BOOK VI.
DEFINITIONS.
1
Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.
2
[
Reciprocally related figures.
See note.]
3
A straight line is said to have been
cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.
4
The
height of any figure is the perpendicular drawn from the vertex to the base.
5
BOOK VI. PROPOSITIONS.
PROPOSITION 1.
Triangles and parallelograms which are under the same height are to one another as their bases.
Let
ABC,
ACD be triangles and
EC,
CF parallelograms under the same height;
I say that, as the base
BC is to the base
CD, so is the triangle
ABC to the triangle
ACD, and the parallelogram
EC to the parallelogram
CF.
For let
BD be produced in both directions to the points
H,
L and let [any number of straight lines]
BG,
GH be
made equal to the base
BC, and any number of straight lines
DK,
KL equal to the base
CD; let
AG,
AH,
AK,
AL be joined.
Then, since
CB,
BG,
GH are equal to one another,
the triangles ABC, AGB, AHG are also equal to one another. [I. 38]
Therefore, whatever multiple the base
HC is of the base
BC, that multiple also is the triangle
AHC of the triangle
ABC.
For the same reason,
whatever multiple the base
LC is of the base
CD, that multiple also is the triangle
ALC of the triangle
ACD; and, if the base
HC is equal to the base
CL, the triangle
AHC is also equal to the triangle
ACL, [
I. 38] if the base
HC is in excess of the base
CL, the triangle
AHC
is also in excess of the triangle
ACL, and, if less, less.
Thus, there being four magnitudes, two bases
BC,
CD and two triangles
ABC,
ACD, equimultiples have been taken of the base
BC and the
triangle
ABC, namely the base
HC and the triangle
AHC, and of the base
CD and the triangle
ADC other, chance, equimultiples, namely the base
LC and the triangle
ALC;
and it has been proved that, if the base
HC is in excess of the base
CL, the triangle
AHC
is also in excess of the triangle
ALC; if equal, equal; and, if less, less.
Therefore, as the base
BC is to the base
CD, so is the triangle
ABC to the triangle
ACD. [
V. Def. 5]
Next, since the parallelogram
EC is double of the triangle
ABC, [
I. 41] and the parallelogram
FC is double of the triangle
ACD, while parts have the same ratio as the same multiples of them, [
V. 15] therefore, as the triangle
ABC is to the triangle
ACD, so is
the parallelogram
EC to the parallelogram
FC.
Since, then, it was proved that, as the base
BC is to
CD, so is the triangle
ABC to the triangle
ACD, and, as the triangle
ABC is to the triangle
ACD, so is the parallelogram
EC to the parallelogram
CF,
therefore also, as the base
BC is to the base
CD, so is the parallelogram
EC to the parallelogram
FC. [
V. 11]
Therefore etc. Q. E. D.
1
2
3
PROPOSITION 2.
If a straight line be drawn parallel to one of the sides of a triangle,
it will cut the sides of the triangle proportionally; and,
if the sides of the triangle be cut proportionally,
the line joining the points of section will be parallel to the remaining side of the triangle.
For let
DE be drawn parallel to
BC, one of the sides of the triangle
ABC; I say that, as
BD is to
DA, so is
CE to
EA.
For let
BE,
CD be joined.
Therefore the triangle
BDE is equal to the triangle
CDE; for they are on the same base
DE and in the same parallels
DE,
BC. [
I. 38]
And the triangle
ADE is another area.
But equals have the same ratio to the same; [
V. 7] therefore, as the triangle
BDE is to the triangle
ADE, so is the triangle
CDE to the triangle
ADE.
But, as the triangle
BDE is to
ADE, so is
BD to
DA; for, being under the same height, the perpendicular drawn from
E to
AB, they are to one another as their bases. [
VI. 1]
For the same reason also, as the triangle
CDE is to
ADE, so is
CE to
EA.
Therefore also, as
BD is to
DA, so is
CE to
EA. [
V. 11]
Again, let the sides
AB,
AC of the triangle
ABC be cut proportionally, so that, as
BD is to
DA, so is
CE to
EA; and let
DE be joined.
I say that
DE is parallel to
BC.
For, with the same construction, since, as
BD is to
DA, so is
CE to
EA, but, as
BD is to
DA, so is the triangle
BDE to the triangle
ADE, and, as
CE is to
EA, so is the triangle
CDE to the triangle
ADE, [
VI. 1] therefore also,
as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11]
Therefore each of the triangles
BDE,
CDE has the same ratio to
ADE.
Therefore the triangle
BDE is equal to the triangle
CDE; [
V. 9] and they are on the same base
DE.
But equal triangles which are on the same base are also in the same parallels. [
I. 39]
Therefore
DE is parallel to
BC.
Therefore etc. Q. E. D.
PROPOSITION 3.
If an angle of a triangle be bisected and the straight line cutting the angle cut the base also,
the segments of the base will have the same ratio as the remaining sides of the triangle; and,
if the segments of the base have the same ratio as the remaining sides of the triangle,
the straight line joined from the vertex to the point of section will bisect the angle of the triangle.
Let
ABC be a triangle, and let the angle
BAC be bisected by the straight line
AD; I say that, as
BD is to
CD, so is
BA to
AC.
For let
CE be drawn through
C parallel to
DA, and let
BA
be carried through and meet it at
E.
Then, since the straight line
AC falls upon the parallels
AD,
EC,
the angle ACE is equal to the angle CAD. [I. 29]
But the angle
CAD is by hypothesis equal to the angle
BAD; therefore the angle
BAD is also equal to the angle
ACE.
Again, since the straight line
BAE falls upon the parallels
AD,
EC,
the exterior angle BAD is equal to the interior angle AEC. [I. 29]
But the angle
ACE was also proved equal to the angle
BAD;
therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6]
And, since
AD has been drawn parallel to
EC, one of the sides of the triangle
BCE, therefore, proportionally, as
BD is to
DC, so is
BA to
AE.
But
AE is equal to
AC; [
VI. 2] therefore, as
BD is to
DC, so is
BA to
AC.
Again, let
BA be to
AC as
BD to
DC, and let
AD be joined; I say that the angle
BAC has been bisected by the straight line
A.D.
For, with the same construction, since, as
BD is to
DC, so is
BA to
AC, and also, as
BD is to
DC, so is
BA to
AE : for
AD has been drawn parallel to
EC, one of the sides of the triangle
BCE: [
VI. 2] therefore also, as
BA is to
AC, so is
BA to
AE. [
V. 11]
Therefore
AC is equal to
AE, [
V. 9] so that the angle
AEC is also equal to the angle
ACE. [
I. 5]
But the angle
AEC is equal to the exterior angle
BAD, [
I. 29] and the angle
ACE is equal to the alternate angle
CAD; [
id.]
therefore the angle BAD is also equal to the angle CAD.
Therefore the angle
BAC has been bisected by the straight line
AD.
Therefore etc. Q. E. D.
PROPOSITION 4.
