to the 0ABC, and from A its

point of contact the chord AB is drawn,

.'. the L GAB = the L ACB in tlie alt. segment: iii. 32.

.'. the L ACB = the l DFE. Constr.

Similarly the z. HAC = the l ABC, in the alt. segment:

.*. the z. ABC = the z. DEF. Constr.

Hence the third l BAG = the third z. EDF,

for the three angles in each triangle are together equal to

two rt. angles, i. 32.

.'. the A ABC is equiangular to the A DEF, and it is

inscribed in the 0ABC.

Q. E. F.

BOOK IV. PROP. 3. 253

Proposition 3. Problem.

About a given circle to circumscribe a triangle equi-

angular to a given triangle.

G E F H

M B

Let ABC be the given circle, and DEF the given A :

it is required to circumscribe about the Â©ABC a triangle

equiangular to the A DEF.

Produce EF both ways to G and H.

Find K the centre of the Â©ABC, iii. 1.

and draw any radius KB.

At K make the l BKA equal to the ^ DEG ; i. 23,

and make the z. BKG equal to the ^ DFH.

Through A, B, C draw LM, MN, NL perp. to KA, KB, KG.

Then LMN shall be the triangle required.

Because LM, MN, NL are drawn perp. to radii at their

extremities,

.'. LM, MN, NL are tangents to the circle. lii. 16.

And because the four angles of the quadrilateral AKBM

together = four rt. angles ; i. 32. Cor.

and of these, the l^ KAM, KBM, are rt. angles; Constr.

.'. the L^ AKB, AMB, together = two rt. angles.

But the L^ DEG, DEF together = two rt. angles; i. 13.

.*. the L^ AKB, AMB -the L^ DEG, DEF;

and of these, the l AKB = the ^ DEG ; Constr.

:. the ^AMB=.the L DEF.

Similarly it may be shewn that the ^ LNM = the lDFE.

.â€¢. the third z. MLN =the third l EDF. i. 32.

.'. the A LMN is equiangular to the A DEF, and it is

circumscribed about the Â©ABC. (J.B. f.

â€¢io4

KtiCl.lDS Kl.hi.MKN

Proposition 4. Problem.

To inscribe a circle in a give^i tricvagle.

Let ABC be the given triangle :

it is required to inscribe a circle in the A ABC.

Bisect the l ^ ABC, ACB by the st. lines Bl, CI, which

intersect at I. i. 9.

From I draw IE, IF, IG perp. to AB, BC, CA. i. 12.

Then in the AÂ« EIB, FIB,

i the z. EBI = the z. FBI ; Constr.

Because -land the z. BEI = the z. BFI, being rt. angles ;

( and Bl is common ;

.-. IE=:|F. I. 26.

Similarly it may be shewn that IF = IG.

.'. IE, IF, IG are all equal.

From centre I, with radius IE, describe a circle:

this circle must pass through the points E, F, G ;

and it will be inscribed in the A ABC.

For since IE, IF, IG are radii of the EFG ;

and since the z. ^ at E, F, G are rt. angles ;

.'. the 0EFG is touched at these points by AB, BC, CA:

III. 16.

.'. the 0EFQ is inscribed in the A ABC.

Q. E. F.

NotÂ«. From page 103 it is seen tliat if A I be joined, then A I

bisects the angle BAC.

BOOK IV. PROP. 4.

255

Hence it follows that the bisectors of the angles of a triangle are

concurrent, the point of intersection being the centre of the inscribed

circle.

The centre of the circle inscribed in a triangle is sometimes called

its in-centre.

Definitiox.

From Ij draw I^G

Because

A circle whicli touches one side of a triangle and the

other two sides produced is said to be an escribed circle of

* the triangle.

To draio an escribed circle of a given triangle.

Let ABC be the given triangle, of which

the two sides AB, AC are produced to E

and F:

it is required to describe a circle touching

BC, and AB, AC produced.

Bisect the z ** CBE, BCF by the st. lines

which intersect at I^, i. 9.

I^H, I^K perp. to AE,

BC, AF. I. 12.

Then in the a' I^BG, I^BH,

I- the z ljBG = the Z liBH, Coiistr.

and the z I^GB^the Z I^HB,

being rt. angles;

{ also IjB is common ;

.-. liG=:liH.

Similarly it may be shewn that I^H = Ij^K ;

.-. IjG, liH, IjK are all equal.

From centre 1^ with radius I^G, describe a circle:

this circle must pass through the points G, H, K :

and it will be an escribed circle of the a ABC.

