Euclid.

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

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Proposition 10.

If four magnitiules of the same kind are proportionals^
the first is greater than, equal to, or less than tJie third,
according as the second is greater thati, equal to, or less than
tlie fourth.

Let A, B, C, D be four magnitudes of the same kind such
that

A : B :: C : D;
then A > = or < C
according as B > = or < D.

If B > D, then A : B < A : D : v. 5.

but A : B :: C : D;
.*. C : D< A : D;
.*. A : D>C : D:

.â€˘. A> C. V. 7.

Similarly it may be shewn that

if B < D, then A < C,
and if B ^ D, then A â€” C.

Proposition 11.

If four magnitudes of the same kind are proportionals,
they are also proportionals when taken alternately.

Let A, B, C, D be four magnitudes of the same kind such
that

then shall

A : B :: C : D;

A : C :: B : D.

Because A : B : : mA : mB,

V. 8.

and C : D :: r^C : wD;

.'. mA : TwB :: nC : nD.

V. 1.

:. mA>-=or <nC

according asmB>-or <nD:

V. 10.

,nd m ai^d n are any whole numbers;

.-. A : C :: B : D.

Def 4.

Note, This inference is usually referred to as altemando or
alternately.

proofs of the propositions of book v. 303

Proposition 12.

If any number of magnitudes of the same kind are 'pro-
portionals^ as one of the antecedents is to its consequent, so
is the sum of the antecedents to the sum of the consequents.

Let A, B, C, D, E, F, . . . be magnitudes of the same kind
such that

A:B::C:D::E:F:: ;

then shall A:B::A + C + E + ... :B + D+F+....

Because A:B::C:D::E:F::...,
.â– . according as mA> = or <:nE,
so is mC>=or <nD,
and mE > = or < nF,

.*. so is mk + mC + mE + ... >=^ov <nB +nD + nf + ...
or m(A+ C + E+...)> = or <7i(B 4- D+ F+ ...);
and wt and n are any whole numbers;

.â€˘. A : B ::A+C + E+... : B+D+F+.... Def 4.

Note. This inference is usually referred to as addendo.

Proposition 13,

If four magnitudes are proportionals, the sudi or differ-
ence of the first and second is to the second as the sum, or
difference of the third and fourth is to the fourth.

Let A : B :

:C:D;

shall

A+B : B :

: C + D : D,

and A ~ B : B :

: C ~ D : D.

If m be any whole number, it is possible to find another
number n such that mA = nB, or lies between wB and
(7^+l)B,

.'. mA + mB = mB -f wB, or lies between mB -f- nB and

wiB + {n + 1) B.

:X)4 EUCLIIVS ELEMENTS.

But DiA 1 mB = m(A + B), and 7Â«B + nS =- {m + w) B ;

.*. m(A + B) - (/M + n) B, or lies between {^tn + w) B

and (m + ?i + 1 ) B.

Also because A : B : : C : D,

.'. mC = ?iD, or lies between nD and (n+ 1)D; Def. 4.

.'. m(C + D) = {m + n) D or lies between (m + n) D and

{m + n + \)D;

that is, the multiples of C + D are distributed among those

of D in the same way as the multiples of A + B among

those of B;

.'. A+ 8 : B :: C + D : D.

In the same Avay it may be proved that
A-B: B::C-D:D,
or B - A : B : : D â€” C : D,
according as A is > or < B.

Note. These inferences are referred to as componendo and divi-
dendo respectively.

Proposition 14.

If there are ttuo sets of magnitudes, such that the first is
to the second of the first set as the first to the second of the
otlier set, and the second to the third of the first set as the
second to the third of tlm other, and so on to the last nia(jni-
tude : then the first is to the last of the first set as tJie fi/rst to
the last of the other.

First, let there be three magnitudes A, B, C, of one set
and tliree, P, Q, R, of another set,

and let A : B : : P : Q,

and B : C :

: Q : R;

then shall A : C :

: P : R.

Because A : B :

: P: Q,

.". mA : mB :

: mP : mQ;

V. 8, Cor.

and because B : C :

:Q: R,

.'. mB : nC :

: 7)1Â®. : 7i,R,

V. 9.

invertendv, nC : mB :

: nR : mQ.

V. 3.

PROOFS OF I'HK PROPOSITIONS OF ROOK \. 305

Now, if mA > nC,

then mA : mB>7iC : mB; v. 5.

.". 77iP : mQ> nR : mQ,

and .'. mP>7iR. v. 7.

