BOOK II.
DEFINITIONS.
1
Any rectangular parallelogram is said to be
contained by the two straight lines containing the right angle.
2
And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a
gnomon.
BOOK II. PROPOSITIONS.
Proposition 1.
If there be two straight lines,
and one of them be cut into any number of segments whatever,
the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.
Let
A,
BC be two straight lines, and let
BC be cut at random at the points
D,
E; I say that the rectangle contained by
A,
BC is equal to the rectangle contained by
A,
BD, that contained by
A,
DE and
that contained by
A,
EC.
For let
BF be drawn from
B at right angles to
BC; [
I. 11] let
BG be made equal to
A, [
I. 3] through
G let
GH be drawn
parallel to
BC, [
I. 31] and through
D,
E,
C let
DK,
EL,
CH be drawn parallel to
BG.
Then
BH is equal to
BK,
DL,
EH.
Now
BH is the rectangle
A,
BC, for it is contained by
GB,
BC, and
BG is equal to
A;
BK is the rectangle
A,
BD, for it is contained by
GB,
BD, and
BG is equal to
A;
and DL is the rectangle A, DE, for DK, that is BG [I. 34], is equal to A.
Similarly also
EH is the rectangle
A,
EC.
Therefore the rectangle
A,
BC is equal to the rectangle
A,
BD, the rectangle
A,
DE and the rectangle
A,
EC.
Therefore etc. Q. E. D.
1
Proposition 2.
If a straight line be cut at random,
the rectangle contained by the whole and both of the segments is equal to the square on the whole.
For let the straight line
AB be cut at random at the point
C; I say that the rectangle contained by
AB,
BC together with the rectangle contained by
BA,
AC is equal to the square on
AB.
For let the square
ADEB be described on
AB [
I. 46], and let
CF be drawn through
C parallel to either
AD or
BE. [
I. 31]
Then
AE is equal to
AF,
CE.
Now
AE is the square on
AB;
AF is the rectangle contained by
BA,
AC, for it is contained by
DA,
AC, and
AD is equal to
AB;
and CE is the rectangle AB, BC, for BE is equal to AB.
Therefore the rectangle
BA,
AC together with the rectangle
AB,
BC is equal to the square on
AB.
Therefore etc. Q. E. D.
Proposition 3.
If a straight line be cut at random,
the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment.
For let the straight line
AB be cut at random at
C; I say that the rectangle contained by
AB,
BC is equal to the rectangle contained by
AC,
CB together with the square on
BC.
For let the square
CDEB be described on
CB; [
I. 46] let
ED be drawn through to
F, and through
A let
AF be drawn parallel to either
CD or
BE. [
I. 31]
Then
AE is equal to
AD,
CE.
Now
AE is the rectangle contained by
AB,
BC, for it is contained by
AB,
BE, and
BE is equal to
BC;
AD is the rectangle
AC,
CB, for
DC is equal to
CB;
and DB is the square on CB. Therefore the rectangle contained by
AB,
BC is equal to the rectangle contained by
AC,
CB together with the square on
BC.
Therefore etc. Q. E. D.
Proposition 4.
If a straight line be cut at random,
the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.
For let the straight line
AB be cut at random at
C;
I say that the square on
AB is equal to the squares on
AC,
CB and twice the rectangle contained by
AC,
CB.
For let the square
ADEB be described on
AB, [
I. 46]
let
BD be joined; through
C let
CF be drawn parallel to either
AD or
EB, and through
G let
HK be drawn parallel to either
AB or
DE. [
I. 31]
Then, since
CF is parallel to
AD, and
BD has fallen on them, the exterior angle
CGB is equal to the interior and opposite angle
ADB. [
I. 29]
But the angle
ADB is equal to the angle
ABD,
since the side BA is also equal to AD; [I. 5] therefore the angle
CGB is also equal to the angle
GBC, so that the side
BC is also equal to the side
CG. [
I. 6]
But
CB is equal to
GK, and
CG to
KB; [
I. 34]
therefore GK is also equal to KB; therefore CGKB is equilateral.
I say next that it is also right-angled.
For, since
CG is parallel to
BK,
the angles KBC, GCB are equal to two right angles. [I. 29]
But the angle
KBC is right;
therefore the angle BCG is also right, so that the opposite angles CGK, GKB are also right. [I. 34]
Therefore
CGKB is right-angled; and it was also proved equilateral;
therefore it is a square; and it is described on CB.
