Euclid.

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

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Online LibraryEuclidA text-book of Euclid's Elements : for the use of schools : Books I-VI and XI → online text (page 9 of 27)
Font size a number. Accordingly it will be found in the course of Book ll.
that several theorems relating to the areas of rectangles and
squares are analogous to W'ell-knowu algebraical formulae.

In view of these principles the rectangle contained by two
straight lines AB, BC is sometimes expressed in the form of a
product, as AB . BC, and the square described on AB as AB'^.
This notation, together with the signs + and â€” , w^ll be employed
in the additional matter appended to this book; but it is not
admitted into Euclid's text because it is desirable in the first
instance to emphasize the distinction between geometrical mag-
nitudes themselves and the numerical equivalents by which they
may be expressed arithmetically.

PiioPosiTioN 1. Theorem.

If there are two straight lines, one of which is divided
into any number of parts, the rectangle contained by the
two straight lines is equal to the sum of the rectangles con-
tained by the undivided straight line and the several parts
of the divided line.

Let P and AB be two straight lines, and let AB be
divided into any number of parts AC, CD, DB :

then shall the rectangle contained by P, AB be equal
to the sum of tlie rectangles contained by P, AC, by P, CD,
and by P, DB.

feOOK II. PROP. 1.

A C D B

K

- H

123

From A draw AF perp. to AB ; i. 11.

and make AG equal to P. i. 3.

Through G draw GH par^ to AB ; I. 31.
and through C, D, B draw CK, DL, BH par^ to AG.

Now the fig. AH is made up of the figs. AK, CL, DH :
and of these,

the fig. AH is the rectangle contained by P, AB;
for the fig. AH is contained by AG, AB ; and AG = P
and the fig. AK is the rectangle contained by P, AC
for the fig. AK is contained by AG, AC; and AG = P
also the fig. CL is the rectangle contained by P, CD

for the fig. CL is contained by CK, CD ;

and CK = the opp. side AG, and AG = P : i. 34.

similarly the fig. DH is the rectangle contained by P, DB.

.*. the rectangle contained by P, AB is equal to the
sum of the rectangles contained by P, AC, by P, CD, and
by P, DB. Q.E.D.

CORRESPONDING ALGEBRAICAL FORMULA.

In accordance with the principles explained on page 122, the result
of this proposition may be written thus :

P.AB-P. AC+P.CD + P.DB.

Now if the line P contains p units of length, and if AC, CD, DB
contain a, b, c units respectively,

and we have

then AB = a + 6 + c,
2>{a + b + c) =pa +])b -tiJc.

124

euclid s elements.
Proposition 2. Theorem.

If a straight line is divided into any two parts ^ the
square on the whole line is equal to the sum of the rectangles
contained by the lohole line and each of the parts.

Let the straight line AB be divided at C into the two
parts AC, CB :

then shall the sq. on AB be equal to the sum of tlic
rects. contained by AB, AC, and by AB, BC.

On AB describe the square ADEB. i. iC.

Through C draw CF par^ to AD. i. 31.

Now the fig. AE is made up of the figs. AF, CE :
and of these,

the fig. AE is the sq. on AB : Constr.

and the fig. AF is the rectangle contained by AB, AC ;
for the fig. AF is contained by AD, AC ; and AD = AB ;
also the fig. CE is the rectangle contained by AB, BC ;
for the fig. CE is contained by BE, BC; and BE = AB.
.-. the sq. on AB =^ the sum of the rects. contained by
AB, AC, and by AB, BC. q.e.d.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written
AB- = AB. AC + AB.BC.
Let AC contain a units of length, and let CB contain h units,
then AB = a + 6,
and we have {a + fc)2 = (a + 6) a + (a + h) h.

book ii. prop. 3.
Proposition 3. Theorem.

125

If a straight line is divided into any two parts, the
rectangle co7itained hy the whole and one of the: parts is
equal to the squai'e on that part together with the rectangle
contained hy the tivo parts.

Let the straight line AB be divided at C into the two
parts AC, CB:

then shall the rect. contained by AB, AC be equal to tlie
sq. on AC together with the rect. contained by AC, CB.

On AC describe the square AFDC ; i. 46.

and through B draw BE parHo AF, meeting FD produced in E.

I. 31.
Now the fig. AE is made up of the figs. AD, CE ;
and of these,
the fig. AE = the rect. contained by AB, AC ;
for AF = AC;
and the fig. AD is the sq. on AC ; Constr.

also the fig. CE is the rect. contained by AC, CB ;
for CD = AC.