Similarly it may be shewn that mP == or < 7iR,
according as mA = or < nC,

.'. A : C :: P : R. J)e/. 4.

Secondly, let there be any number of magnitudes, A, B,
C, ... L, M, of one set, and the same number P, Q, R, ...Y, Z,
of another set, such that

A : B : : P : Q,

B : C :: Q : R,

L : M :: Y : Z;
then shall A : M :: P : Z.

For A : C :: P : R, Proved.

and C : D :: R : S; Hyp.

.'. by the first case A : D : : P : S,
and so on, until finally

A : M = P : Z.
Note. This inference is referred to as ex sequali.

Corollary. If A : B :: P : Q,

and B : C :: R : P;

then A : C :: R : Q.

PROPOSITIOIf 15.

If A : B :: C : D,

and E : B :: F : D;

then shall A+E:B::C+F:D.

For since E : B :: F : D,

IIl/p.

.*. , invertendo, B : E :: D : F.

V. 3.

Also A : B :: C : D,

.'. , ex ceqicali, A : E :: C : F,

V. U.

H. E.

20

30(5 kuclid's elements.

.'. , componendo, A + E : E :: C + F : F. v. 13.

Again, E : B :: F : D, Hyp.

.". , ex cequali, A + E : B : : C 4- F : D. v. 14.

Proposition 16.

Jf two ratios are equal, their duplicate ratios are equal;
and conversely/, if the duplicate ratios of two ratios are equal,
the ratios t/iemselves are equal.

Let A : B : : C : D ;
then shall the duplicate ratio of A to B be equal to that of
C to D.

Let X be a third proportional to A and B, and Y a third
proportional to C and D,

so that A : B :: B : X, and C : D :: D : Y;
then because A : B : : C : D,
.'. B : X :: D : Y;
.'. , ex cequali, A : X :: C : Y.

But A : X and C : Y are respectively the duplicate ratios of
A : B and C : D, Def 13.

.". the duplicate ratio of A : B = that of C : D.

Conversely, let the duplicate ratio of A : B = that of C : D;
then shall A : B :: C : D.

Let P be such that A : B : : C

: P,

.'. , invertendo, B : A :: P :

: C.

Also, by liypothesis, A : X :: C

: Y,

.". , ex cequali, B : X :: P

:Y;

but A : B :: B

:X,

.-. A : B :: P

: Y; V. 1.

.-. C : P :: P ;

: Y; V. 1.

that is, P is the mean proportional

between C and Y.

.-. P=D,

.'. A : B :: C : D.

BOOK YL

Definitions.

1. Two rectilineal figures are said to be equiangular
when the angles of the first, taken in order, are equal
respectively to those of the second, taken in order. Each
angle of. the first figure is said to correspond to the angle
to which it is equal in the second figure, and sides adjacent
to corresponding angles are called corresponding sides.

2. Rectilineal figures are said to be similar when they
are equiangular and have the sides about the equal angles
proportionals, the corresponding sides being homologous.

[See Def. 5, page 288.]

Thus the two quadrilaterals ABCD, EFGH are similar if the
angles at A, B, C, D are respec-
tively equal to those at E, F, G, H,
and if the following proportions
hold

AB : BC :: EF : FG,

BC : CD:: FG : GH,

CD: DA :: GH : HE,

DA : AB :: HE : EF.

3. Two figures are said to have their sides about two
of their angles reciprocally proportional when a side of the
first is to a side of the second as the remaining side of the
second is to the remaining side of the first.

4. A straight line is said to be divided in extreme
and mean ratio when the whole is to the greater segment
as the greater segment is to the less.

5. Two similar rectilineal figures are said to be similarly
situated with respect to two of their sides when these
sides are homologous.

20-2

B

3UÂ«

EUCLID S ELEMENTS.

Proposition 1. Theorem.

The areas of triaru/les of the same altitude are to one
anotlter as t/teir bases.

Let ABC, ACD be two triangles of the same altitude,
namely the perpendicular from A to BD:

then shall the A A^ : the A ACD :: BC : CD.
Produce BD both ways,
and from CB produced cut oft* any number of parts BG, GH,
each equal to BC ;

and from CD produced cut off" any number of parts DK,
KL, LM each equal to CD.

Join AH, AG, AK, AL, AM.
Then the A"* ABC, ABG, AgH are equal in area, for they
are of the same altitude and stand on the equal bases
CB, BG, GH, I. 38.

