Euclid. # Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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them he a right angle ; the triangles shall be equiangular ^ and shall have those

angles equal about which the sides are proportionals.

Let the two triangles ABC, DUFhawe one angle in the one equal

to one angle in the other,

viz. the angle BA C to the angle EDF, and the sides about two other

angles ABC, DJE^i^ proportionals,

so that AB is to BC, as BF to FF;

and in the first case, let each of the remaining angles at C, F be less

than a right angle.

The triangle ABC shall be equiangular to the triangle DFF,

viz. the angle AB C shall be equal to the angle DEF,

and the remaining angle at C equal to the remaining angle at F.

^'

L F

For if the angles ABC, DFF he not equal,

one of them must be greater than the other :

let ABC he the greater,

and at the point B, in the straight line AB,

make the angle ABG equal to the angle DFF-, (l. 23.)

and because the angle at A is equal to the angle at D, (hyp.)

and the angle ^J5 6^ to the angle DFF;

the remaining angle AGB is equal to the remaining angle DFF:

(I. 32.)

therefore the triangle ABG is equiangular to the triangle DFF:

wherefore as AB is to BG, so is DF to FF: (vi. 4.)

but as DF to FF, so, by h)q)othesis, is AB to BC;

therefore as AB to BC, so is AB to BG: (y. 11.)

and because AB has the same ratio to each of the lines BC, BG,

BCis equalto BG; (V. 9.)

and therefore the angle BGC is equal to the angle BCG : (l. 5.)

but the angle BCG is, by hypothesis, less than a right angle ;

therefore also the angle BGC is less than a right angle ;

and therefore the adjacent angle AGB must be greater than a right

angle; (I. 13.)

but it was proved that the angle AGB is equal to the angle at F;

therefore the angle at F is greater than a right angle ;

but, by the hypothesis, it is less than a right angle ; which is absurd.

Therefore the angles ABC, DFF sue not unequal,

â– â– l^ that is, they are equal :

IHK. and the angle at A is equal to the angle at D : (hyp.)

^^Wlierefore the remaining angle at C is equal to the remaming angle at

F: (I. 32.)

therefore the triangle ABC is equiangular to the triangle DFF,

n2

268

Next, let each of the angles at C, F be not less than a right angle.

Then the triangle -4<BC shall also in this case be equiangular to the

triangle DBF,

^o

A

D

E F

The same construction being made,

it may be proved in like manner that BCh equal to BGy

and therefore the angle at C equal to the angle BGC:

but the angle at C is not less than a right angle ; (hyp.)

therefore the angle BGCis not less than a right angle :

wherefore two angles of the triangle BGC are together not less than

two right angles :

which is impossible ; (l. 1 7.)

and therefore the triangle ABC may be proved to be equiangular to

the triangle DBF, as in the first case.

Lastly, let one of the angles at C, F, viz. the angle at C, be a right

angle : in this case likewise the triangle ABC shall be equiangular

to the triangle DFF.

For, if they be not equiangular,

at the point B in the straight line AB make the angle ABG equal

to the angle DBF;

then it may be proved, as in the first case, that BG is equal to BC:

and therefore the angle BCG equal to the angle BGC: (i. 5.)

but the angle BCG is a right angle, (hyp.)

therefore the angle BGCis also a right angle ; (ax. 1.)

whence two of the angles of the triangle BGC are together not less

than two right angles ;

which is impossible : (I. 17.)

tnerefore the triangle ABC is equiangular to the triangle DFF.

Wherefore, if two triangles, &c. q.e.d.

PROPOSITION VIII. THEOREM.

In a right' angled triangle, if a perpendicular he drawn from the right-

angle to the base ; the triangles on each side of it are similar to the whole

triangle, and to one another,

JjBt ABC he a right angled-triangle, having the right angle BAC;

and from the point A let AD be drawn pei-pendicular to the base BC.

Then the triangles ABD, -4 DC shall be similar to the whole tri-

angle ABC, and to one another.

BOOK VI. PROP. VllI, IX. 269

A

D

Because the angle JBACis equal to the angle ABB, each of them

being a right angle, (ax. 11.)

and that the angle at B is common to the two triangles ABC, ABD:

the remaining angle A CB is equal to the remaining angle BAD j

(I. 32.)

therefore the triangle ABC is equiangular to the triangle ABD,

and the sides about their equal angles are proportionals ; (vi. 4.)

â– wherefore the triangles are similar: (VI. def. 1.)

in the like manner it may be demonstrated, that the triangle ADC

is equiangular and similar to the triangle ^^(7.

And the triangles ABD, A CD, being both equiangular and similar

to ABC, are equiangular and similar to each other.

Therefore, in a right-angled, &c. q.e.d.

Cor. From this it is manifest, that the perpendicular drawn from

the right angle of a right-angled triangle to the base, is a mean propor-

tional between the segments of the base ; and also that each of the

sides is a mean proportional between the base, and the segment of it

adjacent to that side : because in the triangles BDA, ADC-, BD is to

DA, as DA to DC; (vi. 4.)

and in the triangles ABC, DBA ; BC is to B A, as BA to BD : (vi.4.)

and in the triangles ABC, A CD-, B Cis to CA, as CA to CD, (VI. 4 . )

PROPOSITION IX. PROBLEM.

From a given straight line to cut off any part required.

Let AB be the given straight line.

It is required to cut off any part from it.

From the point A draw a straight line A C, making any angle with AB)

and in A C take any point D,

and take A C the same multiple of AD, that AB is of the part

which is to be cut off from it ;

join BC, and draw DE parallel to CB.

Then AE shall be the part required to be cut off.

Because ED is parallel to BC, one of the sides of the triangle ABC

as CD is to DA, so is BE to EA ; (vi. 2.)

and by composition, CA is to AD, as BA to AE: (v. 18.)

270 EUCLID'S ELEMENTS.

but CA is a multiple of AD-, (constr.)

therefore JBA is the same multiple of AJS: (v. D.)

whatever part therefore AD is of A C, AE is the same part of AB:

wherefore, from the straight line AB the part required is cut off.

Q.E.F.

PROPOSITION X. PROBLEM.

To divide a given straight line similarly to a given divided straight line^

that iSy into parts that shall have the same ratios to one another which the

parts of the divided given straight line have.