For since I^H, I^G, I^K are radii of the HGK,

and since the angles at H, G, K are rt. angles,

.-, the GHK is touched at these points by BC, and by AB, AC

jjroduced :

.'. the GHK is an escribed circle of the a ABC. q.e.f.

It is clear that every triangle has three escribed circles.

Note. From page 104 it is seen that if Alj be joined, then Alj

bisects the angle BAG : hence it follows that

The bisectors of tico exterior angles of a triangle and the bisector of

the third angle are concurrent, the point of intersection being the centre

of an escribed circle.

25G euclid'8 elements.

Proposition 5. Problem.

To circumscribe a circle about a given tTricmgle.

'^>, /^?^^ bA^o

Let ABC be the given triangle :

it is required to circumscribe a circle about the A ABC.

Draw DS bisecting AB at rt. angles; i. 11.

and draw ES bisecting AC at rt. angles; .

llien since AB, AC are neither par', nor in the same st. line,

.'. DS and ES must meet at some point S.

Join SA ;

and if S be not in BC, join SB, SC.

Then in the AÂ« ADS, BDS,

I AD=BD

Because <and DS is common to both ;

(and the z. ADS = the ^ BDS, being rt. angles ;

.'. SA = SB.

Similarly it may be shewn that SC = SA.

.". SA, SB, SC are all equal.

From centre S, with radius SA, describe a circle :

this circle must pass through the points A, B, C, and is

therefore circumscribed about the A ABC. q.e.f.

It follows that

(i) when the centre of the circumscribed circle falls

within the triangle, each of its angles must be acute, for

each angle is then in a segment greater than a semicircle :

(ii) when the centre falls on one of the sides of the

triangle, the angle opposite to this side must be a right

angle, for it is the angle in a semicircle ;

BOOK IV. PROP. 5, 257

(iii) when the centre falls without the triangle, the

angle opposite to the side beyond which the centre falls,

must be obtuse, for it is the angle in a segment less than a

semicircle.

Therefore, conversely, if the given triangle be acute-angled,

the centre of the circumscribed circle falls within it : if it be

a right-angled triangle, the centre falls on the hypotenuse :

if it be an obtuse-atigled triangle, the centre falls without the

triangle.

Note. From page 103 it is seen that if S be joined to the middle

point of BC, then the joining Hne is perpendicular to BC.

Hence the perpendiculars draivn to the sides of a triangle from their

middle points are concurrent^ the point of intersection being the centre

of the circle circumscribed about the triangle.

The centre of the circle circumscribed about a triangle is some-

times called its circum-centrd.

EXERCISES.

On thk Inscribed, Circumscribed, and Escribed CiRCiiES of a

Triangle.

1. An equilateral triangle is inscribed in a circle, and tangents

are drawn at its vertices, prove that

(i) the resulting figure is an equilateral triangle:

(ii) its area is four times that of the given triangle.

2. Describe a circle to touch two parallel straight lines and a

third straight line which meets them. Shew that two such circles

can be drawn, and that they are equal.

3. Triangles lohich have equal bases and equal vertical angles

have equal circumscribed circles.

4. I is the centre of the circle inscribed in the triangle ABC, and

Ij is the centre of the circle which touches BC and AB, AC produced:

shew that A, i, Ij are collinear.

5. If the inscribed and circumscribed circles of a triangle are con-

centric, shew that the triangle is equilateral; and that the diameter of

the circumscribed circle is double that of the inscribed circle.

6. ABC is a triangle; and I, S are the centres of the inscribed

and circumscribed circles; if A, I, S are collinear, shew that AB = AC.

H. E. 17

2f)S EUCLID S ELEMENTS.

7. The sum of the diameters of the inscribed and circumscribed

circles of a right-angled triangle is equal to the sum of the sides

containing the right angle.

8. If the circle inscribed in a triangle ABC touches the sides at

D, E, F, shew that the triangle DEF is acute-angled; and express its

angles in terms of the angles at A, B, C.

9. If I is the centre of the circle inscribed in the triangle ABC,

and Ij the centre of the escribed circle which touches BC ; shew that

I, B, Ij, C are concyclic.

10. In any triangle the difference of two sides is equal to the dif-

ference of the segments into which the third side is divided at the

point of contact of the inscribed circle.

11. In the triangle ABC the bisector of the angle BAC meets the

base at D, and from I the centre of the inscribed circle a perpendicular

I E is drawn to BC : shew that the angle Bl D is equal to the angle CI E.

12. In the triangle ABC, I and S are the centres of the inscribed

and circumscribed circles : shew that I S subtends at A an angle equal

to half the difference of the angles at the base of the triangle.

13. In a triangle ABC, I and S are the centres of the inscribed

and circumscribed circles, and AD is drawn perpendicular to BC:

shew that A I is the bisector of the angle DAS.