.*. the A AHC is the same multiple of the A ABC that HC
is of BC ;

Similarly the A ACM is the same multiple of ACD that CM
is of CD.

And if HC = CM,'

the A AHC = the A ACM; I. 38.

and if HC is greater than CM,

the A AHC is greater than the A ACM; i. 38, Cor.

and if HC is less than CM,

the A AHC is less than the A ACM. i. 38, Cor.

Now since there are four magnitudes, namely, the

A* ABC, ACD, and the bases BC, CD; and of the antecedents,

any equimultiples have been taken, namely, the A AHC

BOOK XL PROP. 1. 309

and the base HC ; and of the consequents, any equi-
multiples have been taken, namely the A ACM and the
base CM; and since it has been shewn that the A AHC is
greater than, equal to, or less than the A ACM, according
as HC is greater than, equal to, or less than CM;
.'. the four original magnitudes are proportionals, v. Bef. 4.

that is,

the A ABC : the AACD :: the base BC : the base CD. q.e.d.

Corollary. TJie areas of parallelograms of the same
altitude are to one another as their bases.

Let EC, CF be par'"' of the same altitude;
then shall the par"" EC : the par"' CF :: BC : CD.
Then the A ABC : the AACD :: BC : CD; Proved.
but the par"* EC is double of the A ABC,
and the parâ„˘ CF is double of the AACD;
.'. the par"* EC : the par"* CF :: BC : CD. v. 8.

Note. Two straight lines are cut proportionally when the seg-
ments of one line are in the same ratio as the corresponding segments
of the other. [See definition, page 131.]

Fig.l Fig.2
A X B A B X

YD

Thus AB and CD are cut proportionally at X and Y, if
AX : XB :: CY : YD.
And the same definition applies equally whether X and Y divide AB,
CD internally as in Fig. 1 or externally as in Fig. 2.

310

'euclid's elements.
Proposition 2. Theorem.

If a straight line he drawn parallel to one side of a
triangle, it shall cut the other sides, or those sides produced,
proportionally:

Conversely, if the sides or the sides produced he cut p7'o-
portionally, the straight line which joins the points of section,
shall he parallel to the remaining side of the triangle.

A A Y X

Let XY be drawn par' to BC, one of the sides of the
A ABC:
then shall BX : XA : : CY : YA.

Join BY, ex.
Then the A BXY = the A CXY, being on the same base XY
and between the same parallels XY, BC; i. 37.

and AXY is another triangle;
.'. the A BXY : the A AXY :: the A CXY : the A AXY. V. 4.
But the A BXY : the A AXY :: BX : XA, vi. 1.

and the A CXY : the A AXY : : CY : YA,

.-. BX : XA :: CY : YA. v. 1.

Conversely, let BX : XA :: CY : YA, and let XY be joined:
then shall XY be par' to BC.
As before, join BY, CX.
By hypothesis BX : XA :: CY : YA;

but BX : XA :: tlie A BXY : the A AXY, vi. 1.

and CY : YA :: the ACXY : the AAXY ;

.-. the A BXY : the A AXY :: the A CXY : the A AXY. v. 1.

,-. the A BXY -the ACXY; v. 6.

and they are triangles on the same base and on the same

side of it.

.-. XY is par' to BC. â€˘ I. 39.

Q.E.D.

BOOK VI. PROP. 2. 311

EXERCISES.

1. Shew that every quadrilateral is divided by its diagonals into
four triangles proportional to each other.

2. If any tico straight lines are cut by three parallel straight lines,
they are cut proportionally.

3. From a point E in the common base of two triangles ACB,
ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD at
F, G : shew that FG is parallel to CD.

4. In a triangle ABC the straight line DEF meets the sides
BC, CA, AB at the points D, E, F respectively, and it makes
equal angles with AB and AC: prove that

BD : CD :: BF : CE.

5. If the bisector of the angle B of a triangle ABC meets AD at
right angles, shew that a line through D parallel to BC will bisect
AC.

6. From B and C, the extremities of the base of a triangle ABC,
lines BE, CF are drawn to the opposite sides so as to intersect on
the median from A: shew that EF is parallel to BC.

7. From P, a given point in the side AB of a triangle ABC,
draw a straight line to AC produced, so that it will be bisected
by BC.

8. Find a point within a triangle such that, if straight lines be
drawn from it to the three angular points, the triangle will be divided
into three equal triangles.

312 EUCLID'S ELEMENTS.

Proposition 3. Theorem.