Let AB be the straight line given to be divided, and AC the divided

line.

It is required to divide AB similarly to A C.

A

Let ^Cbe divided in the points D, E-,

and let AB, AC he placed so as to contain any angle, and join BC,

and through the points D, E draw DF, EG parallels to BC. (l. 31.)

Then AB shall be divided in the points F, G, similarly to A C.

Tnrough D draw DHK parallel to AB :

therefore each of the figures, FH, JIB is a parallelogram ;

wherefore EH is equal to FG, and HK to GB : (i. 34.)

and because ITE is parallel to KC, one of the sides of the triangle

EKC,

as C^to ED, so is ^^to HD: (vi. 2.)

but KHk equal to BG, and IID to GF-,

therefore, as CE is to ED, sokBG to GF: (V. 7.)

again, because FD is parallel to GE, one of the sides of the triangle

AGE,

as ED is to DA, so is G'i^'to FA : (vi. 2.)

therefore, as has been proved, as CE is to ED, so is J? 6^ to GF,

and as ED is to DA, so is GF to FA :

therefore the given straight line AB is divided similarly to -4 (7. Q.E.F.

PROPOSITION XI. PROBLEM.

To find a third proportional to two given straight lines.

Let AB, AC he the two given straight lines.

It is required to find a third proportional to AB, A C.

A

S

D E

Let AB, AC he placed so as to contain any angle :

produce AB, AC to the points D, E;

^H ^ BOOK VI. PROP. XI, XII, XIII. S71

Up and make BD equal to A C;

join J5(7, and through D, draw Z)^ parallel to BC. (l. 31.)

Then CE shall be a third proportional to ^ J? and ^ C.

Because BCh parallel to DE, a side of the triangle ADE,

AB is to BD, as AC to CE: (vi. 2.)

but BD is equal to ^ C;

therefore as AB is to ^ C, so is ^ C to CE. (v. 7.) _

Wherefore, to the two given straight lines AB, A C, a third pro-

portional C^ is found. Q.E.F.

PROPOSITION XII. PROBLEM.

To find a fourth proportional to three given straight lines.

Let A, B, Cbe the three given straight lines.

It is required to find a fourth proportional to A, B, C.

Take two straight lines DE, DF, containing any angle EDF:

and upon these make DG equal to Ay GE equal to B, and DH equal

to C; (I. 3.)

D A

/\ â– -^^

C

G/ \n

L \

E F

join GH, and through E draw ^i^ parallel to it. (l. 31.)

Then ^i*^ shall be the fourth proportional to A, B, C.

Because GR is parallel to EF, one of the sides of the triangle DEF,

DG is to GE, as DIT to HF; (vi. 2.)

but DG is equal to A, GE to B, and DIZ to C;

therefore, as A is to B, so is C to IIF. (v. 7.)

Wherefore to the three given straight lines A, B, C, a fourth

proportional E[F is found. Q. E. F.

PROPOSITION XIII. PROBLEM.

To find a mean proportional betioeen ttco given straight lines.

Let AB, BChe the two given straight lines.

It is required to find a mean proportional between them.

L::

A

Place AB, BC in a. straight line, and upon AC describe the semi-

circle ADC,

and from the point B draw BD at right angles to A C. (l. 11.)

Then BD shall be a mean proportional between AB and BC.

Join AD, DC,

212

And because the angle ADC in a semicircle is a right angle, (lii. 31.)

and because in the right-angled triangle ADC, BD is drawn from

the right angle perpendicular to the base,

DB is a mean proportional between AB, BC the segments of the

base : (vi. 8. Cor.)

therefore between the two given straight lines AB, BC, di. mean

proportional DB is found, q.e.f.

PROPOSITION XIV. THEOREM.

Equal parallelograms, which have one angle of the one equal to one

angle of the other, have their sides about the equal angles reciprocally pro-

portional: and conversely, parallelograms that have one angle of the one

equal to one angle of the other, and their sides about the equal angles reci-

procally proportional, are equal to one another.

Let AB, BChe equal parallelograms, which have the angles at B

equal.

The sides of the parallelograms AB, BC about the equal angles,

shall be reciprocally proportional ;

that is, DB shall be to BB, as GB to BF.

G C

Let the sides DB, BB be placed in the same straight line ;

wherefore also FB, BG are in one straight line : (i. 14.)

complete the parallelogram FF.

And because the parallelogram AB is equal to BC, and that FF

is another parallelogram,

AB is to FE, as B C to FF: (v. 7.)

but as ^^ to FF so is the base DB to BF, (vi. 1.)

and as ^C to FF, so is the base GB to BF;

therefore, as DB to BF, so is GB to BF: (v. 11.)

Wherefore, the sides of the parallelograms AB, BC about their

equal angles are reciprocally proportional.

Next, let the sides about the equal angles be reciprocally proportional,

viz. as DB to BF, so GB to BF:

the parallelogram AB shall be equal to the parallelogram BC.

Because, as DB to BF, so is GB to BF;

and as DB to BF, so is the parallelogram AB to the parallelogram

FF; (VI. 1.)

and as GB to BF, so is the parallelogram ^Cto the parallelogram FF;

therefore as AB to FF, so B C to FF: (v. 11.)

therefore the parallelogram AB is equal to the parallelogram BC.

(Y. 9.)

Therefore equal parallelograms, &c. q.e.d.

BOOK vr. PROP. XV. 273

PROPOSITION XV. THEOREM.

Equal triangles lohich have one angle of the one equal to one angle of

the other y have their sides about the equal angles reciprocally proportional :

and conversely ^ triangles which have one angle in the one equal to one angle

in the other, and their sides about the equal angles reciprocally proportional,

are equal to one another.

Let -4^ C, ^Z)^ be equal triangles, which have the angle BAC

equal to the angle DAE.

Then the sides about the equal angles of the triangles shall be re-

ciprocally proportional ;

that is, CA shall be to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD be in one

straight line ;

wherefore also EA and AB are in one straight line ; (i. 14.)

and join BD.

Because the ti'iangle ABCis equal to the triangle ADE,

and that ABD is another triangle ;

therefore as the triangle CAB, is to the triangle BAD, so is the

triangle AED to the triangle DAB; (v. 7.)

but as the triangle CAB to the triangle BAD, so is the base CA

to the base AD, (vi. 1.)

and as the triangle EAD to the triangle DAB, so is the base EA

to the base AB ; (vi. 1.)

therefore as CA to AD, so is EA to AB: (v. 11.)

wherefore the sides of the triangles ABC, ADE, about the equal

angles are reciprocally proportional.