In equiangular triangles the sides about the equal angles are proportional,
and those are corresponding sides which subtend the equal angles.
Let
ABC,
DCE be equiangular triangles having the angle
ABC equal to the angle
DCE, the angle
BAC to the angle
CDE, and further the angle
ACB to the angle
CED; I say that in the triangles
ABC,
DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
For let
BC be placed in a straight line with
CE.
Then, since the angles
ABC,
ACB are less than two right angles, [
I. 17] and the angle
ACB is equal to the angle
DEC, therefore the angles
ABC,
DEC are less than two right angles; therefore
BA,
ED, when produced, will meet. [
I. Post. 5]
Let them be produced and meet at
F.
Now, since the angle
DCE is equal to the angle
ABC,
BF is parallel to CD. [I. 28]
Again, since the angle
ACB is equal to the angle
DEC,
AC is parallel to FE. [I. 28]
Therefore
FACD is a parallelogram; therefore
FA is equal to
DC, and
AC to
FD. [
I. 34]
And, since
AC has been drawn parallel to
FE, one side of the triangle
FBE, therefore, as
BA is to
AF, so is
BC to
CE. [
VI. 2]
But
AF is equal to
CD;
therefore, as BA is to CD, so is BC to CE, and alternately, as
AB is to
BC, so is
DC to
CE. [
V. 16]
Again, since
CD is parallel to
BF, therefore, as
BC is to
CE, so is
FD to
DE. [
VI. 2]
But
FD is equal to
AC;
therefore, as BC is to CE, so is AC to DE, and alternately, as
BC is to
CA, so is
CE to
ED. [
V. 16]
Since then it was proved that,
as AB is to BC, so is DC to CE, and, as
BC is to
CA, so is
CE to
ED; therefore,
ex aequali, as
BA is to
AC, so is
CD to
DE. [
V. 22]
Therefore etc. Q. E. D.
PROPOSITION 5.
If two triangles have their sides proportional,
the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.
Let
ABC,
DEF be two triangles having their sides proportional, so that,
as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as
BA is to
AC, so is
ED to
DF; I say that the triangle
ABC is equiangular with the triangle
DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle
ABC to the angle
DEF, the angle
BCA to the angle
EFD, and further the angle
BAC to the angle
EDF.
For on the straight line
EF, and at the points
E,
F on it, let there be constructed the angle
FEG equal to the angle
ABC, and the angle
EFG equal to the angle
ACB; [
I. 23]
therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]
Therefore the triangle
ABC is equiangular with the triangle
GEF.
Therefore in the triangles
ABC,
GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [
VI. 4]
therefore, as AB is to BC, so is GE to EF.
But, as
AB is to
BC, so by hypothesis is
DE to
EF;
therefore, as DE is to EF, so is GE to EF. [V. 11]
Therefore each of the straight lines
DE,
GE has the same ratio to
EF;
therefore DE is equal to GE. [V. 9]
For the same reason
DF is also equal to GF.
Since then
DE is equal to
EG, and
EF is common,
the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle
DEF is equal to the angle
GEF, [
I. 8] and the triangle
DEF is equal to the triangle
GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
DFE is also equal to the angle
GFE,
and the angle EDF to the angle EGF.
And, since the angle
FED is equal to the angle
GEF, while the angle
GEF is equal to the angle
ABC, therefore the angle
ABC is also equal to the angle
DEF.
For the same reason
the angle ACB is also equal to the angle DFE, and further, the angle at
A to the angle at
D;
therefore the triangle ABC is equiangular with the triangle DEF.
Therefore etc. Q. E. D.
PROPOSITION 6.
If two triangles have one angle equal to one angle and the sides about the equal angles proportional,
the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.
Let
ABC,
DEF be two triangles having one angle
BAC equal to one angle
EDF and the sides about the equal angles proportional, so that,
as BA is to AC, so is ED to DF; I say that the triangle
ABC is equiangular with the triangle
DEF, and will have the angle
ABC equal to the angle
DEF, and the angle
ACB to the angle
DFE.
For on the straight line
DF, and at the points
D,
F on it, let there be constructed the angle
FDG equal to either of the angles
BAC,
EDF, and the angle
DFG equal to the angle
ACB; [
I. 23]
therefore the remaining angle at B is equal to the remaining angle at G. [I. 32]
Therefore the triangle
ABC is equiangular with the triangle
DGF.
Therefore, proportionally, as
BA is to
AC, so is
GD to
DF. [
VI. 4]
But, by hypothesis, as
BA is to
AC, so also is
ED to
DF; therefore also, as
ED is to
DF, so is
GD to
DF. [
V. 11]
Therefore
ED is equal to
DG; [
V. 9] and
DF is common;
therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
DFG is equal to the angle
DFE,
and the angle DGF to the angle DEF.
But the angle
DFG is equal to the angle
ACB; therefore the angle
ACB is also equal to the angle
DFE.
And, by hypothesis, the angle
BAC is also equal to the angle
EDF; therefore the remaining angle at
B is also equal to the remaining angle at
E; [
I. 32]
therefore the triangle ABC is equiangular with the triangle DEF.
Therefore etc. Q. E. D.
PROPOSITION 7.
If two triangles have one angle equal to one angle,
the sides about other angles proportional,
and the remaining angles either both less or both not less than a right angle,
the triangles will be equiangular and will have those angles equal,
the sides about which are proportional.
Let
ABC,
DEF be two triangles having one angle equal to one angle, the angle
BAC to the angle
EDF, the sides about other angles
ABC,
DEF proportional, so that, as
AB is to
BC, so is
DE to
EF, and, first, each of the remaining angles at
C,
F less than a right angle; I say that the triangle
ABC is equiangular with the triangle
DEF, the angle
ABC will be equal to the angle
DEF, and the remaining angle, namely the angle at
C, equal to the remaining angle, the angle at
F.
For, if the angle
ABC is unequal to the angle
DEF, one of them is greater.
Let the angle
ABC be greater; and on the straight line
AB, and at the point
B on it, let the angle
ABG be constructed equal to the angle
DEF. [
I. 23]
Then, since the angle
A is equal to
D, and the angle
ABG to the angle
DEF, therefore the remaining angle
AGB is equal to the remaining angle
DFE. [
I. 32]
Therefore the triangle
ABG is equiangular with the triangle
DEF.
Therefore, as
AB is to
BG, so is
DE to
EF [
VI. 4]
But, as
DE is to
EF, so by hypothesis is
AB to
BC; therefore
AB has the same ratio to each of the straight lines
BC,
BG; [
V. 11]
therefore BC is equal to BG, [V. 9] so that the angle at
C is also equal to the angle
BGC. [
I. 5]
But, by hypothesis, the angle at
C is less than a right angle; therefore the angle
BGC is also less than a right angle; so that the angle
AGB adjacent to it is greater than a right angle. [
I. 13]
And it was proved equal to the angle at
F; therefore the angle at
F is also greater than a right angle.
But it is by hypothesis less than a right angle : which is absurd.