14. Shew that the area of a triangle is equal to the rectangle

contained by its semi -perimeter and the radius of the inscribed circle.

15. The diagonals of a quadrilateral ABCD intersect at O: shew

that the centres of the circles circumscribed about the four triangles

AOB, BOC, COD, DOA are at the angular points of a parallelogram.

16. In any triangle ABC, if I is the centre of the inscribed circle,

and if A I is produced to meet the circumscribed circle at O ; shew that

O is the centre of the circle circumscribed about the triangle BIC.

17. Given the base, altitude, and the radius of the circumscribed

circle ; construct the triangle.

18. Describe a circle to intercept equal chords of given length on

three given straight lines.

19. In an equilateral triangle the radii of the circumscribed and

escribed circles are respectively double and treble of the radius of the

inscribed circle.

20. Two circles whose centres are A, B, C touch one another

externally two by two a^ D, E, F : shcwthat the inscribed circle of

the triangle ABC is the circumscribed circle of the triangle DEF.

book iv. prof. 6. 259

Proposition 6. Problem.

To inscribe a square in a given cli'cle.

A

Let A BCD be the given circle :

it is required to inscribe a square in the Â©A BCD.

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

and draw two diameters AC, BD perp. to one another, i. 11.

Join AB, BC, CD, DA.

Then the fig. ABCD shall be the square required.

For in the AÂ« BE A, DEA,

f BE^DE,

Because I and EA is common ;

[and the z. BEA = the L DEA, being rt. angles;

.â€¢. BA ^ DA. I. 4.

Similarly it may be shewn that CD ^^ DA, and that BC â€” CD.

IHb â€¢*â€¢ ^^^Â® â– ^o" ^^^'^ ^'^ equilateral.

^Kl And since BD is a diameter of the 0ABCD,

^^K .'. BAD is a semicircle;

j^^r .". the i_ BAD is a rt. angle. ill. 31.

Similarly the other angles of the hg. ABCD are rt. angles.

.'. the fig. ABCD is a square,

and it is inscribed in the given circle.

Q. E. F.

[Fox Exercibes see page 203.]

17-2

260

EUCLID'S ELEMUNIb.

Proposition 7. Problem.

To circumscribe a square about a given circle.

G A F

Let A BCD be the given circle :

it is required to circumscribe a square about it.

Find E the centre of the 0ABCD : iii. 1.

and draw two diameters AC, BD perp. to one another, i. 11.

Through A, B, C, D draw FG, GH, HK, KF perp. to EA, EB,

EC, ED.

Then the fig. G K shall be the square required.

Because FG, GH, HK, KF are drawn perp. to radii at their

extremities,

.â€¢. FG, GH, HK, KF are tangents to the circle, ill. 16.

And because the i. ^ AEB, EBG are both rt. angles, Constr.

.'. GH is par' to AC. i. 28.

Similarly FK is par' to AC :

and in like manner GF, BD, HK are par'.

Hence the figs. GK, GC, AK, GD, BK, GE are par"'^

.'. GF and HK each = BD ;

also GH and FK each = AC :

but AC - BD ;

.â€¢. GF, FK, KH, HG are all equal :

tliat is, the fig. GK is equilateral.

And since the tig. G E is a par",

.*. the L BGA = the l BEA ;

but the z. BEA is a rt. angle ;

.'. the z. at G is a rt. angle.

Similarly the i. ^ at F, K, H are rt. angles.

.'. the fig. GK is a square, and it has been circumscribed

about the 0ABCD. Q.E.F.

I. 34.

Constr.

BOOK IV. PROP. 8.

261

Proposition 8, Problem.

To inscribe a circle in a given square.

A E D

K

OA

Ey

H

Let A BCD bo the given square :

it is required to inscribe a circle in the sq. A BCD.

Bisect the sides AB, AD at F and E. I. 10.

Through E draw EH par' to A B or DC : i. 31,

and tlirough F draw FK par' to AD or BC, meeting EH at G.

Now AB = AD, being the sides of a square ;

and their halves are equal ; Constr

:. AF = AE. Ax. 7,

But the iig. AG is a par"â„¢; Constr.

:. AF=^GE, and AE = GF;

.-. GE-GF.

Similarly it may be shewn that G E = G K, and G K = G H :

.'. GF, GE, GK, GH are all equal.

From centre G, with radius GE, describe a circle;

this circle must pass through the points F, E, K, H :

and it will be touched by BA, AD, DC, CB; ill

for GF, GE, GK, GH are radii;

and the angles at F, E, K, H are rt. angles. i

Hence the FEKH is inscribed in the sq. A BCD.