Next, let the sides of the triangles ABC, ADE about the equal

angles be reciprocally proportional,

viz. CA to AD as EA to AB.

Then the triangle ^ji5C shall be equal to the triangle ADE.

Join BD as before.

Then because, as CA to AD, so is EA to AB; (hyp.)

and as CA to AD, so is the triangle ABC to the triangle BAD-.

(VI. 1.)

and as EA to AB, so is the triangle EAD to the triangle BAD ;

(VI. 1.)

therefore as the triangle BA C to the triangle BAD, so is the tri-

angle EAD to the triangle BAD; (v. 11.)

that is, the triangles BA C, EAD have the same ratio to the tri-

angle BAD :

wherefore the triangle ABCis equal to the triangle ADE. (v. 9.)

Therefore, equal triangles, &c. q.e.d.

N 5

274

PROPOSITION XVI. THEOREM.

Jf four straight lines be proportionals, the rectangle contained by the

extremes is equal to the rectangle contained by the means: and conversely,

if the rectangle contained by the extremes be equal to the rectangle con-

tained by the means, the four straight lines are proportionals.

Let the four straight lines AB, CD, H, Fhe proportionals,

viz. as AB to CD, so E to F.

The rectangle contained by AB, F, shall be equal to the rectangle

contained by CZ), E.

From the points A, C draw A G, CH at right angles to AB, CD :

(I. 11.)

and make AG equal to F, and CH" equal to E', (i. 3.)

and complete the parallelograms BG, DH. (l. 31.)

Because, as AB to CD, so is E to jP;

and that E is equal to CJS, and F to AG;

AB is to CD as CHto AG: (v. 7.)

therefore the sides of the parallelograms BG, DJI ahout the equal

angles are reciprocally proportional ;

but parallelograms which have their sides about equal angles reci-

procally proportional, are equal to one another : (vi. 14.)

therefore the parallelogram BG is equal to the parallelogram DH:

but the parallelogram. BG is contained by the straight lines AB, F;

because ^G^ is equal to F;

and the parallelogram DH is contained by CD and E;

because CH is equal to E ;

therefore the rectangle contained by the straight lines AB, F, is

equal to that which is contained by CD and E.

And if the rectangle contained by the straight lines AB, F, be

equal to that which is contained by CD, E;

these four lines shall be proportional,

viz. AB shall be to CD, as E to F.

The same construction being made,

because the rectangle contained by the straight lines AB, F, is

equal to that which is contained by CD, E,

and that the rectangle BG is contained by AB, F;

because AG is equal to F;

and the rectangle DH hy CD, E; because CH is equal to E;

therefore the parallelogram BG is equal to the parallelogram DH;

(ax. 1.)

and they are equiangular:

but the sides about the equal angles of equal parallelograms are

reciprocally proportional : (vi. 14.)

wherefore, as AB to CD, so is CHto AG.

BOOK VI. PROP. XVI, XVIT, XVIII. 275

But Cllh equal to JE, and AG to F-,

therefore as ^i? is to CD, so is F to F. (v. 7.)

"WTierefore, if four, &c. q.e.d.

PROPOSITION XVII. THEOREM.

If three straight lines be proportiorials^ the rectangle contained hy the

extremes is equal to the square on the mean ; and eo7ive?-sehjf if the rectangle

contained by the extremes be equal to the square on the mean, the three

straight lines are proportionals.

Let the three straight lines A, B, (7 be proportionals,

viz. as A to B, so B to C.

The rectangle contained by A, C shall be equal to the square on B,

A r â€”

B , ,

D I C L

C A

Take D equal to B.

And because as A to B, so B to C, and that B is equal to Z);

A is to B, as D to C: (v. 7.)

but if four straight lines be proportionals, the rectangle contained

by the extremes is equal to that which is contained by the means ;

(VI. 16.)

therefore the rectangle contained by A, C is equal to that con-

tained hj B, D:

but the rectangle contained by B, D, is the square on B,

because B is equal to D :

therefore the rectangle contained by A, C, is equal to the square on B.

And if the rectangle contained hy A, C, be equal to the square on B,

then A shall be to B, as B to C.

The same construction being raade,

because the rectangle contained by A, Cis equal to the square on B,

and the square on B is equal to the rectangle contained by B, Z>,

because B is equal to I) ;

therefore the rectangle contained by A, C, is equal to that contained

by J5, D:

but if the rectangle contained by the extremes be equal to that con-

tained by the means, the four straight lines are proportionals : (vi. 16.)

therefore A is to ^, as Z> to C:

but B is equal to D ;

â– wherefore, as A to B, so B to C.

Therefore, if three straight lines, &c. Q.E.D.

PROPOSITION XVIII. PROBLEM.

Upon a given straight line to describe a rectilineal Jigure similar, and

similarly situated, to a given rectilineal Jigure.

Let AB be the given straight line, and CDFF the given rectilineal

figure of four sides.

276 Euclid's elements.

It is required upon the given straight line AB to describe a rectili-

neal figure similar, and similarly situated, to CDEF.

&

Join DF, and at the points A, B in the straight line AB, make the

angle BAG equal to the angle at C, (l. 23.)

and the angle ABG equal to the angle CDF;

therefore the remaining angle A GB is equal to the remaining angle

CFD : (i. 32 and ax. 3.)

therefore the triangle FCD is equiangular to the triangle GAB.

Again, at the points G, B, in the straight line GB, make the angle

j5(rZr equal to the angle DFE, (l. 23.)

and the angle 6r^^ equal to FDF;

therefore the remaining angle GHB is equal to the remaining angle

FFD,

and the triangle FDF equiangular to the triangle GBH:

then, because the angle AGB is equal to the angle CFD, and BGH

to DFE,

the whole angle AGHis, equal to the whole angle CFE-, (ax. 2.)

for the same reason, the angle ABU is, equal to the angle CDE:

also the angle at A is equal to the angle at C, (constr.)

and the angle GHB to FED :

therefore the rectilineal figure ABJIG is equiangular to CDEF:

likewise these figures have their sides about the equal angles pro-

portionals ;

because the triangles GAB, FCD being equiangular,

BA is to A G, as CD to CF; (vi. 4.)

and because ^G^ is to GB, as CF to FD ;

and as GB is to Gil, so is FD to FE,

by reason of the equiangular triangles BGH, DFE,

therefore, ex sequali, AG is to GH, as Ci^ to FE. (v. 22.)