Therefore the angle
ABC is not unequal to the angle
DEF; therefore it is equal to it.
But the angle at
A is also equal to the angle at
D; therefore the remaining angle at
C is equal to the remaining angle at
F. [
I. 32]
Therefore the triangle
ABC is equiangular with the triangle
DEF.
But, again, let each of the angles at
C,
F be supposed not less than a right angle; I say again that, in this case too, the triangle
ABC is equiangular with the triangle
DEF.
For, with the same construction, we can prove similarly that
BC is equal to BG; so that the angle at
C is also equal to the angle
BGC. [
I. 5]
But the angle at
C is not less than a right angle; therefore neither is the angle
BGC less than a right angle.
Thus in the triangle
BGC the two angles are not less than two right angles: which is impossible. [
I. 17]
Therefore, once more, the angle
ABC is not unequal to the angle
DEF; therefore it is equal to it.
But the angle at
A is also equal to the angle at
D; therefore the remaining angle at
C is equal to the remaining angle at
F. [
I. 32]
Therefore the triangle
ABC is equiangular with the triangle
DEF.
Therefore etc. Q. E. D.
PROPOSITION 8.
If in a right-angled triangle a perpendicular be drawn from the right angle to the base,
the triangles adjoining the perpendicular are similar both to the whole and to one another.
Let
ABC be a right-angled triangle having the angle
BAC right, and let
AD be drawn from
A perpendicular to
BC; I say that each of the triangles
ABD,
ADC is similar to the whole
ABC and, further, they are similar to one another.
For, since the angle
BAC is equal to the angle
ADB, for each is right, and the angle at
B is common to the two triangles
ABC and
ABD, therefore the remaining angle
ACB is equal to the remaining angle
BAD; [
I. 32] therefore the triangle
ABC is equiangular with the triangle
ABD.
Therefore, as
BC which subtends the right angle in the triangle
ABC is to
BA which subtends the right angle in the triangle
ABD, so is
AB itself which subtends the angle at
C in the triangle
ABC to
BD which subtends the equal angle
BAD in the triangle
ABD, and so also is
AC to
AD which subtends the angle at
B common to the two triangles. [
VI. 4]
Therefore the triangle
ABC is both equiangular to the triangle
ABD and has the sides about the equal angles proportional.
Therefore the triangle
ABC is similar to the triangle
ABD. [
VI. Def. 1]
Similarly we can prove that the triangle
ABC is also similar to the triangle
ADC; therefore each of the triangles
ABD,
ADC is similar to the whole
ABC.
I say next that the triangles
ABD,
ADC are also similar to one another.
For, since the right angle
BDA is equal to the right angle
ADC, and moreover the angle
BAD was also proved equal to the angle at
C, therefore the remaining angle at
B is also equal to the remaining angle
DAC; [
I. 32] therefore the triangle
ABD is equiangular with the triangle
ADC.
Therefore, as
BD which subtends the angle
BAD in the triangle
ABD is to
DA which subtends the angle at
C in the triangle
ADC equal to the angle
BAD, so is
AD itself which subtends the angle at
B in the triangle
ABD to
DC which subtends the angle
DAC in the triangle
ADC equal to the angle at
B, and so also is
BA to
AC, these sides subtending the right angles; [
VI. 4] therefore the triangle
ABD is similar to the triangle
ADC. [
VI. Def. 1]
Therefore etc.
PORISM.
From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base. Q. E. D.
PROPOSITION 9.
From a given straight line to cut off a prescribed part.
Let
AB be the given straight line; thus it is required to cut off from
AB a prescribed part.
Let the third part be that prescribed.
Let a straight line
AC be drawn through from
A containing with
AB any angle; let a point
D be taken at random on
AC, and let
DE,
EC be made equal to
AD. [
I. 3]
Let
BC be joined, and through
D let
DF be drawn parallel to it. [
I. 31]
Then, since
FD has been drawn parallel to
BC, one of the sides of the triangle
ABC, therefore, proportionally, as
CD is to
DA, so is
BF to
FA. [
VI. 2]
But
CD is double of
DA;
therefore BF is also double of FA; therefore BA is triple of AF.
Therefore from the given straight line
AB the prescribed third part
AF has been cut off. Q. E. F.
4
PROPOSITION 10.
To cut a given uncut straight line similarly to a given cut straight line.
Let
AB be the given uncut straight line, and
AC the straight line cut at the points
D,
E; and let them be so placed as to contain any angle; let
CB be joined, and through
D,
E let
DF,
EG be drawn parallel to
BC, and through
D let
DHK be drawn parallel to
AB. [
I. 31]
Therefore each of the figures
FH,
HB is a parallelogram; therefore
DH is equal to
FG and
HK to
GB. [
I. 34]
Now, since the straight line
HE has been drawn parallel to
KC, one of the sides of the triangle
DKC, therefore, proportionally, as
CE is to
ED, so is
KH to
HD. [
VI. 2]
But
KH is equal to
BG, and
HD to
GF; therefore, as
CE is to
ED, so is
BG to
GF.
Again, since
FD has been drawn parallel to
GE, one of the sides of the triangle
AGE, therefore, proportionally, as
ED is to
DA, so is
GF to
FA. [
VI. 2]
But it was also proved that,
as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA.
Therefore the given uncut straight line
AB has been cut similarly to the given cut straight line
AC. Q. E. F.
PROPOSITION 11.
To two given straight lines to find a third proportional.
Let
BA,
AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to
BA,
AC.
For let them be produced to the points
D,
E, and let
BD be made equal to
AC; [
I. 3] let
BC be joined, and through
D let
DE be drawn parallel to it. [
I. 31]
Since, then,
BC has been drawn parallel to
DE, one of the sides of the triangle
ADE, proportionally, as
AB is to
BD, so is
AC to
CE. [
VI. 2]
But
BD is equal to
AC; therefore, as
AB is to
AC, so is
AC to
CE.
Therefore to two given straight lines
AB,
AC a third proportional to them,
CE, has been found. Q. E. F.
5
PROPOSITION 12.
To three given straight lines to find a fourth proportional.
Let
A,
B,
C be the three given straight lines; thus it is required to find a fourth proportional to
A,
B,
C.
Let two straight lines
DE,
DF be set out containing any angle
EDF; let
DG be made equal to
A,
GE equal to
B, and further
DH equal to
C; let
GH be joined, and let
EF be drawn through
E parallel to it. [
I. 31]
Since, then,
GH has been drawn parallel to
EF, one of the sides of the triangle
DEF, therefore, as
DG is to
GE, so is
DH to
HF. [
VI. 2]
But
DG is equal to
A,
GE to
B, and
DH to
C; therefore, as
A is to
B, so is
C to
HF.
Therefore to the three given straight lines
A,
B,
C a fourth proportional
HF has been found. Q. E. F.
PROPOSITION 13.
To two given straight lines to find a mean proportional.
Let
AB,
BC be the two given straight lines; thus it is required to find a mean proportional to
AB,
BC.