.-. the rect. contained by AB, AC is equal to the sq. on
AC together with the rect. contained by AC, CB. q.e.d.

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written AB . AC = AC2+ AC . CB.
Let AC, CB contain a and h units of length respectively,
then AB = a + &,
and we have {a + l^a^a- + ah.

Note. It should be observed that Props. 2 and 3 are special cases
of Prop. 1.

126

kuclid's elements.
Proposition 4. Theorem.

If a straight line is divided into any two parts, the
square on the lohole line is equal to the sum of the squares
on the two parts together with twice the rectangle contained
hy the two parts.

7

/

G

Let the straight line AB be divided at C into tlio
two parts AC, CB :

then shall the sq, on AB be equal to the sum of th(^
sqq. on AC, CB, together with twice the rect. AC, CB.

On AB describe the square ADEB ; i. 40.

and join BD.
Through C draw CF par^ to BE, meeting BD in G. i. 31,
Through G draw HGK par^ to AB.

It is first required to shew that the fig. CK is the
sq. on BC.

Because the straight line BGD meets the par^^ CG, AD,
.-. theext. angle CGB = the int. opp. angle ADB. i. 29.
But AB = AD, being sides of a square ;
.-. the angle ADB = the angle A BD ;
.-. the angle CGB = the angle CBG.
.*. CB=-CG.
And the opp. sides of the par*" CK are equal
.'. the fig, CK is equilateral ;
and the angle CBK is a riglit angle ;
.'. CK is a square, and it is described on BC. :

Similarly the fig. HF is the sq. on HG, that is, the sq.
on AC,

for HG - the opp. side AC, I. 34.

I. T).

I. G.
I. 34.

Def 28.
40, Cor.

BOOK II. PROP. 4. ]27

Again, the complement AG -the complement GE. i. 43.

But the fig. AG = the rect. AC, CB ; for CG - CB.
.-. the two figs. AG, GE = twice the rect. AC, CB.

*Now the sq. on AB = the fig. AE

= the figs. HF, CK, AG, GE
= the sqq. on AC, CB together with
twice the rect. AC, CB.
.". tlie sq. on AB = the sum of the sqq. on AC, CB witli
twice the rect. AC, CB. q.e.d.

â– ^ For the purpose of oral work, this step of the proof
may conveniently be arranged as follows :

Kow the sq. on AB is equal to the fig. AE,

that is, to the figs. HF, CK, AG, GE;
that is, to the sqq. on AC, CB togetlier
with twice the rect. AC, CB.

Corollary. Parallelograms about the diagonals of a
square are themselves squares.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this important Proposition may be written tlius ;

AB-^ = AC-^ + CB2 + 2AC.CB.
Let AC = a, and CB = ?;;

then AB = Â« + ?>,
and \\Q liave (a + Vf = a- -\- h^ + lah.

128

euclid's elements.
Proposition 5. Theorem.

If a straight line is divided equally and also unequally,
the rectangle contained hy the unequal parts, and tlie square
on the line between tlie points of section, are together equal to
tlie square on lialf the line.

Q B

/

G H

/

F

Let the straight line AB be divided equally at P, and
unequally at Q :

then the rect. AQ, QB and tlie sq. on PQ shall be to-
gether equal to the sq. on PB.

On PB describe the square PCDB.

Join BC.

Through Q draw QE par^ to BD, cutting BC in F

Through F draw LFHG par^ to AB.

Through A draw AG par^ to BD,

Now the complement PF = the complement FD :

to each add the fig. QL;

I. 46.

31.

I. 43

then the fig. PL

the fig. QD.

But the fig. PL = the fig. AH, for they are par^'" on
equal bases and between the same par^^ i. 36.

the fig. AH

the fig. QD.

PF;

then the fig. AF = the gnomon PLE.

QF

II. 4.

Now the fig. AF = the rect. AQ, QB, for QB

.'. the rect. AQ, QB = the gnomon PLE.
To each add the sq. on PQ, that is, the fig, HE;
then the rect. AQ, QB with the sq. on PQ

= the gnomon PLE with the fig. HE
= the whole fig. PD,
which is the sq. on PB.

BOOK II. PROP. 5. 129

That is, the rect. AQ, QB and the sq. on PQ are together
equal to the sq. on PB. q.e.d.

Corollary. From this Proposition it follows that the
difference of the squares on two straight lines is equal to the
rectangle contained hy their sum and difference.