In the same manner it may be proved that AB is to BH, as CD

to DE:

and GH is to HB, as FE to ED. (vi. 4.)

"Wherefore, because the rectilineal figures ABHG, CDEF are

equiangular,

and have their sides about the equal angles proportionals,

they are similar to one another, (vi. def. 1.)

Kext, let it be required to describe upon a given straight line AB,

a rectilineal figure similar, and similarly situated, to the rectilineal

figure CDKEF of five sides.

Join DE, and upon the given straight line AB describe the rectili-

neal figure ABHG similar, and similarly situated, to the quadrilateral

figure CDEF, by the former case :

and at the points B, H, in the straight line BH, make the angle

HBL equal to the angle EDK,

and the angle BHL equal to the angle DEK;

BOOK vr. PROP, xviii, XIX. 277

therefore the remaining angle at L is equal to the remaining angle

at K. (I. 32, and ax. 3.)

And because the figures ABIIG, CDEF are similar,

the angle GHBis equal to the angle FED: (vi. def. 1.)

and BHL is equal to DEK;

wherefore the whole angle GIIL is equal to the whole angle FEK

for the same reason the angle ABL is equal to the angle CDK

therefore the five-sided figures A GHLB, CFEKJD are equiangular

and because the figures A GHB, CFED are similar,

GHh to HB, as FE to ED; (vi. def. 1.)

but as IIB to HL, so is ED to EK-, (vi. 4.)

therefore, ex sequali, G^^is to HL, as FE to EK-. (v. 22.)

for the same reason, ^ J5 is to BL, as CD to DKx

and BL is to LH, as DK to KE, (vi. 4.)

because the triangles BLII, DKE are equiangular :

therefore because the five-sided figures A GHLB, CFEKD are equi-

angular,

and have their sides about the equal angles proportionals,

they are similar to one another.

In the same manner a rectilineal figure of six sides may be described

upon a given straight line similar to one given, and so on. Q. E. f.

PROPOSITION XIX. THEOREM.

Similar triangles are to one another in the duplicate ratio of their homo-

logous sides.

Let ABC, DEF he similar triangles, having the angle B equal to

the angle E,

and let AB be to BC, as DE to EF,

so that the side J?Cmay be homologous to EF. (v. def. 12.)

Then the triangle ABC ^hdll have to the triangle DEF the dupli-

cate ratio of that which ^Chas to EF.

A D

Take BG ^ third proportional to BC, EF, (vi. 11.)

so that -SCmay be to EF, as EF to BG, and join GA,

Then, because as AB to BC, so DE to EF',

alternatelv, AB is to DE, as BC to EF: (v. 16.)

but as BC to EF, so is EF to BG ; (constr.)

therefore, as AB to DE, so is ^i^to BG: (v. 11.)

therefore the sides of the triangles ABG, DEF, which are about the

equal angles, are reciprocally proportional :

but triangles, which have the sides about two equal angles recipro-

Ically proportional, are equal to one another: (vi. 15.)

therefore the triangle ABG is equal to the triangle DEF:

^ and because as BCis to EF^ so EF to BG j

I

278 EUCLlD^S

and that if three straight lines be proportionals, the first is said to

have to the third, the duplicate ratio of that which it has to the second :

(V. def. 10.)

therefore JBChas toBG the duplicate ratio of that which ^Chas to EF:

but as ^Cis to^G^,so isthetriangle^^Ctothe triangle^^G^; (vi.l.)

therefore the triangle ABCh^^ to the triangle ABG, the duplicate

ratio of that which ^Chas to EF:

but the triangle ABG is equal to the triangle DEF-,

therefore also the triangle ^J?C has to the triangle DEF^ the dupli-

cate ratio of that which ^Chas to EF.

Therefore similar triangles, &c. Q. E. D,

Cor. From this it is manifest, that if three straight lines be pro-

portionals, as the first is to the third, so is any triangle upon the first,

to a similar and similarly described triangle upon the second.

PROPOSITION XX. THEOREM.

Similar polygons may be divided into the same number of similar tri-

angles, having the same ratio to one another that the polygons have ; and the

polygons have to one another the duplicate ratio of that which their homo-

logous sides have.

Let ABODE, FGHKL be similar polygons and let AB be the

side homologous to FG :

the polygons ABODE, FGHKL may be divided into the same

number of similar triangles, whereof each shall have to each the same

ratio which the polygons have ;

and the polygon ABODE shall have to the polygon FGHKL the

duplicate ratio of that which the side AB has to the side FG.

L^

M

C

Join BE, EC, GL, LH.

And because the polygon ABODE is similar to the polygon FGHKLy

the angle BAE is equal to the angle GEL, (vi. def. 1.)

and BA is to AE, as G^J^to FL : (vi. def. 1.)

therefore, because the triangles ABE, FGL have an angle in one,

equal to an angle in the other, and their sides about these equal angles

proportionals,

the triangle ABE is equiangular to the triangle FGL : (vi. 6.)

and therefore similar to it ; (vi. 4.)

wherefore the angle ABE is equal to the angle FGL :

and, because the polygons are similar,

the whole angle ABO is equal to the whole angle FGH; (vi. def. 1.)

therefore the remaining angle EBO is equal to the remaining angle

LGH: (I. 32. and ax. 3.)

and because the triangles ABE, FGL are similar,

EB is to BA, as LG to GF; (vi. 4.)

BOOK VI. PROP. XX. 279

and also, because the polygons are similar,

AÂ£ is to JBC, as FG to GIT; (vi. def. 1.)

therefore, ex sequali, ^j5 is to Â£C, as ZG to GH; (v. 22.)

that is, the sides about the equal angles EEC, LGHzxq proportionals ;

therefore, the triangle BBC is equiangular to the triangle LGH,

(vi. 6.) and similar to it ; (vi. 4.)

for the same reason, the triangle BCD likewise is similar to the tri-

angle ZiTX:

therefore the similar polygons ABODE, FGHKL are divided into

angles equal about which the sides are proportionals.