For the same reason
HF is also a square; and it is described on HG, that is AC. [I. 34]
Therefore the squares
HF,
KC are the squares on
AC,
CB.
Now, since
AG is equal to
GE, and
AG is the rectangle
AC,
CB, for
GC is equal to
CB,
therefore GE is also equal to the rectangle AC, CB.
Therefore
AG,
GE are equal to twice the rectangle
AC,
CB.
But the squares
HF,
CK are also the squares on
AC,
CB; therefore the four areas
HF,
CK,
AG,
GE are equal to the squares on
AC,
CB and twice the rectangle contained by
AC,
CB.
But
HF,
CK,
AG,
GE are the whole
ADEB,
which is the square on
AB.
Therefore the square on
AB is equal to the squares on
AC,
CB and twice the rectangle contained by
AC,
CB.
Therefore etc. Q. E. D.
2
3
4
Proposition 5.
If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.
For let a straight line
AB be cut into equal segments at
C and into unequal segments at
D; I say that the rectangle contained by
AD,
DB together with the square on
CD is equal to the square on
CB.
For let the square
CEFB be described on
CB, [
I. 46] and let
BE be joined; through
D let
DG be drawn parallel to either
CE or
BF, through
H again let
KM be drawn parallel to either
AB or
EF, and again through
A let
AK be drawn parallel to either
CL or
BM. [
I. 31]
Then, since the complement
CH is equal to the complement
HF, [
I. 43] let
DM be added to each;
therefore the whole CM is equal to the whole DF.
But
CM is equal to
AL,
since AC is also equal to CB; [I. 36] therefore AL is also equal to DF. Let
CH be added to each;
therefore the whole AH is equal to the gnomon NOP.
But
AH is the rectangle
AD,
DB, for
DH is equal to
DB,
therefore the gnomon NOP is also equal to the rectangle AD, DB.
Let
LG, which is equal to the square on
CD, be added to each;
therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD.
But the gnomon
NOP and
LG are the whole square
CEFB, which is described on
CB;
therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.
Therefore etc. Q. E. D.
5
Proposition 6.
If a straight line be bisected and a straight line be added to it in a straight line,
the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.
For let a straight line
AB be bisected at the point
C, and let a straight line
BD be added to it in a straight line;
I say that the rectangle contained by
AD,
DB together with the square on
CB is equal to the square on
CD.
For let the square
CEFD be described on
CD, [
I. 46] and let
DE be joined; through the point
B let
BG be drawn parallel to either
EC or
DF, through the point
H let
KM be drawn parallel to either
AB or
EF, and further through
A let
AK be drawn parallel to either
CL or
DM. [
I. 31]
Then, since AC is equal to CB, AL is also equal to CH. [I. 36] But
CH is equal to
HF. [
I. 43]
Therefore AL is also equal to HF.
Let
CM be added to each;
therefore the whole AM is equal to the gnomon NOP.
But
AM is the rectangle
AD,
DB,
for DM is equal to DB; therefore the gnomon NOP is also equal to the rectangle AD, DB.
Let
LG, which is equal to the square on
BC, be added to each;
therefore the rectangle contained by AD, DB together with the square on CB is equal to the gnomon NOP and LG.
But the gnomon
NOP and
LG are the whole square
CEFD, which is described on
CD;
therefore the rectangle contained by AD, DB together with the square on CB is equal to the square on CD.
Therefore etc. Q. E. D.
Proposition 7.
If a straight line be cut at random,
the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment.
For let a straight line
AB be cut at random at the point
C;
I say that the squares on
AB,
BC are equal to twice the rectangle contained by
AB,
BC and the square on
CA.
For let the square
ADEB be described on
AB, [
I. 46] and let the figure be drawn.
Then, since
AG is equal to
GE, [
I. 43] let
CF be added to each;
therefore the whole AF is equal to the whole CE.
Therefore
AF,
CE are double of
AF.
But
AF,
CE are the gnomon
KLM and the square
CF; therefore the gnomon
KLM and the square
CF are double of
AF.
But twice the rectangle
AB,
BC is also double of
AF; for
BF is equal to
BC; therefore the gnomon
KLM and the square
CF are equal to twice the rectangle
AB,
BC.
Let
DG, which is the square on
AC, be added to each; therefore the gnomon
KLM and the squares
BG,
GD are equal to twice the rectangle contained by
AB,
BC and the square on
AC.