For let X and Y be the given a P Q g

st. lines, of which X is the greater. ' ^- ^

Draw AP equal to X, and pro- X
duce it to B, making PB equal to Y
AP, that is to X.

From PB cut off PQ equal to Y.

Then AQ is equal to the sum of X and Y,

and QB is equal to the difference of X and Y.

Now because AB is divided equally at P and unequally at Q,

.-.the rect. AQ, QB with sq. on PQ = the sq. on PB; ii. 5.
that is, the difference of the sqq. on PB, PQ^the rect. AQ, QB,
or, the difference of the sqq. on X and Y = the rect. contained by the
sum and the difference of X and Y.

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written

AQ.QB + PQ2=pB2.

Let AB = 2a; and let PQ = &;

then AP and PB each = a.

AlsoAQ = a + ?;; andQB = a-ft.

Hence we have

(a + 6) (a-fo) + ^2_^2^

or (a + &)(a-6) = a2-62.

EXERCISE.

In the above figure shew that AP is half the sitm of AQ and QB :
and that PQ is half their difference.

II. E.

130

Euclid's elements.
Proi'osition G. Theorem.

If a straight line is bisected and produced to any point,
the rectangle contained hy the wliole line thus produced, and
the part of it produced, together mlth the square on half
the line bisected, is equal to the square on the straight line
made up of the half and the part produced.

A P B Q

/

G H

/

F

Let the straight line AB be bisected at P, and pro-
duced to Q :

then the rect. AQ, QB and the sq. on PB shall be to-
gether equal to the sq. on PQ.

On PQ describe the square PCDQ. i. 4G.

Join QC.
Through B draw BE par^ to QD, meeting QC in F. i. 31.
Through F draw LFHG par^ to AQ.
Through A draw AG par^ to QD.

Now the complement PF = the complement FD. i. 43.
But the fig. PF = the fig. AH ; for they are par"" on
equal bases and between the same par^^ i. 30.

.-. the fig. AH = the fig. FD.
To each add the fig. PL;
then the fig. AL = the gnomon PLE.
Now the fig. AL = the rect. AQ, QB, for QB = QL ;

.'. the rect. AQ, QB = the gnomon PLE.
To each add the sq. on PB, that is, the fig. HE ;
then tlie rect. AQ, QB with the sq. on PB

= the gnomon PLE with the fig. HE
^ the whole fig. PD,
which is the square on PQ.
That is, the rect. AQ, QB and the sq. on PB are together
equal to the sq. on PQ, 9.E.D,

BOOK II. PROP. 6. 131

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written

AQ.QB+PB2r=PQ2.

Let AB = 2a ; and let PQ. = b;

then AP and PB each = a.
AlsoAQ = rt + 6; andQB = &-a.
Hence we have

{a + h){b-a)+a^=b'^,
or {b + a){b~a) = b^ - a^.

Definition. If a point X is taken in a straight line AB, or in A B
produced, the distances of the

point of section from the ex- A X b
tremities of AB are said to be
the segments into which AB is
divided at X. . r> y

In the former case AB is
divided internally, in the latter case externally.

Thus in the annexed figures the segments into which AB is
divided at X are the lines XA and XB.

This definition enables us to include Props. 5 and 6 in a single
Enunciation.

If a straight line is bisected, and also divided {Internally or ex-
ternally) into ttco unequal segments, the rectangle contained by the un-
equal segments is equal to the difference of the squares on half the line,
and on the line between the points of section.

EXERCISE.

Shew that the Enunciations of Props. 5 and 6 may take the
following form :

The rectangle contained by tioo straight lines is equal to the differ-
ence of the squares on half their sum and on half their difference.

[See Ex., p. 129.]

i:^2

EUCLID Â« ELEMENTS.

Proposition

Theorem.

If a straight line is divided into any trco parts, the sum
of tlie squares on the whole line and on one of the parts
is equal to tivice the rectangle contained by the whole and,
that part, together with the square on the other part.

/_

G

Let the straight line AB be divided at C into the two
parts AC, CB :

then shall the sum of the sqq. on AB, BC be equal to
twice the rect. AB, BC together with the sq. on AC.

On AB describe the square ADEB. l 4G.

Join BD.
Through C draw CF par^ to BE, meeting BD in G. i. 31.
Through G draw HGK par^ to AB.

Now the complement AG =the complement GE; i. 43.

to each add the fig. CK:

then the fig. AK = the fig. CE.