Let the two triangles ABC, DUFhawe one angle in the one equal

to one angle in the other,

viz. the angle BA C to the angle EDF, and the sides about two other

angles ABC, DJE^i^ proportionals,

so that AB is to BC, as BF to FF;

and in the first case, let each of the remaining angles at C, F be less

than a right angle.

The triangle ABC shall be equiangular to the triangle DFF,

viz. the angle AB C shall be equal to the angle DEF,

and the remaining angle at C equal to the remaining angle at F.

^'

L F

For if the angles ABC, DFF he not equal,

one of them must be greater than the other :

let ABC he the greater,

and at the point B, in the straight line AB,

make the angle ABG equal to the angle DFF-, (l. 23.)

and because the angle at A is equal to the angle at D, (hyp.)

and the angle ^J5 6^ to the angle DFF;

the remaining angle AGB is equal to the remaining angle DFF:

(I. 32.)

therefore the triangle ABG is equiangular to the triangle DFF:

wherefore as AB is to BG, so is DF to FF: (vi. 4.)

but as DF to FF, so, by h)q)othesis, is AB to BC;

therefore as AB to BC, so is AB to BG: (y. 11.)

and because AB has the same ratio to each of the lines BC, BG,

BCis equalto BG; (V. 9.)

and therefore the angle BGC is equal to the angle BCG : (l. 5.)

but the angle BCG is, by hypothesis, less than a right angle ;

therefore also the angle BGC is less than a right angle ;

and therefore the adjacent angle AGB must be greater than a right

angle; (I. 13.)

but it was proved that the angle AGB is equal to the angle at F;

therefore the angle at F is greater than a right angle ;

but, by the hypothesis, it is less than a right angle ; which is absurd.

Therefore the angles ABC, DFF sue not unequal,

â– â– l^ that is, they are equal :

IHK. and the angle at A is equal to the angle at D : (hyp.)

^^Wlierefore the remaining angle at C is equal to the remaming angle at

F: (I. 32.)

therefore the triangle ABC is equiangular to the triangle DFF,

n2

268

Next, let each of the angles at C, F be not less than a right angle.

Then the triangle -4<BC shall also in this case be equiangular to the

triangle DBF,

^o

A

D

E F

The same construction being made,

it may be proved in like manner that BCh equal to BGy

and therefore the angle at C equal to the angle BGC:

but the angle at C is not less than a right angle ; (hyp.)

therefore the angle BGCis not less than a right angle :

wherefore two angles of the triangle BGC are together not less than

two right angles :

which is impossible ; (l. 1 7.)

and therefore the triangle ABC may be proved to be equiangular to

the triangle DBF, as in the first case.

Lastly, let one of the angles at C, F, viz. the angle at C, be a right

angle : in this case likewise the triangle ABC shall be equiangular

to the triangle DFF.

For, if they be not equiangular,

at the point B in the straight line AB make the angle ABG equal

to the angle DBF;

then it may be proved, as in the first case, that BG is equal to BC:

and therefore the angle BCG equal to the angle BGC: (i. 5.)

but the angle BCG is a right angle, (hyp.)

therefore the angle BGCis also a right angle ; (ax. 1.)

whence two of the angles of the triangle BGC are together not less

than two right angles ;

which is impossible : (I. 17.)

tnerefore the triangle ABC is equiangular to the triangle DFF.

Wherefore, if two triangles, &c. q.e.d.

PROPOSITION VIII. THEOREM.

In a right' angled triangle, if a perpendicular he drawn from the right-

angle to the base ; the triangles on each side of it are similar to the whole

triangle, and to one another,

JjBt ABC he a right angled-triangle, having the right angle BAC;

and from the point A let AD be drawn pei-pendicular to the base BC.

Then the triangles ABD, -4 DC shall be similar to the whole tri-

angle ABC, and to one another.

BOOK VI. PROP. VllI, IX. 269

A

D

Because the angle JBACis equal to the angle ABB, each of them

being a right angle, (ax. 11.)

and that the angle at B is common to the two triangles ABC, ABD:

the remaining angle A CB is equal to the remaining angle BAD j

(I. 32.)

therefore the triangle ABC is equiangular to the triangle ABD,

and the sides about their equal angles are proportionals ; (vi. 4.)

â– wherefore the triangles are similar: (VI. def. 1.)

in the like manner it may be demonstrated, that the triangle ADC

is equiangular and similar to the triangle ^^(7.

And the triangles ABD, A CD, being both equiangular and similar

to ABC, are equiangular and similar to each other.

Therefore, in a right-angled, &c. q.e.d.

Cor. From this it is manifest, that the perpendicular drawn from

the right angle of a right-angled triangle to the base, is a mean propor-

tional between the segments of the base ; and also that each of the

sides is a mean proportional between the base, and the segment of it

adjacent to that side : because in the triangles BDA, ADC-, BD is to

DA, as DA to DC; (vi. 4.)

and in the triangles ABC, DBA ; BC is to B A, as BA to BD : (vi.4.)

and in the triangles ABC, A CD-, B Cis to CA, as CA to CD, (VI. 4 . )

PROPOSITION IX. PROBLEM.

From a given straight line to cut off any part required.

Let AB be the given straight line.

It is required to cut off any part from it.

From the point A draw a straight line A C, making any angle with AB)

and in A C take any point D,

and take A C the same multiple of AD, that AB is of the part

which is to be cut off from it ;

join BC, and draw DE parallel to CB.

Then AE shall be the part required to be cut off.

Because ED is parallel to BC, one of the sides of the triangle ABC

as CD is to DA, so is BE to EA ; (vi. 2.)

and by composition, CA is to AD, as BA to AE: (v. 18.)

270 EUCLID'S ELEMENTS.

but CA is a multiple of AD-, (constr.)

therefore JBA is the same multiple of AJS: (v. D.)

whatever part therefore AD is of A C, AE is the same part of AB:

wherefore, from the straight line AB the part required is cut off.

Q.E.F.

PROPOSITION X. PROBLEM.

To divide a given straight line similarly to a given divided straight line^

that iSy into parts that shall have the same ratios to one another which the

parts of the divided given straight line have.

Let AB be the straight line given to be divided, and AC the divided

line.