Let them be placed in a straight line, and let the semicircle
ADC be described on
AC; let
BD be drawn from the point
B at right angles to the straight line
AC, and let
AD,
DC be joined.
Since the angle
ADC is an angle in a semicircle, it is right. [
III. 31]
And, since, in the right-angled triangle
ADC,
DB has been drawn from the right angle perpendicular to the base, therefore
DB is a mean proportional between the segments of the base,
AB,
BC. [
VI. 8, Por.]
Therefore to the two given straight lines
AB,
BC a mean proportional
DB has been found. Q. E. F.
PROPOSITION 14.
In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
Let
AB,
BC be equal and equiangular parallelograms having the angles at
B equal, and let
DB,
BE be placed in a straight line;
therefore FB, BG are also in a straight line. [I. 14]
I say that, in
AB,
BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as
DB is to
BE, so is
GB to
BF.
For let the parallelogram
FE be completed.
Since, then, the parallelogram
AB is equal to the parallelogram
BC,
and FE is another area, therefore, as
AB is to
FE, so is
BC to
FE. [
V. 7]
But, as
AB is to
FE, so is
DB to
BE, [
VI. 1] and, as
BC is to
FE, so is
GB to
BF. [
id.] therefore also, as
DB is to
BE, so is
GB to
BF. [
V. 11]
Therefore in the parallelograms
AB,
BC the sides about the equal angles are reciprocally proportional.
Next, let
GB be to
BF as
DB to
BE; I say that the parallelogram
AB is equal to the parallelogram
BC.
For since, as
DB is to
BE, so is
GB to
BF, while, as
DB is to
BE, so is the parallelogram
AB to the parallelogram
FE, [
VI. 1] and, as
GB is to
BF, so is the parallelogram
BC to the parallelogram
FE, [
VI. 1] therefore also, as
AB is to
FE, so is
BC to
FE; [
V. 11] therefore the parallelogram
AB is equal to the parallelogram
BC. [
V. 9]
Therefore etc. Q. E. D.
PROPOSITION 15.
In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle,
and in which the sides about the equal angles are reciprocally proportional,
are equal.
Let
ABC,
ADE be equal triangles having one angle equal to one angle, namely the angle
BAC to the angle
DAE; I say that in the triangles
ABC,
ADE the sides about the equal angles are reciprocally proportional, that is to say, that,
as CA is to AD, so is EA to AB.
For let them be placed so that
CA is in a straight line with
AD; therefore
EA is also in a straight line with
AB. [
I. 14]
Let
BD be joined.
Since then the triangle
ABC is equal to the triangle
ADE, and
BAD is another area, therefore, as the triangle
CAB is to the triangle
BAD, so is the triangle
EAD to the triangle
BAD. [
V. 7]
But, as
CAB is to
BAD, so is
CA to
AD, [
VI. 1] and, as
EAD is to
BAD, so is
EA to
AB. [
id.]
Therefore also, as
CA is to
AD, so is
EA to
AB. [
V. 11]
Therefore in the triangles
ABC,
ADE the sides about the equal angles are reciprocally proportional.
Next, let the sides of the triangles
ABC,
ADE be reciprocally proportional, that is to say, let
EA be to
AB as
CA to
AD; I say that the triangle
ABC is equal to the triangle
ADE.
For, if
BD be again joined, since, as
CA is to
AD, so is
EA to
AB, while, as
CA is to
AD, so is the triangle
ABC to the triangle
BAD, and, as
EA is to
AB, so is the triangle
EAD to the triangle
BAD, [
VI. 1] therefore, as the triangle
ABC is to the triangle
BAD, so is the triangle
EAD to the triangle
BAD. [
V. 11]
Therefore each of the triangles
ABC,
EAD has the same ratio to
BAD.
Therefore the triangle
ABC is equal to the triangle
EAD. [
V. 9]
Therefore etc. Q. E. D.
PROPOSITION 16.
If four straight lines be proportional,
the rectangle contained by the extremes is equal to the rectangle contained by the means; and,
if the rectangle contained by the extremes be equal to the rectangle contained by the means,
the four straight lines will be proportional.
Let the four straight lines
AB,
CD,
E,
F be proportional, so that, as
AB is to
CD, so is
E to
F; I say that the rectangle contained by
AB,
F is equal to the rectangle contained by
CD,
E.
Let
AG,
CH be drawn from the points
A,
C at right angles to the straight lines
AB,
CD, and let
AG be made equal to
F, and
CH equal to
E.
Let the parallelograms
BG,
DH be completed.
Then since, as
AB is to
CD, so is
E to
F, while
E is equal to
CH, and
F to
AG, therefore, as
AB is to
CD, so is
CH to
AG.
Therefore in the parallelograms
BG,
DH the sides about the equal angles are reciprocally proportional.
But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [
VI. 14] therefore the parallelogram
BG is equal to the parallelogram
DH.
And
BG is the rectangle
AB,
F, for
AG is equal to
F; and
DH is the rectangle
CD,
E, for
E is equal to
CH; therefore the rectangle contained by
AB,
F is equal to the rectangle contained by
CD,
E.
Next, let the rectangle contained by
AB,
F be equal to the rectangle contained by
CD,
E; I say that the four straight lines will be proportional, so that, as
AB is to
CD, so is
E to
F.
For, with the same construction, since the rectangle
AB,
F is equal to the rectangle
CD,
E, and the rectangle
AB,
F is
BG, for
AG is equal to
F, and the rectangle
CD,
E is
DH, for
CH is equal to
E,
therefore BG is equal to DH.
And they are equiangular
But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [
VI. 14]
Therefore, as
AB is to
CD, so is
CH to
AG.
But
CH is equal to
E, and
AG to
F; therefore, as
AB is to
CD, so is
E to
F.
Therefore etc. Q. E. D.
PROPOSITION 17
If three straight lines be proportional,
the rectangle contained by the extremes is equal to the square on the mean;
and,
if the rectangle contained by the extremes be equal to the square on the mean,
the three straight lines will be proportional.
Let the three straight lines
A,
B,
C be proportional, so that, as
A is to
B, so is
B to
C; I say that the rectangle contained by
A,
C is equal to the square on
B.
Let
D be made equal to
B.
Then, since, as
A is to
B, so is
B to
C, and
B is equal to
D, therefore, as
A is to
B, so is
D to
C.
But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [
VI. 16]
Therefore the rectangle
A,
C is equal to the rectangle
B,
D.
But the rectangle
B,
D is the square on
B, for
B is equal to
D; therefore the rectangle contained by
A,
C is equal to the square on
B.
Next, let the rectangle
A,
C be equal to the square on
B; I say that, as
A is to
B, so is
B to
C.
For, with the same construction, since the rectangle
A,
C is equal to the square on
B, while the square on
B is the rectangle
B,
D, for
B is equal to
D, therefore the rectangle
A,
C is equal to the rectangle
B,
D.
But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [
VI. 16]
Therefore, as
A is to
B, so is
D to
C.