But the fig. AK = the rect. AB, BC ; for BK = BC.

.-. the two figs. AK, CE = twice the rect. AB, BC.

But the two figs. AK, CE make up the gnomon AKF and the

fig. CK :
.'. the gnomon AKF wit-h the fig. CK = twice the rect. AB, BC.

To each add the fig. HF, which is the sq. on AC :
then the gnomon AKF with the figs. CK, HF

= twice the rect. AB, BC with tlie sq. on AC.

Now the sqq. on AB, BC = the figs. AE, CK

= the gnomon AKF with tli-

figs. CK, HF
^ twice the rect. AB, BC with
the sq. on AC.

BOOK 11. PROP.

133

CORRESPONDING ALGEBRAICAL FORMULA. ^

The result of this proposition may be written
AB2 + BC2 = 2AB. BC + AC-.
Let AB = rt, and BC = &; then AC = a- 6.
Hence we have a^ + h^= 2a6 + (Â«-&)-,

or {a-bf = a"-2ah + h".

y Proposition 8. Theorem.
If a straight li7ie he divided into any two i)arts^ four
times the rectangle contained by the whole line and one of
the parts, together with the square on the other part, is equal
to the square on the straight line which is made up of the
whole and that part.

[As this proposition is of little importance we merely give the
figure, and the leading points in Euclid's proof.]

Let A B be divided at C. A C B D

Produce AB to D, making BD equal
to BC.

On AD describe the square AEFD;
and complete the construction as in-
dicated in the figure.

Euclid then proves (i) that the figs.
CK, BN, GR, KO are all equal.

(ii) that the figs. AG, MP, PL, RF are all equal.

Hence the eight figures named above are four times the
sum of the figs. AG, CK; that is, four times the fig. AK;
that is, four times the rect. AB, BC.

But the whole fig. AF is made up of these eight figures,
together with the fig. X H , which is the sq. on AC :

hence the sq. on AD = four times the rect. AB, BC,
together with the sq. on AC. q.e.d.

G

K

/

P

/

/

R

H L F

B

The accompanying figure will suggest a
less cumbrous proof, which we leave as an
Exercise to the student.

134

Euclid's klements.
Proposition 9. Theorem.

]f a straight line is divided equally and also uihequally,
the sum of the squares on the two unequal parts is twice
the sum of the squares on half the line and on tlie line
between the points of section.

Let the straight line AB be divided equally at P, and
unequally at Q :

then shall the sum of the sqq. on AQ, QB be twice the
sum of the sqq. on AP, PGl

At P draw PC at rt. angtes to AB ;
and make PC equal to AP or PB.

Join AC, BC.

Through Q draw QD par^ to PC;

and through D draw DE par^ to AB.

Then since PA = PC,

â€˘. the angle PAC = the angle PCA.

1. 11.
I. 3.

I. 31.

Constr.

I- 5-

And since, in the triangle A PC, the angle A PC is a rt.
angle, Constr.

.' . the sum of the angles PAC, PCA is a rt. angle: i. 32.
hence each of the angles PAC, PCA is half a rt. angle.
So also, each of the angles PBC, PCB is half a rt. angle.
.'. the whole angle ACB is a rt. angle.
Again, the ext. angle CED = the int. opp. angle CPB, i. 29.
.â€˘. the angle CED is a rt. angle :
and the angle ECD is half a rt. angle. Proved.
.'. also the angle EDC is half a rt. angle; i. 32.
.*. the angle ECD = the angle EDC;

.-. EC = ED. I. G.

BOOK II. PROP. 9. 135

Again, the ext. angle DQB = the int. opp. angle CPB. i. 29.
.â€˘, the angle DQB is a rt. angle.
And the angle QBD is half a rt. angle ; Proved.
.-. also the angle QDB is half a rt. angle : i. 32.
.'. the angle QBD = the angle QDB ;

.-. QD = QB. I. 6.

Now the sq. on AP = the sq. on PC ; for AP = PC. Gonstr.
But the sq. on AC = the sum of the sqq. on AP, PC,

for the angle A PC is a rt. angle. i. 47.

.â€˘. the sq. on AC is twice the sq. on A P.

So also, the sq. on CD is twice the sq. on ED, that is, twice
the sq. on the opp. side PQ. i. 34.

Now the sqq. on AQ, QB = the sqq. on AQ, QD

= the sq. on AD, for AQD is a rt.

angle; i. 47.

= the sum of the sqq. on AC, CD,

for AC D is a rt. angle; i. 47.