It is required to divide AB similarly to A C.

A

Let ^Cbe divided in the points D, E-,

and let AB, AC he placed so as to contain any angle, and join BC,

and through the points D, E draw DF, EG parallels to BC. (l. 31.)

Then AB shall be divided in the points F, G, similarly to A C.

Tnrough D draw DHK parallel to AB :

therefore each of the figures, FH, JIB is a parallelogram ;

wherefore EH is equal to FG, and HK to GB : (i. 34.)

and because ITE is parallel to KC, one of the sides of the triangle

EKC,

as C^to ED, so is ^^to HD: (vi. 2.)

but KHk equal to BG, and IID to GF-,

therefore, as CE is to ED, sokBG to GF: (V. 7.)

again, because FD is parallel to GE, one of the sides of the triangle

AGE,

as ED is to DA, so is G'i^'to FA : (vi. 2.)

therefore, as has been proved, as CE is to ED, so is J? 6^ to GF,

and as ED is to DA, so is GF to FA :

therefore the given straight line AB is divided similarly to -4 (7. Q.E.F.

PROPOSITION XI. PROBLEM.

To find a third proportional to two given straight lines.

Let AB, AC he the two given straight lines.

It is required to find a third proportional to AB, A C.

A

S

D E

Let AB, AC he placed so as to contain any angle :

produce AB, AC to the points D, E;

^H ^ BOOK VI. PROP. XI, XII, XIII. S71

Up and make BD equal to A C;

join J5(7, and through D, draw Z)^ parallel to BC. (l. 31.)

Then CE shall be a third proportional to ^ J? and ^ C.

Because BCh parallel to DE, a side of the triangle ADE,

AB is to BD, as AC to CE: (vi. 2.)

but BD is equal to ^ C;

therefore as AB is to ^ C, so is ^ C to CE. (v. 7.) _

Wherefore, to the two given straight lines AB, A C, a third pro-

portional C^ is found. Q.E.F.

PROPOSITION XII. PROBLEM.

To find a fourth proportional to three given straight lines.

Let A, B, Cbe the three given straight lines.

It is required to find a fourth proportional to A, B, C.

Take two straight lines DE, DF, containing any angle EDF:

and upon these make DG equal to Ay GE equal to B, and DH equal

to C; (I. 3.)

D A

/\ â– -^^

C

G/ \n

L \

E F

join GH, and through E draw ^i^ parallel to it. (l. 31.)

Then ^i*^ shall be the fourth proportional to A, B, C.

Because GR is parallel to EF, one of the sides of the triangle DEF,

DG is to GE, as DIT to HF; (vi. 2.)

but DG is equal to A, GE to B, and DIZ to C;

therefore, as A is to B, so is C to IIF. (v. 7.)

Wherefore to the three given straight lines A, B, C, a fourth

proportional E[F is found. Q. E. F.

PROPOSITION XIII. PROBLEM.

To find a mean proportional betioeen ttco given straight lines.

Let AB, BChe the two given straight lines.

It is required to find a mean proportional between them.

L::

A

Place AB, BC in a. straight line, and upon AC describe the semi-

circle ADC,

and from the point B draw BD at right angles to A C. (l. 11.)

Then BD shall be a mean proportional between AB and BC.

Join AD, DC,

212

And because the angle ADC in a semicircle is a right angle, (lii. 31.)

and because in the right-angled triangle ADC, BD is drawn from

the right angle perpendicular to the base,

DB is a mean proportional between AB, BC the segments of the

base : (vi. 8. Cor.)

therefore between the two given straight lines AB, BC, di. mean

proportional DB is found, q.e.f.

PROPOSITION XIV. THEOREM.

Equal parallelograms, which have one angle of the one equal to one

angle of the other, have their sides about the equal angles reciprocally pro-

portional: and conversely, parallelograms that have one angle of the one

equal to one angle of the other, and their sides about the equal angles reci-

procally proportional, are equal to one another.

Let AB, BChe equal parallelograms, which have the angles at B

equal.

The sides of the parallelograms AB, BC about the equal angles,

shall be reciprocally proportional ;

that is, DB shall be to BB, as GB to BF.

G C

Let the sides DB, BB be placed in the same straight line ;

wherefore also FB, BG are in one straight line : (i. 14.)

complete the parallelogram FF.

And because the parallelogram AB is equal to BC, and that FF

is another parallelogram,

AB is to FE, as B C to FF: (v. 7.)

but as ^^ to FF so is the base DB to BF, (vi. 1.)

and as ^C to FF, so is the base GB to BF;

therefore, as DB to BF, so is GB to BF: (v. 11.)

Wherefore, the sides of the parallelograms AB, BC about their

equal angles are reciprocally proportional.

Next, let the sides about the equal angles be reciprocally proportional,

viz. as DB to BF, so GB to BF:

the parallelogram AB shall be equal to the parallelogram BC.

Because, as DB to BF, so is GB to BF;

and as DB to BF, so is the parallelogram AB to the parallelogram

FF; (VI. 1.)

and as GB to BF, so is the parallelogram ^Cto the parallelogram FF;

therefore as AB to FF, so B C to FF: (v. 11.)

therefore the parallelogram AB is equal to the parallelogram BC.

(Y. 9.)

Therefore equal parallelograms, &c. q.e.d.

BOOK vr. PROP. XV. 273

PROPOSITION XV. THEOREM.

Equal triangles lohich have one angle of the one equal to one angle of

the other y have their sides about the equal angles reciprocally proportional :

and conversely ^ triangles which have one angle in the one equal to one angle

in the other, and their sides about the equal angles reciprocally proportional,

are equal to one another.

Let -4^ C, ^Z)^ be equal triangles, which have the angle BAC

equal to the angle DAE.

Then the sides about the equal angles of the triangles shall be re-

ciprocally proportional ;

that is, CA shall be to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD be in one

straight line ;

wherefore also EA and AB are in one straight line ; (i. 14.)

and join BD.

Because the ti'iangle ABCis equal to the triangle ADE,

and that ABD is another triangle ;

therefore as the triangle CAB, is to the triangle BAD, so is the

triangle AED to the triangle DAB; (v. 7.)

but as the triangle CAB to the triangle BAD, so is the base CA

to the base AD, (vi. 1.)

and as the triangle EAD to the triangle DAB, so is the base EA

to the base AB ; (vi. 1.)

therefore as CA to AD, so is EA to AB: (v. 11.)

wherefore the sides of the triangles ABC, ADE, about the equal

angles are reciprocally proportional.