If the vertical angle of a triangle he bisected by a straight
line which cuts tlie base^ the segments of the base shall have
to one another the same ratio as the remaining sides of the
triangle:

Conversely, if the base he divided so tJiat its segments
have to one another the same ratio as the remxiining sides of
tlie triangle Imve, tlie straight line drawn from the vertex to
the point of section shall bisect the vertical angle.

In the A ABC let the z. BAG be bisected by AX, which
meets the base at X ;
then shall BX : XC : : BA : AC.

Through C draw CE par' to XA, to meet BA produced

at E. I. 31.
Then because XA and CE are par',

.'. the L BAX = the int. opp. l AEC, i. 29.

and the l XAC = the alt. i. ACE. i. 29.

But the I. BAX = the l XAC; Hyp.
.'. the ^ AEC = the ^ ACE;

AC = AE. 1.6.

Again, because XA is par' to CE, a side of the A BCE,

.'. BX : XC :: BA : AE; VI. 2.

that is, BX : XC : : BA : AC.

BOOK VI. PROP. 3. 313

Conversely, let BX : XC :: BA : AC; and let AX be joined:

then shall the z. BAX ^ z. XAC.
For, with the same construction as before,

because XA is par' to CE, a side of the A BCE,

.-. BX : XC :: BA : AE. VI. 2.

But by hypothesis BX : XC :: BA : AC;

.-. BA : AE :: BA : AC; v. 1.

.'. AE = AC;
.-. the L ACE = the l AEC. i. 5.

But because XA is par' to CE,
.-. the /_ XAC = the alt. z. ACE. i. 29.

and the ext. z_ BAX = the int. opp. l AEC; i. 29.
.-. the L BAX = the l XAC.

Q.E.D.

EXERCISES.

1. The side BC of a triangle ABC is bisected at D, and the angles
ADB, ADC are bisected by the straight lines DE, DF, meeting AB,
AC at E, F respectively: shew that EF is parallel to BC.

2. Apply Proposition 3 to trisect a given finite straight line.

3. If the line bisecting the vertical angle of a triangle be divided
into parts which are to one another as the base to the sum of the
sides, the point of division is the centre of the inscribed circle.

4. ABCD is a quadrilateral: shew that if the bisectors of the
angles A and C meet in the diagonal BD, the bisectors of the angles
B and D will meet on AC.

5. Construct a triangle having given the base, the vertical angle,
and the ratio of the remaining sides.

6. Employ this proposition to shew that the bisectors of the
angles of a triangle are concurrent.

7. AB is a diameter of a circle,, CD is a chord at right angles to
it, and E any point in CD: AE and BE are drawn and produced to
cut the circle in F and G : shew that the quadrilateral CFDG has any
two of its adjacent sides in the same ratio as the remaining two.

314 EUCLID'S ELEMENTS.

Proposition A. Theorem.

If one side of a triangle he produced, and the exterior
angle so formed he hisected hi/ a straight line which cuts the
base produced, the segments between the bisector and the
extremities of the base shall have to one another the same
ratio as the remaining sides of the triangle Jiave:

Conversely, if the segments of the base produced liave to
one another the same ratio as the remaining sides of the tri-
angle have, the straight line drawn from the vertex to the
point of section shall bisect the exterior vertical angle.

In the A ABC let BA be produced to F, and let tlie
exterior z. CAF be bisected by AX which meets the base
produced at X :
then shall BX : XC : : BA : AC.

Through C draw CE par' to XA, i. 31.

and let CE meet BA at E.

Then because AX and CE are par',
.'. the ext. L FAX ='the int. opp. L AEC,

and the L XAC = the alt. i. ACE. i. 29.

But the L FAX = the l XAC; ffi/p.

.'. the z. AEC -the ^ACE;

.-. AC = AE. I. 6.

Again, because XA is par' to CE, a side of the A BCE,

Co7istr.
.". BX : XC :: BA : AE; VL 2.

that is, BX : XC : : BA : AC.

BOOK VI. PROP. A. 315

Conversely, let BX : XC :: BA : AC, and let AX be joined:

then shall the l FAX - the l XAC.
For, with the same construction as before,

because AX is par' to CE, a side of the A BCE,

.*. BX : XC :: BA : AE. vi. 2.

But by hypothesis BX : XC :: BA : AC;

.-. BA : AE :: BA : AC; v. 1.

.'. AE = AC,
.-. the :LACE = the ;L AEC. i. 5.