-â– = twice the sq. on AP with twice

the sq. on PQ. Proved.

That is,

the sum of the sqq. on AQ, QB = twice the sum of the sqq.

Q.E.D.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written
AQ2+QB2 = 2(AP2+PQ2).
LetAB = 2a; andPQ=Z*;

then AP and PB each = a.
AlsoAQ = a + &; andQB = a-Z>.
Hence we have

136

v^^:;

,v , EUCLID 8 ELEMENTS.

^^'' Proposition 10. Theorem.

If a straight line is bisected and produced to any jjoint^
the sum of the squares on the whole line thus produced, and
on tJte part produced, is ttoice the sum of the squares on ludj'
the line bisected and on the line made u^p of the half and the
part produced.

(X

2a +W

Let the St. line AB be bisected at P, and produced to Q :
then shall the sum of the sqq. on AQ, QB be twice tlu^
sum of the sqq. on AP, PQ.

At P draw PC at right angles to AB ; i. 11.

and make PC equal to PA or PB. i. 3.

Join AC, BC.

Through Q draw QD par^ to PC, to

in D;
and through

in E.

meet CB produced

I. 31.

draw DE par^ to AB, to meet CP produced

Then since PA ~ PC, Consfr.

. â€˘. the angle PAC = the angle PCA. i. 5.

And since in the triangle A PC, the angle A PC is a rt. angle,
.-. the sum of the angles PAC, PCA is a rt. angle. i. 32.
Hence eacli of the angles PAC, PCA is half a rt. angle.
So also, each of the angles PBC, PCB is half a rt. angle.
.*. the whole angle AC B is a rt. angle.

Again, tJie ext. angle CPB = the int. opp. angle CED : i. 29.

.'. the angle CED is a rt. angle :

md the angle ECD is half a rt. angle.

Proved.

.-. the angle EDC is half a rt. angle.

I. 32.

. â€˘. the angle ECD = the angle EDC ;

.-. EC=ED.

1.6.

BOOK II. PROP. 10. 137

Again, tlie angle DQB ^ the alt. angle CPB. i. 29.

.-. the angle DGIB is a rt. angle.
Also the angle QBD = the vert. opp. angle CBP ; i. 15.
that is, the angle QBD is half a rt. angle.

.â€˘. the angle QDB is half a rt. angle : i. 32.

.-. the angle QBD = the angle QDB ;

.-. QB = QD. I. G.

Now the sq. on AP == the sq. on PC j for AP = PC. Constr.
But the sq. on AC = the sum of the sqq. on AP, PC,

for the angle APC is a rt. angle. i. 47.

.*. the sq. on AC is twice the sq. on AP.
So also, the sq. on CD is twice the sq. on ED, that is,
twice the sq. on the opp. side PQ. i. 34.

Now the sqq. on AQ, QB = the sqq. on AQ, QD

= the sq. on AD, for AQD is a rt.

angle; i. 47.

= the sum of the sqq. on AC, CD,

for AC D is a rt. angle ; I. 47.

= twice tlie sq. on AP with twice

the sq. on PQ. Proved.

That is,

the sum of the sqq. on AQ, QB is twice the sum of the sqq.

on AP, PQ. Q.E.D.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written
AQ2+ BQ2 = 2 (AP-+ PQ2).
Let AB = 2rt; and PQ = 6 ;

then AP and PB each = a.
Also AQ=a + 6; and BQ=6 - a.
Hence we have

EXERCISE.

Shew that the enunciations of Props. 9 and 10 may take the
following form :

The sum of the squares on two straigJtt lines is equal to tioice the
sum of the squares on half their sum and on half their difference.

I :'>f

JiUCLlD'S ELEMENTS.

f. * Proposition 11. Problem.

' ^-'^^^0 divide a given straight line into two 2yctrt8, so that
the rectangle contained hy the whole and one "part may he
equal to the square on th^ otlier part.

Let AB be the given straight line.
It is required to divide it into two parts, so tliat the
rectangle contained by the whole and one part may be
equal to the square on the other part.

46.
10.

On AB describe the square ACDB.
Bisect AC at E.
Join EB.
Produce CA to F, making EF equal to EB. I. 3.

On AF describe the square AFGH. I. 46.

Then shall AB be divided at H, so that the rect. AB, BH is
equal to the sq. on AH.

Produce GH to meet CD in K.

Then because CA is bisected at E, and produced to F,
.-. tlie rect. CF, FA with the sq. on AE = the sq. on FE ii. 6.