But
B is equal to
D;
therefore, as A is to B, so is B to C.
Therefore etc. Q. E. D.
PROPOSITION 18.
On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.
Let
AB be the given straight line and
CE the given rectilineal figure; thus it is required to describe on the straight line
AB a rectilineal figure similar and similarly situated to the rectilineal figure
CE.
Let
DF be joined, and on the straight line
AB, and at the points
A,
B on it, let the angle
GAB be constructed equal to the angle at
C, and the angle
ABG equal to the angle
CDF. [
I. 23]
Therefore the remaining angle
CFD is equal to the angle
AGB; [
I. 32]
therefore the triangle FCD is equiangular with the triangle GAB.
Therefore, proportionally, as
FD is to
GB, so is
FC to
GA, and
CD to
AB.
Again, on the straight line
BG, and at the points
B,
G on it, let the angle
BGH be constructed equal to the angle
DFE, and the angle
GBH equal to the angle
FDE. [
I. 23]
Therefore the remaining angle at
E is equal to the remaining angle at
H; [
I. 32]
therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4]
But it was also proved that, as
FD is to
GB, so is
FC to
GA, and
CD to
AB;
therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB.
And, since the angle
CFD is equal to the angle
AGB, and the angle
DFE to the angle
BGH, therefore the whole angle
CFE is equal to the whole angle
AGH.
For the same reason
the angle CDE is also equal to the angle ABH.
And the angle at
C is also equal to the angle at
A,
and the angle at E to the angle at H.
Therefore
AH is equiangular with
CE; and they have the sides about their equal angles proportional;
therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1]
Therefore on the given straight line
AB the rectilineal figure
AH has been described similar and similarly situated to the given rectilineal figure
CE. Q. E. F.
PROPOSITION 19.
Similar triangles are to one another in the duplicate ratio of the corresponding sides.
Let
ABC,
DEF be similar triangles having the angle at
B equal to the angle at
E, and such that, as
AB is to
BC, so
is
DE to
EF, so that
BC corresponds to
EF; [
V. Def. 11] I say that the triangle
ABC has to the triangle
DEF a ratio duplicate of that which
BC has to
EF.
For let a third proportional
BG be taken to
BC,
EF, so that, as
BC is to
EF, so is
EF to
BG; [
VI. 11]
and let
AG be joined.
Since then, as
AB is to
BC, so is
DE to
EF, therefore, alternately, as
AB is to
DE, so is
BC to
EF. [
V. 16]
But, as
BC is to
EF, so is
EF to
BG; therefore also, as
AB is to
DE, so is
EF to
BG. [
V. 11]
Therefore in the triangles
ABG,
DEF the sides about the equal angles are reciprocally proportional.
But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [
VI. 15]
therefore the triangle
ABG is equal to the triangle
DEF.
Now since, as
BC is to
EF, so is
EF to
BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [
V. Def. 9] therefore
BC has to
BG a ratio duplicate of that which
CB
has to
EF.
But, as
CB is to
BG, so is the triangle
ABC to the triangle
ABG; [
VI. 1] therefore the triangle
ABC also has to the triangle
ABG a ratio duplicate of that which
BC has to
EF.
But the triangle
ABG is equal to the triangle
DEF; therefore the triangle
ABC also has to the triangle
DEF a ratio duplicate of that which
BC has to
EF.
Therefore etc.
PORISM.
From this it is manifest that, if three straight
lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second. Q. E. D.
6
PROPOSITION 20.
Similar polygons are divided into similar triangles,
and into triangles equal in multitude and in the same ratio as the wholes,
and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding
side.
Let
ABCDE,
FGHKL be similar polygons, and let
AB correspond to
FG; I say that the polygons
ABCDE,
FGHKL are divided into similar triangles, and into triangles equal in multitude and in
the same ratio as the wholes, and the polygon
ABCDE has to the polygon
FGHKL a ratio duplicate of that which
AB has to
FG.
Let
BE,
EC,
GL,
LH be joined.
Now, since the polygon
ABCDE is similar to the polygon
FGHKL, the angle
BAE is equal to the angle
GFL;
and, as BA is to AE, so is GF to FL. [VI. Def. 1]
Since then
ABE,
FGL are two triangles having one angle equal to one angle and the sides about the equal angles
proportional, therefore the triangle
ABE is equiangular with the triangle
FGL; [
VI. 6]
so that it is also similar; [VI. 4 and Def. 1] therefore the angle
ABE is equal to the angle
FGL.
But the whole angle
ABC is also equal to the whole angle
FGH because of the similarity of the polygons; therefore the remaining angle
EBC is equal to the angle
LGH.
And, since, because of the similarity of the triangles
ABE,
FGL,
as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons,
as AB is to BC, so is FG to GH, therefore,
ex aequali, as
EB is to
BC, so is
LG to
GH; [
V. 22]
that is, the sides about the equal angles
EBC,
LGH are proportional; therefore the triangle
EBC is equiangular with the triangle
LGH, [
VI. 6]
so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1]
For the same reason the triangle
ECD is also similar to the triangle
LHK.
Therefore the similar polygons
ABCDE,
FGHKL have been divided into similar triangles, and into triangles equal in
multitude.
I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and
ABE,
EBC,
ECD are antecedents, while
FGL,
LGH,
LHK are their consequents, and that the polygon
ABCDE
has to the polygon
FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is
AB to
FG.
For let
AC,
FH be joined.
Then since, because of the similarity of the polygons,
the angle
ABC is equal to the angle
FGH, and, as
AB is to
BC, so is
FG to
GH,
the triangle ABC is equiangular with the triangle FGH; [VI. 6] therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF.
And, since the angle
BAM is equal to the angle
GFN, and the angle
ABM is also equal to the angle
FGN, therefore the remaining angle
AMB is also equal to the remaining angle
FNG; [
I. 32] therefore the triangle
ABM is equiangular with the triangle
FGN.
Similarly we can prove that the triangle
BMC is also equiangular with the triangle
GNH.
Therefore, proportionally, as
AM is to
MB, so is
FN to
NG,
and, as
BM is to
MC, so is
GN to
NH; so that, in addition,
ex aequali,
as AM is to MC, so is FN to NH.
But, as
AM is to
MC, so is the triangle
ABM to
MBC, and
AME to
EMC; for they are to one another as their
bases. [
VI. 1]
Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [
V. 12] therefore, as the triangle
AMB is to
BMC, so is
ABE to
CBE.
But, as
AMB is to
BMC, so is
AM to
MC; therefore also, as
AM is to
MC, so is the triangle
ABE to the triangle
EBC.
For the same reason also, as
FN is to
NH, so is the triangle
FGL to the triangle
GLH.
And, as
AM is to
MC, so is
FN to
NH; therefore also, as the triangle
ABE is to the triangle
BEC, so is the triangle
FGL to the triangle
GLH; and, alternately, as the triangle
ABE is to the triangle
FGL,
so is the triangle
BEC to the triangle
GLH.