But because AX is par^ to CE,
.". the z. XAC = the alt. L ACE,
and the ext. l FAX = the int. opp. ^ AEC ; i. 29.
.'. the ^ FAX = the /.XAC. q.e.d.

Propositions 3 and A may be both included in one enunciation
as follows :

If the interior or exterior vertical angle of a triangle he bisected
htj a straight line which also cuts the base, the base shall be divided
internally or externally into segments which have the same ratio as
the sides of the triangle :

Conversely, if the base be divided internally or externally into seg-
ments which have the same ratio as the sides of the triangle, the straight
line drawn from the point of division to tJie vertex will bisect the
interior or exterior vertical angle.

EXERCISES.

1. In the circunaference of a circle of which AB is a diameter, n
point P is taken ; straight lines PC, PD are drawn equally inclined
to AP and on opposite sides of it, meeting AB in C and D ;

shew that AC : CB :: AD : DB.

2. From a point A straight lines are drawn making the angles
BAC, CAD, DAE, each equal to half a right angle, and they are cut
by a straight line BCDE, which makes BAE an isosceles triangle:
shew that BC or DE is a mean proportional between BE and CD.

3. By means of Propositions 3 and A, prove that the straight
lines bisecting one angle of a triangle internally, and the other two
externally, are concurrent.

316 Euclid's elements.

Proposition 4. Theorem.

If two triangles he equiangular to one anotlier, the sides
about the equal angles shall be proportionals, those sides
which are opposite to equal angles being homologous.

B C E

Let the A ABC be equiangular to the A DCE, having the
L ABC equal to the L DCE, the z_ BCA equal to the L CED,
and consequently the ^i. CAB equal to the z. EDC: i. 32.
then shall the sides about these equal angles be propor-
tionals, namely

AB

BC :

: DC

:CE,

BC

CA :

: CE

ED,

and AB

AC :

: DC

DE.

Let the A DCE be placed so that its side CE may be
contiguous to BC, and in the same straight line with it.

Then because the z. * ABC, ACB are together less than

two rt. angles, l 17.

and the z. ACB = the z. DEC; Ilyp-

.'. the z.* ABC, DEC are together less than two rt. angles;

.'. BA and ED will meet if produced. Ax. 12.

Let them be produced and meet at F.

Then because the z. ABC = the z. DCE, -ffy/>.

.*. BF is par' to CD; I. 28.

and because the z. ACB = the z. DEC, ffyp-

.'. AC is par" to FE, I. 28.

.". FACD is a par'";

.'. AF = CD, and AC ^ FD. i. 34.

BOOK VI, PROP. 4. 317

Again, because CD is par' to BF, a side of the A EBF,

.-. BC : CE :: FD : DE; VI. 2.

but FD== AC;
.-. BC : CE :: AC : DE;

and, alternately, BC : CA :: CE : ED. v. 11.

Again, because AC is par' to FE, a side of the A FBE,

.-. BA : AF :: BC : CE; y;, 2.
but AF = CD;

.-. BA

: CD :

: BC :

: CE;

and, alternately, AB :

: BC :

: DC :

: CE.

V. 11.

Also BC ;

: CA :

: CE ;

: ED;

Proved.

.'., ex cequali, AB ;

: AC :

: DC :

: DE.

V. 14.

Q. E. D.

[For Alternative Proof see Page 320.]

EXERCISES.

1. If one of the parallel sides of a trapezium is double the
other, shew that the diagonals intersect one another at a point of
trisection.

2. In the side AC of a triangle ABC any point D is taken : shew
that if AD, DC, AB, BC are bisected in E, F, G, H respectively,
then EG is equal to HF.

3. AB and CD are two parallel straight lines; E is the middle
point of CD ; AC and BE meet at F, and AE and BD meet at G :
shew that FG is parallel to AB.

4. ABCDE is a regular pentagon, and AD and BE intersect in F :
shew that AF : AE :: AE : AD.

5. In the figure of i. 43 shew that EH and GF are parallel, and
that FH and GE will meet on CA produced.

6. Chords AB and CD of a circle are produced towards B and
D respectively to meet in the point E, and through E, the line EF is
drawn parallel to AD to meet CB produced in F. Prove that EF is a
mean proportional between FB and FC.

318 KUCLID's ELEMENTb.

Proposition 5. Theorem.

//* the sides of two triangles^ taken in order about each of
their angles^ he proportionals, the triangles shall he equi-
angular to one another, having those angles equal which are
02)posite to the homologous sides.