Next, let the sides of the triangles ABC, ADE about the equal

angles be reciprocally proportional,

viz. CA to AD as EA to AB.

Then the triangle ^ji5C shall be equal to the triangle ADE.

Join BD as before.

Then because, as CA to AD, so is EA to AB; (hyp.)

and as CA to AD, so is the triangle ABC to the triangle BAD-.

(VI. 1.)

and as EA to AB, so is the triangle EAD to the triangle BAD ;

(VI. 1.)

therefore as the triangle BA C to the triangle BAD, so is the tri-

angle EAD to the triangle BAD; (v. 11.)

that is, the triangles BA C, EAD have the same ratio to the tri-

angle BAD :

wherefore the triangle ABCis equal to the triangle ADE. (v. 9.)

Therefore, equal triangles, &c. q.e.d.

N 5

274

PROPOSITION XVI. THEOREM.

Jf four straight lines be proportionals, the rectangle contained by the

extremes is equal to the rectangle contained by the means: and conversely,

if the rectangle contained by the extremes be equal to the rectangle con-

tained by the means, the four straight lines are proportionals.

Let the four straight lines AB, CD, H, Fhe proportionals,

viz. as AB to CD, so E to F.

The rectangle contained by AB, F, shall be equal to the rectangle

contained by CZ), E.

From the points A, C draw A G, CH at right angles to AB, CD :

(I. 11.)

and make AG equal to F, and CH" equal to E', (i. 3.)

and complete the parallelograms BG, DH. (l. 31.)

Because, as AB to CD, so is E to jP;

and that E is equal to CJS, and F to AG;

AB is to CD as CHto AG: (v. 7.)

therefore the sides of the parallelograms BG, DJI ahout the equal

angles are reciprocally proportional ;

but parallelograms which have their sides about equal angles reci-

procally proportional, are equal to one another : (vi. 14.)

therefore the parallelogram BG is equal to the parallelogram DH:

but the parallelogram. BG is contained by the straight lines AB, F;

because ^G^ is equal to F;

and the parallelogram DH is contained by CD and E;

because CH is equal to E ;

therefore the rectangle contained by the straight lines AB, F, is

equal to that which is contained by CD and E.

And if the rectangle contained by the straight lines AB, F, be

equal to that which is contained by CD, E;

these four lines shall be proportional,

viz. AB shall be to CD, as E to F.

The same construction being made,

because the rectangle contained by the straight lines AB, F, is

equal to that which is contained by CD, E,

and that the rectangle BG is contained by AB, F;

because AG is equal to F;

and the rectangle DH hy CD, E; because CH is equal to E;

therefore the parallelogram BG is equal to the parallelogram DH;

(ax. 1.)

and they are equiangular:

but the sides about the equal angles of equal parallelograms are

reciprocally proportional : (vi. 14.)

wherefore, as AB to CD, so is CHto AG.

BOOK VI. PROP. XVI, XVIT, XVIII. 275

But Cllh equal to JE, and AG to F-,

therefore as ^i? is to CD, so is F to F. (v. 7.)

"WTierefore, if four, &c. q.e.d.

PROPOSITION XVII. THEOREM.

If three straight lines be proportiorials^ the rectangle contained hy the

extremes is equal to the square on the mean ; and eo7ive?-sehjf if the rectangle

contained by the extremes be equal to the square on the mean, the three

straight lines are proportionals.

Let the three straight lines A, B, (7 be proportionals,

viz. as A to B, so B to C.

The rectangle contained by A, C shall be equal to the square on B,

A r â€”

B , ,

D I C L

C A

Take D equal to B.

And because as A to B, so B to C, and that B is equal to Z);

A is to B, as D to C: (v. 7.)

but if four straight lines be proportionals, the rectangle contained

by the extremes is equal to that which is contained by the means ;

(VI. 16.)

therefore the rectangle contained by A, C is equal to that con-

tained hj B, D:

but the rectangle contained by B, D, is the square on B,

because B is equal to D :

therefore the rectangle contained by A, C, is equal to the square on B.

And if the rectangle contained hy A, C, be equal to the square on B,

then A shall be to B, as B to C.

The same construction being raade,

because the rectangle contained by A, Cis equal to the square on B,

and the square on B is equal to the rectangle contained by B, Z>,

because B is equal to I) ;

therefore the rectangle contained by A, C, is equal to that contained

by J5, D:

but if the rectangle contained by the extremes be equal to that con-

tained by the means, the four straight lines are proportionals : (vi. 16.)

therefore A is to ^, as Z> to C:

but B is equal to D ;

â– wherefore, as A to B, so B to C.

Therefore, if three straight lines, &c. Q.E.D.

PROPOSITION XVIII. PROBLEM.

Upon a given straight line to describe a rectilineal Jigure similar, and

similarly situated, to a given rectilineal Jigure.

Let AB be the given straight line, and CDFF the given rectilineal

figure of four sides.

276 Euclid's elements.

It is required upon the given straight line AB to describe a rectili-

neal figure similar, and similarly situated, to CDEF.

&

Join DF, and at the points A, B in the straight line AB, make the

angle BAG equal to the angle at C, (l. 23.)

and the angle ABG equal to the angle CDF;

therefore the remaining angle A GB is equal to the remaining angle

CFD : (i. 32 and ax. 3.)

therefore the triangle FCD is equiangular to the triangle GAB.

Again, at the points G, B, in the straight line GB, make the angle

j5(rZr equal to the angle DFE, (l. 23.)

and the angle 6r^^ equal to FDF;

therefore the remaining angle GHB is equal to the remaining angle

FFD,

and the triangle FDF equiangular to the triangle GBH:

then, because the angle AGB is equal to the angle CFD, and BGH

to DFE,

the whole angle AGHis, equal to the whole angle CFE-, (ax. 2.)

for the same reason, the angle ABU is, equal to the angle CDE:

also the angle at A is equal to the angle at C, (constr.)

and the angle GHB to FED :

therefore the rectilineal figure ABJIG is equiangular to CDEF:

likewise these figures have their sides about the equal angles pro-

portionals ;

because the triangles GAB, FCD being equiangular,

BA is to A G, as CD to CF; (vi. 4.)

and because ^G^ is to GB, as CF to FD ;

and as GB is to Gil, so is FD to FE,

by reason of the equiangular triangles BGH, DFE,

therefore, ex sequali, AG is to GH, as Ci^ to FE. (v. 22.)