Similarly we can prove, if
BD,
GK be joined, that, as the triangle
BEC is to the triangle
LGH, so also is the triangle
ECD to the triangle
LHK.
And since, as the triangle
ABE is to the triangle
FGL,
so is
EBC to
LGH, and further
ECD to
LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [
V. 12 therefore, as the triangle
ABE is to the triangle
FGL, so is the polygon
ABCDE to the polygon
FGHKL.
But the triangle
ABE has to the triangle
FGL a ratio duplicate of that which the corresponding side
AB has to the corresponding side
FG; for similar triangles are in the duplicate ratio of the corresponding sides. [
VI. 19]
Therefore the polygon
ABCDE also has to the polygon
FGHKL a ratio duplicate of that which the corresponding side
AB has to the corresponding side
FG.
Therefore etc.
PORISM.
Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the
corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Q. E. D.
7
8
PROPOSITION 21.
Figures which are similar to the same rectilineal figure are also similar to one another.
For let each of the rectilineal figures
A,
B be similar to
C; I say that
A is also similar to
B.
For, since
A is similar to
C, it is equiangular with it and has the sides about the equal angles proportional. [
VI. Def. 1]
Again, since
B is similar to
C, it is equiangular with it and has the sides about the equal angles proportional.
Therefore each of the figures
A,
B is equiangular with
C and with
C has the sides about the equal angles proportional;
therefore A is similar to B. Q. E. D.
PROPOSITION 22.
If four straight lines be proportional,
the rectilineal figures similar and similarly described upon them will also be proportional;
and,
if the rectilineal figures similar and similarly described upon them be proportional,
the straight lines will themselves also be proportional.
Let the four straight lines
AB,
CD,
EF,
GH be proportional, so that, as
AB is to
CD, so is
EF to
GH, and let there be described on
AB,
CD the similar and similarly situated rectilineal figures
KAB,
LCD, and on
EF,
GH the similar and similarly situated rectilineal figures
MF,
NH; I say that, as
KAB is to
LCD, so is
MF to
NH.
For let there be taken a third proportional
O to
AB,
CD, and a third proportional
P to
EF,
GH. [
VI. 11]
Then since, as
AB is to
CD, so is
EF to
GH,
and, as CD is to O, so is GH to P, therefore,
ex aequali, as
AB is to
O, so is
EF to
P. [
V. 22]
But, as
AB is to
O, so is
KAB to
LCD, [
VI. 19, Por.]
and, as EF is to P, so is MF to NH; therefore also, as
KAB is to
LCD, so is
MF to
NH. [
V. 11]
Next, let
MF be to
NH as
KAB is to
LCD; I say also that, as
AB is to
CD, so is
EF to
GH.
For, if
EF is not to
GH as
AB to
CD,
let EF be to QR as AB to CD, [VI. 12] and on
QR let the rectilineal figure
SR be described similar and similarly situated to either of the two
MF,
NH. [
VI. 18]
Since then, as
AB is to
CD, so is
EF to
QR, and there have been described on
AB,
CD the similar and similarly situated figures
KAB,
LCD, and on
EF,
QR the similar and similarly situated figures
MF,
SR, therefore, as
KAB is to
LCD, so is
MF to
SR.
But also, by hypothesis,
as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11]
Therefore
MF has the same ratio to each of the figures
NH,
SR;
therefore NH is equal to SR. [V. 9]
But it is also similar and similarly situated to it;
therefore GH is equal to QR.
And, since, as
AB is to
CD, so is
EF to
QR, while
QR is equal to
GH, therefore, as
AB is to
CD, so is
EF to
GH.
Therefore etc. Q. E. D.
PROPOSITION 23.
Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides.
Let
AC,
CF be equiangular parallelograms having the angle
BCD equal to the angle
ECG;
I say that the parallelogram
AC has to the parallelogram
CF the ratio compounded of the ratios of the sides.
For let them be placed so that
BC is in a straight line with
CG;
therefore DC is also in a straight line with CE.
Let the parallelogram
DG be completed; let a straight line
K be set out, and let it be contrived that,
as BC is to CG, so is K to L, and, as
DC is to
CE, so is
L to
M. [
VI. 12]
Then the ratios of
K to
L and of
L to
M are the same
as the ratios of the sides, namely of
BC to
CG and of
DC to
CE.
But the ratio of
K to
M is compounded of the ratio of
K to
L and of that of
L to
M; so that
K has also to
M the ratio compounded of the ratios
of the sides.
Now since, as
BC is to
CG, so is the parallelogram
AC to the parallelogram
CH, [
VI. 1] while, as
BC is to
CG, so is
K to
L, therefore also, as
K is to
L, so is
AC to
CH. [
V. 11]
Again, since, as
DC is to
CE, so is the parallelogram
CH to
CF, [
VI. 1] while, as
DC is to
CE, so is
L to
M, therefore also, as
L is to
M, so is the parallelogram
CH to the parallelogram
CF. [
V. 11]
Since then it was proved that, as
K is to
L, so is the parallelogram
AC to the parallelogram
CH, and, as
L is to
M, so is the parallelogram
CH to the parallelogram
CF, therefore,
ex aequali, as
K is to
M, so is
AC to the parallelogram
CF.
But
K has to
M the ratio compounded of the ratios of the sides;
therefore AC also has to CF the ratio compounded of the ratios of the sides.
Therefore etc. Q. E. D.
9
10
PROPOSITION 24.
In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.
Let
ABCD be a parallelogram, and
AC its diameter, and let
EG,
HK be parallelograms about
AC; I say that each of the parallelograms
EG,
HK is similar both to the whole
ABCD and to the other.
For, since
EF has been drawn parallel to
BC, one of the sides of the triangle
ABC,
proportionally, as BE is to EA, so is CF to FA. [VI. 2]
Again, since
FG has been drawn parallel to
CD, one of the sides of the triangle
ACD,
proportionally, as CF is to FA, so is DG to GA. [VI. 2]
But it was proved that,
as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore,
componendo,
as BA is to AE, so is DA to AG, [V. 18] and, alternately,
as BA is to AD, so is EA to AG. [V. 16]
Therefore in the parallelograms
ABCD,
EG, the sides about the common angle
BAD are proportional.
And, since
GF is parallel to
DC,
the angle AFG is equal to the angle DCA; and the angle
DAC is common to the two triangles
ADC,
AGF;
therefore the triangle ADC is equiangular with the triangle AGF.
For the same reason
the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram
ABCD is equiangular with the parallelogram
EG.
Therefore, proportionally,
as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as
CB is to
BA, so is
FE to
EA.
And, since it was proved that,
as DC is to CA, so is GF to FA, and, as
AC is to
CB, so is
AF to
FE, therefore,
ex aequali, as
DC is to
CB, so is
GF to
FE. [
V. 22]
Therefore in the parallelograms
ABCD,
EG the sides about the equal angles are proportional; therefore the parallelogram
ABCD is similar to the parallelogram
EG. [
VI. Def. 1]
For the same reason the parallelogram
ABCD is also similar to the parallelogram
KH; therefore each of the parallelograms
EG,
HK is similar to
ABCD.