= the sq. on EB. Constr.
But tlie sq. on EB = the sum of the sqq. on AB, AE,

for the angle EAB is a rt. angle. i. 47.

.*. the rect. CF, FA with the sq. on AE = the sum of the
sqq. on AB, AE.

From these take the sq. on AE :
then the rect. CF, FA = the sq. on AB.

Book il prop. 11. 139

But the rect. CF, FA = the fig. FK ; for FA = FG ;
and the sq. on AB = the fig. AD. Constr.

.-. the fig. FK = the fig. AD.
From these take the common fig. AK,
then the remaining fig. FH = tkfe remaining fig. HD.
But the fig. HD - the rect. AB, BH ; for BD - AB;
and the fig. FH is the sq. on AH.
.-. the rect. AB, BH = the sq. on AH. q.e.f.

Definition. A straight line is said to be divided in Medial Section
when the rectangle contained by the given line and one of its segments
is equal to the square on the other segment.

The student should observe that this division may be internal or
external.

Thus if the straight line AB is divided internally at H, and ex-
ternally at H', so that

(i) AB.BH=AH2, , A H B
(ii) AB.BH' = AH'2, U ^ L] Â°

we shall in either case consider that AB is divided in medial section.

The case of internal section is alone given in Euchd ii. 11 ; but a
straight line may be divided externally in medial section by a similar
process. See Ex. 21, p. 146.

ALGEBRAICAL ILLUSTRATION.

It is required to find a point H in AB, or AB produced, such that

AB.BH = AH2.
Let AB contain a units of length, and let AH contain x units;
then H B = a - .r :
ivud X must be such that a [a- x)=x",
or x'^ + ax-a'^ = 0.

Thus the construction for dividing a straight line in medial section
corresi^onds to the algebraical solution of this quadratic equation.

EXERCISES.

In the figure of ii. 11, shew that
(i) if CH is produced to meet BF at L, CL is at right angles

to BF:
(ii) if BE and CH meet at O, AO is at right angles to CH ;
(iii) the lines BG, DF, AK are parallel :
(iv)' CF is divided in medial section at A.

i

110

EUCLID S KLEMEXTS.

â–  Proposition 12. Theorem,

In an obtuse- angled tt-iangle, if a perpendicular is
drawn from either of the acute angles to the opposite side
lyroduced^ tlie squai'e on the S'lde subtending the obticse angle is
greater than the squares on the sides containing the obtuac
angle, bg twice the rectangle contained by the side on which,
wlien. produced, tJie perpendicular falls, a7id the line inter-
cepted witJwut the triangle, between the perpendicular and
the obtuse angle.

A

Let ABC be an obtuse-angled triangle, having the obtuse
angle at C ; and let AD be drawn from A perp. to BC
produced :

then shall the sq. on AB be greater than the sqq. on
BC, CA, by twice the rect. BC, CD.

Because BD is divided into two parts at C,
.'. the sq. on BD =the sum of the sqq. on BC, CD, with twice
the rect. BC, CD. II. 4.

To each add the sq. on DA.
Then the sqq. on BD, DA = the sum of the sqq. on BC, CD,
DA, with twice the rect. BC, CD.
But the sum of the sqq. on BD, DA = the sq. on AB,

for the angle at D is a rt. angle.

I. 4:]

Similarly the sum of the sqq, on CD, DA = the sq. on CA.

.-, the sq. on AB = the sum of the sqq. on BC, CA, with
twice the rect. BC, CD.
That is, the sq. on AB is greater than the sum of tlie
sqq, on BC, CA by twice the rect. BC, CD. q.e.d.

[For alternative Enunciations to Props. 12 and 1,3 and Exercises,
see p. 142.]

BOOK II. PROP. 13.

141

PiiOPOsiTioN 13. Theorem.

In every triangle the square on the side subtending an
acute angle, is less than the squares on the sides containing
that angle, hy twice the rectangle contained by either of these
sides, and the straight line intercepted between the 'perpen-
dicular let fall on it from the opposite angle, and the a,cute
angle.

Let ABC be any triangle having the angle at B an
acute angle ; and let AD be the perp. drawn from A to the
opp. side BC :

then shall the sq. on AC be less than the sum of tlie
sqq. on AB, BC, by twice the rect. CB, BD.
Now AD may fall within the triangle ABC, as in Fig. 1, or

without it, as in Fi

in Fig.

BC is divided into two parts at D,

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