In the same manner it may be proved that AB is to BH, as CD

to DE:

and GH is to HB, as FE to ED. (vi. 4.)

"Wherefore, because the rectilineal figures ABHG, CDEF are

equiangular,

and have their sides about the equal angles proportionals,

they are similar to one another, (vi. def. 1.)

Kext, let it be required to describe upon a given straight line AB,

a rectilineal figure similar, and similarly situated, to the rectilineal

figure CDKEF of five sides.

Join DE, and upon the given straight line AB describe the rectili-

neal figure ABHG similar, and similarly situated, to the quadrilateral

figure CDEF, by the former case :

and at the points B, H, in the straight line BH, make the angle

HBL equal to the angle EDK,

and the angle BHL equal to the angle DEK;

BOOK vr. PROP, xviii, XIX. 277

therefore the remaining angle at L is equal to the remaining angle

at K. (I. 32, and ax. 3.)

And because the figures ABIIG, CDEF are similar,

the angle GHBis equal to the angle FED: (vi. def. 1.)

and BHL is equal to DEK;

wherefore the whole angle GIIL is equal to the whole angle FEK

for the same reason the angle ABL is equal to the angle CDK

therefore the five-sided figures A GHLB, CFEKJD are equiangular

and because the figures A GHB, CFED are similar,

GHh to HB, as FE to ED; (vi. def. 1.)

but as IIB to HL, so is ED to EK-, (vi. 4.)

therefore, ex sequali, G^^is to HL, as FE to EK-. (v. 22.)

for the same reason, ^ J5 is to BL, as CD to DKx

and BL is to LH, as DK to KE, (vi. 4.)

because the triangles BLII, DKE are equiangular :

therefore because the five-sided figures A GHLB, CFEKD are equi-

angular,

and have their sides about the equal angles proportionals,

they are similar to one another.

In the same manner a rectilineal figure of six sides may be described

upon a given straight line similar to one given, and so on. Q. E. f.

PROPOSITION XIX. THEOREM.

Similar triangles are to one another in the duplicate ratio of their homo-

logous sides.

Let ABC, DEF he similar triangles, having the angle B equal to

the angle E,

and let AB be to BC, as DE to EF,

so that the side J?Cmay be homologous to EF. (v. def. 12.)

Then the triangle ABC ^hdll have to the triangle DEF the dupli-

cate ratio of that which ^Chas to EF.

A D

Take BG ^ third proportional to BC, EF, (vi. 11.)

so that -SCmay be to EF, as EF to BG, and join GA,

Then, because as AB to BC, so DE to EF',

alternatelv, AB is to DE, as BC to EF: (v. 16.)

but as BC to EF, so is EF to BG ; (constr.)

therefore, as AB to DE, so is ^i^to BG: (v. 11.)

therefore the sides of the triangles ABG, DEF, which are about the

equal angles, are reciprocally proportional :

but triangles, which have the sides about two equal angles recipro-

Ically proportional, are equal to one another: (vi. 15.)

therefore the triangle ABG is equal to the triangle DEF:

^ and because as BCis to EF^ so EF to BG j

I

278 EUCLlD^S

and that if three straight lines be proportionals, the first is said to

have to the third, the duplicate ratio of that which it has to the second :

(V. def. 10.)

therefore JBChas toBG the duplicate ratio of that which ^Chas to EF:

but as ^Cis to^G^,so isthetriangle^^Ctothe triangle^^G^; (vi.l.)

therefore the triangle ABCh^^ to the triangle ABG, the duplicate

ratio of that which ^Chas to EF:

but the triangle ABG is equal to the triangle DEF-,

therefore also the triangle ^J?C has to the triangle DEF^ the dupli-

cate ratio of that which ^Chas to EF.

Therefore similar triangles, &c. Q. E. D,

Cor. From this it is manifest, that if three straight lines be pro-

portionals, as the first is to the third, so is any triangle upon the first,

to a similar and similarly described triangle upon the second.

PROPOSITION XX. THEOREM.

Similar polygons may be divided into the same number of similar tri-

angles, having the same ratio to one another that the polygons have ; and the

polygons have to one another the duplicate ratio of that which their homo-

logous sides have.

Let ABODE, FGHKL be similar polygons and let AB be the

side homologous to FG :

the polygons ABODE, FGHKL may be divided into the same

number of similar triangles, whereof each shall have to each the same

ratio which the polygons have ;

and the polygon ABODE shall have to the polygon FGHKL the

duplicate ratio of that which the side AB has to the side FG.

L^

M

C

Join BE, EC, GL, LH.

And because the polygon ABODE is similar to the polygon FGHKLy

the angle BAE is equal to the angle GEL, (vi. def. 1.)

and BA is to AE, as G^J^to FL : (vi. def. 1.)

therefore, because the triangles ABE, FGL have an angle in one,

equal to an angle in the other, and their sides about these equal angles

proportionals,

the triangle ABE is equiangular to the triangle FGL : (vi. 6.)

and therefore similar to it ; (vi. 4.)

wherefore the angle ABE is equal to the angle FGL :

and, because the polygons are similar,

the whole angle ABO is equal to the whole angle FGH; (vi. def. 1.)

therefore the remaining angle EBO is equal to the remaining angle

LGH: (I. 32. and ax. 3.)

and because the triangles ABE, FGL are similar,

EB is to BA, as LG to GF; (vi. 4.)

BOOK VI. PROP. XX. 279

and also, because the polygons are similar,

AÂ£ is to JBC, as FG to GIT; (vi. def. 1.)

therefore, ex sequali, ^j5 is to Â£C, as ZG to GH; (v. 22.)

that is, the sides about the equal angles EEC, LGHzxq proportionals ;

therefore, the triangle BBC is equiangular to the triangle LGH,

(vi. 6.) and similar to it ; (vi. 4.)

for the same reason, the triangle BCD likewise is similar to the tri-

angle ZiTX:

therefore the similar polygons ABODE, FGHKL are divided into

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38