But figures similar to the same rectilineal figure are also similar to one another; [
VI. 21] therefore the parallelogram
EG is also similar to the parallelogram
HK.
Therefore etc. Q. E. D.
PROPOSITION 25.
To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.
Let
ABC be the given rectilineal figure to which the figure to be constructed must be similar, and
D that to which it must be equal; thus it is required to construct one and the same figure similar to
ABC and equal to
D.
Let there be applied to
BC the parallelogram
BE equal to the triangle
ABC [
I. 44], and to
CE the parallelogram
CM equal to
D in the angle
FCE which is equal to the angle
CBL. [
I. 45]
Therefore
BC is in a straight line with
CF, and
LE with
EM.
Now let
GH be taken a mean proportional to
BC,
CF [
VI. 13], and on
GH let
KGH be described similar and similarly situated to
ABC. [
VI. 18]
Then, since, as
BC is to
GH, so is
GH to
CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [
VI. 19, Por.] therefore, as
BC is to
CF, so is the triangle
ABC to the triangle
KGH.
But, as
BC is to
CF, so also is the parallelogram
BE to the parallelogram
EF. [
VI. 1]
Therefore also, as the triangle
ABC is to the triangle
KGH, so is the parallelogram
BE to the parallelogram
EF; therefore, alternately, as the triangle
ABC is to the parallelogram
BE, so is the triangle
KGH to the parallelogram
EF. [
V. 16]
But the triangle
ABC is equal to the parallelogram
BE; therefore the triangle
KGH is also equal to the parallelogram
EF.
But the parallelogram
EF is equal to
D; therefore
KGH is also equal to
D.
And
KGH is also similar to
ABC.
Therefore one and the same figure
KGH has been constructed similar to the given rectilineal figure
ABC and equal to the other given figure
D. Q. E. D.
11
PROPOSITION 26.
If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it,
it is about the same diameter with the whole
For from the parallelogram
ABCD let there be taken away the parallelogram
AF similar and similarly situated to
ABCD, and having the angle
DAB common with it; I say that
ABCD is about the same diameter with
AF.
For suppose it is not, but, if possible, let
AHC be the diameter < of
ABCD >, let
GF be produced and carried through to
H, and let
HK be drawn through
H parallel to either of the straight lines
AD,
BC. [
I. 31]
Since, then,
ABCD is about the same diameter with
KG, therefore, as
DA is to
AB, so is
GA to
AK. [
VI. 24]
But also, because of the similarity of
ABCD,
EG,
as DA is to AB, so is GA to AE; therefore also, as
GA is to
AK, so is
GA to
AE. [
V. 11]
Therefore
GA has the same ratio to each of the straight lines
AK,
AE.
Therefore
AE is equal to
AK [
V. 9], the less to the greater : which is impossible.
Therefore
ABCD cannot but be about the same diameter with
AF; therefore the parallelogram
ABCD is about the same diameter with the parallelogram
AF.
Therefore etc. Q. E. D.
PROPOSITION 27.
Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line,
that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.
Let
AB be a straight line and let it be bisected at
C; let there be applied to the straight line
AB the parallelogram
AD deficient by the parallelogrammic figure
DB described on the half of
AB, that is,
CB; I say that, of all the parallelograms applied to
AB and deficient by parallelogrammic figures similar and similarly situated to
DB,
AD is greatest.
For let there be applied to the straight line
AB the parallelogram
AF deficient by the parallelogrammic figure
FB similar and similarly situated to
DB; I say that
AD is greater than
AF.
For, since the parallelogram
DB is similar to the parallelogram
FB,
they are about the same diameter. [VI. 26]
Let their diameter
DB be drawn, and let the figure be described.
Then, since
CF is equal to
FE, [
I. 43] and
FB is common, therefore the whole
CH is equal to the whole
KE.
But
CH is equal to
CG, since
AC is also equal to
CB. [
I. 36]
Therefore
GC is also equal to
EK.
Let
CF be added to each;
therefore the whole AF is equal to the gnomon LMN; so that the parallelogram
DB, that is,
AD, is greater than the parallelogram
AF.
Therefore etc.
PROPOSITION 28.
To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.
Let
AB be the given straight line,
C the given rectilineal figure to which the figure to be applied to
AB is required to be equal, not being greater than the parallelogram described on the half of
AB and similar to the defect, and
D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line
AB a parallelogram equal to the given rectilineal figure
C and deficient by a parallelogrammic figure which is similar to
D.
Let
AB be bisected at the point
E, and on
EB let
EBFG be described similar and similarly situated to
D; [
VI. 18] let the parallelogram
AG be completed.
If then
AG is equal to
C, that which was enjoined will have been done;
for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.
But, if not, let
HE be greater than
C.
Now
HE is equal to
GB;
therefore GB is also greater than C.
Let
KLMN be constructed at once equal to the excess by which
GB is greater than
C and similar and similarly situated to
D. [
VI. 25]
But
D is similar to
GB;
therefore KM is also similar to GB. [VI. 21]
Let, then,
KL correspond to
GE, and
LM to
GF.
Now, since
GB is equal to
C,
KM,
therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM.
Let
GO be made equal to
KL, and
GP equal to
LM; and let the parallelogram
OGPQ be completed;
therefore it is equal and similar to KM.
Therefore
GQ is also similar to
GB; [
VI. 21] therefore
GQ is about the same diameter with
GB. [
VI. 26]
Let
GQB be their diameter, and let the figure be described.
Then, since
BG is equal to
C,
KM, and in them
GQ is equal to
KM, therefore the remainder, the gnomon
UWV, is equal to the remainder
C.
And, since
PR is equal to
OS,
let QB be added to each; therefore the whole
PB is equal to the whole
OB.
But
OB is equal to
TE, since the side
AE is also equal to the side
EB; [
I. 36]
therefore TE is also equal to PB.
Let
OS be added to each;
therefore the whole TS is equal to the whole, the gnomon VWU.
But the gnomon
VWU was proved equal to
C;
therefore TS is also equal to C.
Therefore to the given straight line
AB there has been applied the parallelogram
ST equal to the given rectilineal figure
C and deficient by a parallelogrammic figure
QB which is similar to
D. Q. E. F.
PROPOSITION 29.
To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.
Let
AB be the given straight line,
C the given rectilineal figure to which the figure to be applied to
AB is required to be equal, and
D that to which the excess is required to be similar; thus it is required to apply to the straight line
AB a parallelogram equal to the rectilineal figure
C and exceeding by a parallelogrammic figure similar to
D.
Let
AB be bisected at
E; let there be described on
EB the parallelogram
BF similar and similarly situated to
D; and let
GH be constructed at once equal to the sum of
BF,
C and similar and similarly situated to
D. [
VI. 25]
Let
KH correspond to
FL and
KG to
FE.