Let the A^ ABC, DEF have their sides proportionals,
so that AB : BC :: DE : EF,

BC : CA :: EF : FD,
nnd consequently, ex cequali,

AB : CA :: DE : FD.

Then shall the triangles be equiangular.

At E in FE make the lFEG equal to the /.ABC;

and at F in EF make the l EFG equal to the l BCA; i. 23.

then the remaining ;L EG F = the remaining ;L BAG. 1.32.

.". the A GEF is equiangular to the A ABC;

.-. GE : EF :: AB : BC. VI. 4.

But AB : BC :: DE : EF; Hyp.

.*. GE : EF :: DE : EF; V. 1.

.-. GE= DE.
Similarly GF = DF.
Then in the triangles GEF, DEF
( GE=DE,

Because < GF=DF,

(and EF is common;
.-. the ^GEF = the ^ DEF, i. 8.

and the l GFE =the l DFE,
and the z.EGF=the ^ EDF.

But the ÂŁ. GEF = the ^ ABC; Constr.

.". the ^ DEF -the ^ ABC.
Similarly, the ^ EFD = the l BCA,

BOOK VI. PROP. 6. 319

.'. the remaining L FDE = the remaining l. CAB; i. 32.
that is, the A DEF is equiangular to the A ABC.

Q.E.D.

Propositiox 6. Theorem.

If two triangles have one angle of the one equal to one
angle of the other, and the sides about the equal angles pro-
portionals, the triangles shall he similar.

In the A^ ABC, DEF let the z. BAC = the z. EDF,
and let BA : AC :: ED : DF.

Then shall the A^ ABC, DEF be similar.
At D in FD make the l FDG equal to one of the ^ ^ EDF, BAC :
at F in DF make the L DFG equal to the L ACB; i. 23.
.'. the remaining z. FGD = the remaining l. ABC. i. 32.
Then the A ABC is equiangular to the ADGF;

.â€˘, BA : AC :: GD : DF. vi. 4.

But BA : AC :: ED : DF; Hyp.

.'. GD : DF :: ED : DF,
.-. GD-ED.
Then in the A^ GDF, EDF,
[ GD=.ED,

Because -| and DF is common;

[and the z. GDF =- the z. EDF; Constr.

.'. the A^ GDF, EDF are equal in all respects, i. 4.
so that the A EDF is equiangular to the A GDF;
but the A GDF is equiangular to the ABAC; Constr.
.', the A EDF is equiangular to the ABAC;
.'. their sides about the equal angles are proportionals, vi. 4.
that is, the A^ ABC, DEF are similar.

Q. E. D.

:i'2() kuclid's elements.

Note 1. From Definition 2 it is seen that two conditions are
necessary for similarity of rectilineal figures, namely (1) the figures
must be equiangular, and (2) the sides about the equal angles must
be proportionals. In the case of triangles we learn from Props. 4
and 5 that each of these conditions follows from the other : this how-
ever is not necessarily the case with rectilineal figures of more than
three sidles.

Note 2. We have given Euclid's demonstrations of Propositions
4, 5, 6 ; but these propositions also admit of easy proof by the method
of superposition.
, As an illustration, we will apply this method to Proposition 4.

Proposition 4. [Alternative Proof.]

If two triangles be equiangular to one another, the sides about the
equal angles shall be proportionals, those sides which are opposite to
equal angles being homologous.

Let the A ABC be equiangular to the a DEF, having the Z ABC
equal to the Z DEF, the Z BCA equal to the z EFD, and conse-
quently the z CAB equal to the z FDE : i. 32.

then shall the sides about these equal angles be proportionals.

Apply the a ABC to the a DEF, so that B falls on E and BA
along ED:

then BC will fall along EF, since the Z ABC = the z DEF. Hyp.
Let G and H be the points in ED and EF, on which A and C fall.
Join GH.
Then because the z EGH = the z EDF, Hyp.

:. GH is par' to DF:
.-. DG : GE:: FH : HE;
.â€˘. , componendo, DE : GE :: FE : HE, v. 13.

.'., alternately, DE : FE :: GE : HE, v. 11.

that is, DE : EF :: AB : BC.

Similarly by applying the a ABC to the a DEF, so that the point
C may fall on F, it may be proved that

EF : FD :: BC : CA.
.-. , ex cequali, DE : DF :: AB : AC.

Q. E. D.

BOOK VI. PROP. 7.

321

Proposition 7. Theorkm.

If two triangles have one angle of the one equal to one
angle of the other and the sides about one other angle in each
^proportional, so that the sides opposite to the equal angles are

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