But the gnomon
KLM and the squares
BG,
GD are the whole
ADEB and
CF,
which are squares described on AB, BC; therefore the squares on
AB,
BC are equal to twice the rectangle contained by
AB,
BC together with the square on
AC.
Therefore etc. Q. E. D.
Proposition 8.
If a straight line be cut at random,
four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line.
For let a straight line
AB be cut at random at the point
C;
I say that four times the rectangle contained by
AB,
BC together with the square on
AC is equal to the square described on
AB,
BC as on one straight line.
For let [the straight line]
BD be produced in a straight line [with
AB], and let
BD be made equal to
CB; let the square
AEFD be described on
AD, and let the figure be drawn double.
Then, since
CB is equal to
BD, while
CB is equal to
GK, and
BD to
KN, therefore
GK is also equal to
KN.
For the same reason
QR is also equal to RP.
And, since
BC is equal to
BD, and
GK to
KN, therefore
CK is also equal to
KD, and
GR to
RN. [
I. 36]
But
CK is equal to
RN, for they are complements of the parallelogram
CP; [
I. 43] therefore
KD is also equal to
GR; therefore the four areas
DK,
CK,
GR,
RN are equal to one another.
Therefore the four are quadruple of CK.
Again, since
CB is equal to
BD, while
BD is equal to
BK, that is
CG, and
CB is equal to
GK, that is
GQ,
therefore CG is also equal to GQ.
And, since
CG is equal to
GQ, and
QR to
RP,
AG is also equal to MQ, and QL to RF. [I. 36]
But
MQ is equal to
QL, for they are complements of the parallelogram
ML; [
I. 43]
therefore AG is also equal to RF; therefore the four areas
AG,
MQ,
QL,
RF are equal to one another.
Therefore the four are quadruple of AG. But the four areas
CK,
KD,
GR,
RN were proved to be quadruple of
CK;
therefore the eight areas, which contain the gnomon STU, are quadruple of AK.
Now, since
AK is the rectangle
AB,
BD, for
BK is equal to
BD, therefore four times the rectangle
AB,
BD is quadruple of
AK.
But the gnomon
STU was also proved to be quadruple of
AK;
therefore four times the rectangle AB, BD is equal to the gnomon STU.
Let
OH, which is equal to the square on
AC, be added to each;
therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH.
But the gnomon
STU and
OH are the whole square
AEFD,
which is described on AD. therefore four times the rectangle
AB,
BD together with the square on
AC is equal to the square on
AD
But
BD is equal to
BC; therefore four times the rectangle contained by
AB,
BC together with the square on
AC is equal to the square on
AD, that is to the square described on
AB and
BC as on one straight line.
Therefore etc. Q. E. D.
Proposition 9.
If a straight line be cut into equal and unequal segments,
the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.
For let a straight line
AB be cut into equal segments at
C, and into unequal segments at
D;
I say that the squares on
AD,
DB are double of the squares on
AC,
CD.
For let
CE be drawn from
C at right angles to
AB, and let it be made equal to either
AC or
CB; let
EA,
EB be joined, let
DF be drawn through
D parallel to
EC, and
FG through
F parallel to
AB, and let
AF be joined.
Then, since
AC is equal to
CE,
the angle EAC is also equal to the angle AEC.
And, since the angle at
C is right,
the remaining angles EAC, AEC are equal to one right angle. [I. 32]
And they are equal;
therefore each of the angles CEA, CAE is half a right angle.
For the same reason
each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right.
And, since the angle
GEF is half a right angle, and the angle
EGF is right, for it is equal to the interior and opposite angle
ECB, [
I. 29]
the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6]
Again, since the angle at
B is half a right angle, and the angle
FDB is right, for it is again equal to the interior and opposite angle
ECB, [
I. 29]
the remaining angle BFD is half a right angle; [I. 32] therefore the angle at B is equal to the angle DFB, so that the side FD is also equal to the side DB. [I. 6]
Now, since
AC is equal to
CE,
the square on AC is also equal to the square on CE; therefore the squares on AC, CE are double of the square on AC.
But the square on
EA is equal to the squares on
AC,
CE, for the angle
ACE is right; [
I. 47]
therefore the square on EA is double of the square on AC.
Again, since
EG is equal to
GF,
the square on EG is also equal to the square on GF; therefore the squares on EG, GF are double of the square on GF.