Q. E. p.

16.

29.

[For Exercises see p. 263.]

262

EUCLID'S ELKMENTS.

Proposition 9. Problem,

To circum8C7'ibe a circle about a give)i square.

A

I. J)ef. 28,

I. Def. 28.

I. 8.

Let ABCD be the given square :

it is required to circumscribe a circle about the sq. ABCD.

Join AC, BD, intersecting at E.

Then in the AÂ« BAC, DAC,

I BA = DA,

Because â€¢< and AC is common ;

( and BC = DC ;

.-. the L BAC = the z. DAC :

that is, the diagonal AC bisects the l BAD.

Similarly the remaining angles of the square are bisected

])y the diagonals AC or BD.

Hence each of the z. ^ EAD, EDA is half a rt. angle ;

.-. the ^ EAD = the ^EDA:

.-. EA = ED. I. 6.

Similarly it may be shewn that ED : - EC, and EC - EB.

.*. EA, EB, EC, ED are all equal.

From centre E, with radius EA, describe a circle :

this circle must pass through the points A, B, C, D, and is

therefore circumscribed about the sq. ABCD. Q. E.P.

BOOK IT. PROP. 9. 263

Definition. A rectilineal figure about which a circle

may be described is said to be Cyclic.

EXERCISES ON PROPOSITIONS 6 â€” 9.

1. If a circle can he inscribed in a quadrilatei'al, slieio that the

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

2. If the sum of one pair of opposite sides of a quadrilateral is

equal to the sxim of the other pair, sheio that a circle may he inscribed

in the figure.

[Bisect two adjacent angles of the figure, and so describe a circle to

touch three of its sides. Then prove indirectly by means of the

last exercise that this circle must also touch the fourth side.]

3. Prove that a rhomhus and a square are the only parallelograms

in which a circle can he inscrihed.

4. All cyclic parallelograms are rectangular.

5. The greatest rectangle which can he inscrihed in a given circle

is a square,

6. Circumscribe a rhombus about a given circle.

7. All squares circumscribed about a given circle are equal.

8. The area of a square circumscribed about a circle is double of

the area of the inscribed square.

9. A BCD is a square inscribed in a circle, and P is any point on

the arc AD : shew that the side AD subtends at P an angle three times

as great as that subtended at P by any one of the other sides.

10. Inscribe a square in a given square A BCD so that one of its

angular points should be at a given point X in AB.

11. In a given square inscribe the square of minimum area.

12. Describe (i) a circle, (ii) a square about a given rectangle.

13. Inscribe (i) a circle, (ii) a square in a given quadrant.

14. In a given circle inscribe a rectangle equal to a given recti-

lineal figure.

15. ABCD is a square inscribed in a circle, and P is any point on

the circumference; shew that the sum of the squares on PA, PB, PC,

PD is double the square on the diameter. [See Ex. 24, p. 147.]

2G4 EUCLID'S ELKMKNTS.

Proposition 10. Problem.

To describe an isosceles triangle having each of ike angles

at the base double of the third angle.

Take any straight line AB.

Divide AB at C, so that the rect. BA, BC -the sq. on AC.

II. 11.

From centre A, with radius AB, describe the BDE ;

and in it place the chord BD equal to AC. i/. I.

Join DA.

Tlien ABD shall be the triangle required.

Join CD ;

and about the AACD circumscribe a circle. iv. 5.

Then the rect. BA, BC = the sq. on AC Constr.

= the sq. on BD. Constr.

Hence BD is a t^angent to the Â©ACD : in. 37.

and from the point of contact D a chord DC is drawn ;

.'. the ^ BDC =. the z_ CAD in the alt. segment, in. 32.

To each of these equals add the lCDA:

then the whole z. BDA =^ the sum of the z. * CAD, CDA.

But the ext. z. BCD = the sum of the z. Â» CAD, CDA ; i. 32.

.'. the L BCD =the z. BDA.

And since AB = AD, being radii of the Â©BDE,

.*. the z. DBA = the z. BDA : l. 5.

.'. the z. DB'C = the L DCB ;

BOOK IV. PROP. 10. 265

.'. DC=: DB; I. 6.

that is, DC = CA : Constr.

:. the L CAD = the l CDA ; i. 5.

.*. the sum of the l ^ CAD, CDA -= twice the angle at A.

But the ^ ADB = the sum of the z. Â« CAD, CDA ; Proved.

.'. each of the l^ ABD, ADB = twice the angle at A.

Q. E. F.

EXERCISES ON PROPOSITION 10.