Now, since
GH is greater than
FB, therefore
KH is also greater than
FL, and
KG than
FE.
Let
FL,
FE be produced, let
FLM be equal to
KH, and
FEN to
KG, and let
MN be completed;
therefore MN is both equal and similar to GH.
But
GH is similar to
EL;
therefore MN is also similar to EL; [VI. 21] therefore
EL is about the same diameter with
MN. [
VI. 26]
Let their diameter
FO be drawn, and let the figure be described.
Since
GH is equal to
EL,
C, while
GH is equal to
MN, therefore
MN is also equal to
EL,
C.
Let
EL be subtracted from each;
therefore the remainder, the gnomon XWV, is equal to C.
Now, since
AE is equal to
EB,
AN is also equal to NB [I. 36], that is, to LP [I. 43].
Let
EO be added to each;
therefore the whole AO is equal to the gnomon VWX.
But the gnomon
VWX is equal to
C;
therefore AO is also equal to C.
Therefore to the given straight line
AB there has been applied the parallelogram
AO equal to the given rectilineal figure
C and exceeding by a parallelogrammic figure
QP which is similar to
D, since
PQ is also similar to
EL [
VI. 24]. Q. E. F.
PROPOSITION 30.
To cut a given finite straight line in extreme and mean ratio.
Let
AB be the given finite straight line;
thus it is required to cut AB in extreme and mean ratio.
On
AB let the square
BC be described; and let there be applied to
AC the parallelogram
CD equal to
BC and exceeding by the figure
AD similar to
BC. [
VI. 29]
Now
BC is a square;
therefore AD is also a square.
And, since
BC is equal to
CD, let
CE be subtracted from each;
therefore the remainder BF is equal to the remainder AD.
But it is also equiangular with it; therefore in
BF,
AD the sides about the equal angles are reciprocally proportional; [
VI. 14]
therefore, as FE is to ED, so is AE to EB.
But
FE is equal to
AB, and
ED to
AE.
Therefore, as
BA is to
AE, so is
AE to
EB.
And
AB is greater than
AE;
therefore AE is also greater than EB.
Therefore the straight line
AB has been cut in extreme and mean ratio at
E, and the greater segment of it is
AE. Q. E. F.
PROPOSITION 31.
In right-
angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.
Let
ABC be a right-angled triangle having the angle
BAC right; I say that the figure on
BC is equal to the similar and similarly described figures on
BA,
AC.
Let
AD be drawn perpendicular.
Then since, in the right-angled triangle
ABC,
AD has been drawn from the right angle at
A perpendicular to the base
BC, the triangles
ABD,
ADC adjoining the perpendicular are similar both to the whole
ABC and to one another. [
VI. 8]
And, since
ABC is similar to
ABD, therefore, as
CB is to
BA, so is
AB to
BD. [
VI. Def. 1]
And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [
VI. 19, Por.]
Therefore, as
CB is to
BD, so is the figure on
CB to the similar and similarly described figure on
BA.
For the same reason also, as
BC is to
CD, so is the figure on
BC to that on
CA; so that, in addition, as
BC is to
BD,
DC, so is the figure on
BC to the similar and similarly described figures on
BA,
AC.
But
BC is equal to
BD,
DC; therefore the figure on
BC is also equal to the similar and similarly described figures on
BA,
AC.
Therefore etc. Q. E. D.
PROPOSITION 32.
If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.
Let
ABC,
DCE be two triangles having the two sides
BA,
AC proportional to the two sides
DC,
DE, so that, as
AB is to
AC, so is
DC to
DE, and
AB parallel to
DC, and
AC to
DE; I say that
BC is in a straight line with
CE.
For, since
AB is parallel to
DC, and the straight line
AC has fallen upon them, the alternate angles
BAC,
ACD are equal to one another. [
I. 29]
For the same reason
the angle CDE is also equal to the angle ACD; so that the angle
BAC is equal to the angle
CDE.
And, since
ABC,
DCE are two triangles having one angle, the angle at
A, equal to one angle, the angle at
D,
and the sides about the equal angles proportional, so that, as
BA is to
AC, so is
CD to
DE,
therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6] therefore the angle ABC is equal to the angle DCE.
But the angle
ACD was also proved equal to the angle
BAC;
therefore the whole angle ACE is equal to the two angles ABC, BAC.
Let the angle
ACB be added to each; therefore the angles
ACE,
ACB are equal to the angles
BAC,
ACB,
CBA.
But the angles
BAC,
ABC,
ACB are equal to two right angles; [
I. 32]
therefore the angles ACE, ACB are also equal to two right angles.
Therefore with a straight line
AC, and at the point
C on it, the two straight lines
BC,
CE not lying on the same side make the adjacent angles
ACE,
ACB equal to two right angles;
therefore BC is in a straight line with CE. [I. 14]
Therefore etc. Q. E. D.
PROPOSITION 33.
In equal circles angles have the same ratio as the circumferences on which they stand,
whether they stand at the centres or at the circumferences.
Let
ABC,
DEF be equal circles, and let the angles
BGC,
EHF be angles at their centres
G,
H, and the angles
BAC,
EDF angles at the circumferences; I say that, as the circumference
BC is to the circumference
EF, so is the angle
BGC to the angle
EHF, and the angle
BAC to the angle
EDF.
For let any number of consecutive circumferences
CK,
KL be made equal to the circumference
BC, and any number of consecutive circumferences
FM,
MN equal to the circumference
EF; and let
GK,
GL,
HM,
HN be joined.
Then, since the circumferences
BC,
CK,
KL are equal to one another, the angles
BGC,
CGK,
KGL are also equal to one another; [
III. 27] therefore, whatever multiple the circumference
BL is of
BC, that multiple also is the angle
BGL of the angle
BGC.
For the same reason also, whatever multiple the circumference
NE is of
EF, that multiple also is the angle
NHE of the angle
EHF.
If then the circumference
BL is equal to the circumference
EN, the angle
BGL is also equal to the angle
EHN; [
III. 27] if the circumference
BL is greater than the circumference
EN, the angle
BGL is also greater than the angle
EHN; and, if less, less.
There being then four magnitudes, two circumferences
BC,
EF, and two angles
BGC,
EHF, there have been taken, of the circumference
BC and the angle
BGC equimultiples, namely the circumference
BL and the angle
BGL, and of the circumference
EF and the angle
EHF equimultiples, namely the circumference
EN and the angle
EHN.
And it has been proved that, if the circumference
BL is in excess of the circumference
EN, the angle
BGL is also in excess of the angle
EHN; if equal, equal; and if less, less.
Therefore, as the circumference
BC is to
EF, so is the angle
BGC to the angle
EHF. [
V. Def. 5]
But, as the angle
BGC is to the angle
EHF, so is the angle
BAC to the angle
EDF; for they are doubles respectively.
Therefore also, as the circumference
BC is to the circumference
EF, so is the angle
BGC to the angle
EHF, and the angle
BAC to the angle
EDF.
Therefore etc. Q. E. D.