But the square on
EF is equal to the squares on
EG,
GF;
therefore the square on EF is double of the square on GF.
But
GF is equal to
CD; [
I. 34] therefore the square on
EF is double of the square on
CD.
But the square on
EA is also double of the square on
AC;
therefore the squares on AE, EF are double of the squares on AC, CD.
And the square on
AF is equal to the squares on
AE,
EF, for the angle
AEF is right; [
I. 47] therefore the square on
AF is double of the squares on
AC,
CD.
But the squares on
AD,
DF are equal to the square on
AF, for the angle at
D is right; [
I. 47] therefore the squares on
AD,
DF are double of the squares on
AC,
CD.
And
DF is equal to
DB; therefore the squares on
AD,
DB are double of the squares on
AC,
CD.
Therefore etc. Q. E. D.
Proposition 10.
If a straight line be bisected,
and a straight line be added to it in a straight line,
the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line.
For let a straight line
AB be bisected at
C, and let a straight line
BD be added to it in a straight line;
I say that the squares on
AD,
DB are double of the squares on
AC,
CD.
For let
CE be drawn from the point
C at right angles to
AB [
I. 11], and let it be made equal to either
AC or
CB [
I. 3]; let
EA,
EB be joined; through
E let
EF be drawn parallel to
AD, and through
D let
FD be drawn parallel to
CE. [
I. 31]
Then, since a straight line
EF falls on the parallel straight lines
EC,
FD,
the angles CEF, EFD are equal to two right angles; [I. 29] therefore the angles FEB, EFD are less than two right angles.
But straight lines produced from angles less than two right angles meet; [
I. Post. 5]
therefore EB, FD, if produced in the direction B, D, will meet.
Let them be produced and meet at
G, and let
AG be joined.
Then, since
AC is equal to
CE, the angle
EAC is also equal to the angle
AEC; [
I. 5] and the angle at
C is right;
therefore each of the angles EAC, AEC is half a right angle. [I. 32]
For the same reason
each of the angles CEB, EBC is also half a right angle; therefore the angle AEB is right.
And, since the angle
EBC is half a right angle, the angle
DBG is also half a right angle. [
I. 15]
But the angle BDG is also right, for it is equal to the angle
DCE, they being alternate; [
I. 29]
therefore the remaining angle DGB is half a right angle; [I. 32] therefore the angle
DGB is equal to the angle
DBG,
so that the side BD is also equal to the side GD. [I. 6]
Again, since the angle
EGF is half a right angle, and the angle at
F is right, for it is equal to the opposite angle, the angle at
C, [
I. 34]
the remaining angle FEG is half a right angle; [I. 32] therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6]
Now, since the square on
EC is equal to the square on
CA, the squares on
EC,
CA are double of the square on
CA.
But the square on
EA is equal to the squares on
EC,
CA; [
I. 47] therefore the square on
EA is double of the square on
AC. [
C. N. 1]
Again, since
FG is equal to
EF, the square on
FG is also equal to the square on
FE; therefore the squares on
GF,
FE are double of the square on
EF.
But the square on
EG is equal to the squares on
GF,
FE; [
I. 47] therefore the square on
EG is double of the square on
EF.
And
EF is equal to
CD; [
I. 34]
therefore the square on EG is double of the square on CD. But the square on
EA was also proved double of the square on
AC; therefore the squares on
AE,
EG are double of the squares on
AC,
CD.
And the square on
AG is equal to the squares on
AE,
EG; [
I. 47] therefore the square on
AG is double of the squares on
AC,
CD. But the squares on
AD,
DG are equal to the square on
AG; [
I. 47] therefore the squares on
AD,
DG are double of the squares on
AC,
CD.
And
DG is equal to
DB; therefore the squares on
AD,
DB are double of the squares on
AC,
CD.
Therefore etc. Q. E. D.
Proposition 11.
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
Let
AB be the given straight line; thus it is required to cut
AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
For let the square
ABDC be described on
AB; [
I. 46] let
AC be bisected at the point
E, and let
BE be joined; let
CA be drawn through to
F, and let
EF be made equal to
BE; let the square
FH be described on
AF, and let
GH be drawn through to
K.
I say that
AB has been cut at
H so as to make the rectangle contained by
AB,
BH equal to the square on
AH.
For, since the straight line
AC has been bisected at
E, and
FA is added to it,
the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [II. 6]
But
EF is equal to
EB;
therefore the rectangle CF, FA together with the square on AE is equal to the square on EB.