1. In an isosceles triangle in which each of the angles at the

base is double of the vertical angle, shew that the vertical angle is

one-fifth of two right angles.

2. Divide a right angle into Jive equal parts.

3. Describe an isosceles triangle whose vertical angle shall be

three times either angle at the base. Point out a triangle of this kind

in the figure of Proposition 10.

4. In the figure of Proposition 10, if the two circles intersect at F,

sheiD that BD = DF.

5. In the figure of Proposition 10, shew that the circle ACD is

equal to the circle circumscribed about the triangle ABD.

6. In the figure of Proposition 10, if the two circles intersect at F,

shew that

(i) BD, DF are sides of a regular decagon inscribed in the

circle EBD.

(ii) AC, CD, DF are sides of a regular pentagon inscribed

in the circle ACD.

7. In the figure of Proposition 10, shew that the centre of the

circle circumscribed about the triangle DBC is the middle point of

the arc CD.

8. In the figure of Proposition 10, if I is the centre of the circle

inscribed in the triangle ABD, and I', S' the centres of the inscribed

and circumscribed circles of the triangle DBC, shew that S'l = 8'!'.

266 euctjd's elements.

Proposition 11. Problem.

To inscribe a regular pentagon in a given circle.

Let ABC be a given circle :

it is required to inscribe a regular pentagon in the Â©ABC.

Describe an isosceles AFGH, having each of the angles

at G and H double of the angle at F. I v. 10.

In the 0ABG inscribe the AACD equiangular to the

AFGH, â€¢ IV. 2.

so that each of the l^ ACD, ADC is double of the l CAD.

Bisect the l ^ ACD, ADC by CE and DB, which meet the

O '^'* at E and B. I. 9.

Join AB, BC, AE, ED.

Then ABCDE shall be the required regular pentagon.

Because each of the l^ ACD, ADC = twice the ^ CAD ;

and because the z. ^ ACD, ADC are bisected by CE, DB,

.-. the five L^ ADB, BDC, CAD, DCE, EGA are all equal.

.*. the five arcs AB, BC, CD, DE, EA are all equal, iii. 26.

.*. the five chords AB, BC, CD, DE, EA are all equal. III. 29.

.'. the pentagon ABCDE is equilateral.

Again the arc AB = the arc DE ; Proved.

to each of these equals add the arc BCD ;

.*. the whole arc ABCD = the whole arc BCDE :

hence the angles at the O^^ which stand upon these

equal arcs are equal ; iii. 27.

that is, the ^ AED = the z_ BAE.

In like manner the remaining angles of the pentagon

may be shewn to be equal ;

.'. the pentagon is equiangular.

Hence the pentagon, being both equilateral and equi-

angular, is regular ; and it is inscribed in the ABC. q.e.f.

BOOK IV. PROP. 12.

267

Proposition 12. Problem.

To circumscribe a regular pentagon about a given circle.

Lot A BCD be the given circle :

it is required to circumscribe a regular pentagon about it.

Inscribe a regular pentagon in the 0ABCD, iv. 11.

and let A, B, C, D, E be its angular points.

At the points A, B, C, D, E draw GH, HK, KL, LM, MG,

tangents to the circle. in. 17.

Then shall GHKLM be the required regular pentagon.

Find F the centre of the A BCD ; iii. 1.

and join FB, FK, FC, FL, FD.

Then in the two AÂ« BFK, CFK,

( BF = CF, being radii of the circle,

J and FK is common :

I and KB == KC, being tangents to tlie circle from

\ the same point K. in. 17. Cor.

Because

.-. the

^BFK =

= the Z.CFK,

I. 8.

also the

Â£.BKF =

= the Z.CKF.

I. 8.

Cor.

Hence the

/.BFC^

^ twice the l

CFK,

and the

^BKC =

- twice the l

CKF.

Similarly it may be shewn

that the l CFD = twice the l CFL,

and that the l CLD = twice the l CLF.

But since the arc BC = the arc CD,

.-. the L BFC - the l CFD ;

and the halves of these angles are equal,

that is, the l CFK = the 4 CFL,

IV. 11.

III. 27,

EUCLID S EI.E.MKNTS.

Then in the AÂ« CFK, CFL,

( the z. CFK = the z. CFL, Proved.

Because -j and the z. FCK = the l FCL,beingrt. angles, hi. 18.

( and FC is common ;

.-. CK = CL, I. 26.

and the z. FKC -the L FLC.

Hence KL is double of KC; similarly HK is double of KB.

And since KC = KB, iii. 17. Cor.

.'. KL = HK.

In the same way it may be shewn that every two con-

secutive sides are equal ; ,

Online Library → Euclid → A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 17 of 27)