But the squares on
BA,
AE are equal to the square on
EB, for the angle at
A is right; [
I. 47]
therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE.
Let the square on
AE be subtracted from each;
therefore the rectangle CF, FA which remains is equal to the square on AB.
Now the rectangle
CF,
FA is
FK, for
AF is equal to
FG; and the square on
AB is
AD;
therefore FK is equal to AD.
Let
AK be subtracted from each;
therefore FH which remains is equal to HD.
And
HD is the rectangle
AB,
BH, for
AB is equal to
BD; and
FH is the square on
AH;
therefore the rectangle contained by AB, BH is equal to the square on HA. therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA. Q. E. F.
Proposition 12.
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle,
namely that on which the
perpendicular falls,
and the straight line cut off outside by the perpendicular towards the obtuse angle.
Let
ABC be an obtuse-angled triangle having the angle
BAC obtuse, and let
BD be drawn from the point
B perpendicular to
CA produced;
I say that the square on
BC is greater than the squares on
BA,
AC by twice the rectangle contained by
CA,
AD.
For, since the straight line
CD has been cut at random at the point
A, the square on
DC is equal to the squares on
CA,
AD and twice the rectangle contained by
CA,
AD. [
II. 4]
Let the square on
DB be added to each; therefore the squares on
CD,
DB are equal to the squares on
CA,
AD,
DB and twice the rectangle
CA,
AD.
But the square on
CB is equal to the squares on
CD,
DB, for the angle at
D is right; [
I. 47]
and the square on AB is equal to the squares on AD, DB; [I. 47] therefore the square on
CB is equal to the squares on
CA,
AB and twice the rectangle contained by
CA,
AD;
so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.
Therefore etc. Q. E. D.
Proposition 13.
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle,
namely that on which the perpendicular falls,
and the straight line cut off within by the perpendicular towards the acutc angle.
Let
ABC be an acute-angled triangle having the angle at
B acute, and let
AD be drawn from the point
A perpendicular to
BC;
I say that the square on
AC is less than the squares on
CB,
BA by twice the rectangle contained by
CB,
BD.
For, since the straight line
CB has been cut at random at
D,
the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7]
Let the square on
DA be added to each; therefore the squares on
CB,
BD,
DA are equal to twice the rectangle contained by
CB,
BD and the squares on
AD,
DC.
But the square on
AB is equal to the squares on
BD,
DA, for the angle at
D is right; [
I. 47] and the square on
AC is equal to the squares on
AD,
DC; therefore the squares on
CB,
BA are equal to the square on
AC and twice the rectangle
CB,
BD,
so that the square on
AC alone is less than the squares on
CB,
BA by twice the rectangle contained by
CB,
BD.
Therefore etc. Q. E. D.
Proposition 14.
To construct a square equal to a given rectilineal figure.
Let
A be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure
A.
For let there be constructed the rectangular parallelogram
BD equal to the rectilineal figure
A. [
I. 45]
Then, if
BE is equal to
ED, that which was enjoined will have been done; for a square
BD has been constructed equal to the rectilineal figure
A.
But, if not, one of the straight lines
BE,
ED is greater.
Let
BE be greater, and let it be produced to
F; let
EF be made equal to
ED, and let
BF be bisected at
G.
With centre
G and distance one of the straight lines
GB,
GF let the semicircle
BHF be described; let
DE be produced
to
H, and let
GH be joined.
Then, since the straight line
BF has been cut into equal segments at
G, and into unequal segments at
E,
the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5]
But
GF is equal to
GH; therefore the rectangle
BE,
EF together with the square on
GE is equal to the square on
GH.
But the squares on
HE,
EG are equal to the square on
GH; [
I. 47]
therefore the rectangle
BE,
EF together with the square on
GE is equal to the squares on
HE,
EG.
Let the square on
GE be subtracted from each;
therefore the rectangle contained by BE, EF which remains is equal to the square on EH.
But the rectangle
BE,
EF is
BD, for
EF is equal to
ED;
therefore the parallelogram BD is equal to the square on HE.
And
BD is equal to the rectilineal figure
A.
Therefore the rectilineal figure
A is also equal to the square
which can be described on
EH.
Therefore a square, namely that which can be described on
EH, has been constructed equal to the given rectilineal figure
A. Q. E. F.
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