Let the following be postulated:
Proposition 1
On a given finite straight line to construct an equilateral triangle.Let AB be the given finite straight line.
Thus it is required to construct an equilateral triangle on the straight line AB.
With centre A and distance
AB let the circle
BCD
be described;
[Post. 3]
again, with centre B
and distance BA let the circle
ACE be described;
[Post. 3]
and from the point C, in which
the circles cut one another, to the points
A, B
let the straight lines CA,
CB be joined.
[Post. 1]
Now since the point A is the
centre of the
circle CDB,
AC is equal to
AB.
[Def. 15]
Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]
But CA was also proved equal
to AB;
therefore each of the straight lines
CA, CB is
equal to AB.
And things which are equal to the same thing are also
equal to one another;
[C.N. 1]
therefore CA is also equal
to CB.
Therefore the three straight lines CA, AB, BC are equal to one another.
Therefore the triangle ABC
is equilateral; and it has been constructed on the given
finite straight line AB.
(Being) what it was required to do.
Proposition 2
To place at a given point (as an extremity) a straight line equal to a given straight line.Let A be the given point, and BC the given straight line.
Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC.
From the point A to the
point B let the straight line
AB be joined;
[Post. 1]
and on it let the equilateral
triangle DAB be constructed.
[I. 1]
Let the straight lines
AE, BF
be produced in a straight
line with DA,
DB;
[Post. 2]
with centre B and distance
BC let the circle
CGH be described;
and again, with centre D and
distance DG let the
circle GKL be described.
[Post. 3]
Then, since the point B is the centre of the circle CGH, BC is equal to BG.
Again, since the point D is the centre of the circle GKL, DL is equal to DG.
And in these DA is equal
to DB;
therefore the remainder AL
is equal to the
remainder BG.
[C.N. 3]
But BC was also proved
equal to BG;
therefore each of the straight
lines AL, BC is
equal to BG.
And things which are equal to the same thing are also equal
to one another;
[C.N. 1]
therefore AL is also
equal to BC.
Therefore at the given point A
the straight line AL is
placed equal to the given straight
line BC.
(Being) what it was required to do.
Proposition 3
Given two unequal straight lines, to cut off from the greater a straight line equal to the less.Let AB, C be the two given unequal straight lines, and let AB be the greater of them.
Thus it is required to cut off from AB the greater a straight line equal to C the less.
At the point A let
AD be placed equal to the straight
line C;
[I. 2]
and with centre A and
distance AD let the circle
DEF be described.
[Post. 3]
Now, since the point A is the
centre of the
circle DEF,
AE is equal
to AD.
[Def. 15]
But C is also equal to AD.
Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]
Therefore, given the two straight
lines AB, C,
from AB
the greater AE has been cut off
equal to C the less.
(Being) what it was required to do.
Proposition 4
If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal angles subtend.Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF, respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF.
I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.
For, if the triangle ABC be
applied to the
triangle DEF,
and if the point A be placed
on the point D
and the straight line AB on
DE,
then the point B will also
coincide with E, because
AB is equal
to DE.
Again, AB coinciding with
DE,
the straight line AC will
also coincide with DF, because the
angle BAC is equal to
the angle EDF;
hence the point C will also coincide
with the point F, because
AC is again equal
to DF.
But B also coincided with
E;
hence the base BC will
coincide with the base EF.
[For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible.
Therefore the base BC will coincide with EF] and will be equal to it. [C.N. 4]
Thus the whole triangle ABC
will coincide with the whole
triangle DEF,
and will be equal to it.
And the remaining angles will also coincide with the remaining
angles and will be equal to them,
the angle ABC to the
angle DEF,
and the angle ACB to
the angle DFE.
Therefore etc.
(Being) what it was required to prove.
Proposition 5
In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further; the angles under the base will be equal to one another.
Let ABC be an isosceles triangle
having the side AB
equal to the side AC;
and let the straight lines
BD, CE,
be produced further in a straight line with
AB,
AC.
[Post. 2]
I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE.
Let a point F be taken at
random on BD;
from AE the greater let
AG be cut off equal to
AF the less;
[I. 3]
and let the straight
lines FC,
GB be joined.
[Post. 1]
Then, since AF is
equal to AG and
AB
to AC,
the two sides FA,
AC are equal to
the two sides GA,
AB
respectively;
and they contain a common angle, the
angle FAG.
Therefore the base FC
is equal to the base GB,
and the triangle AFC is
equal to the triangle AGB,
and the remaining angles will be equal to the remaining angles
respectively, namely those which the equal sides
subtend,
that is, the angle ACF to
the angle ABG,
and the angle AFC to
the angle AGB.
[I. 4]
And, since [I. 4]
Again, let sides subtending equal angles be equal,
as AB
to DE;
I say again that the remaining sides will be equal to
the remaining sides, namely
AC to
DF and
BC to
EF, and further
the remaining angle BAC is equal to
the remaining angle EDF.
For if BC is unequal to EF, one of them is greater.
Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.
Then, since BH is equal
to EF, and
AB to
DE,
the two sides
AB, BH
are equal to the two sides
DE, EF
respectively, and they contain equal angles;
therefore the base AH is equal
to the base DF,
and the triangle ABH is equal
to the triangle DEF,
and the remaining angles will be equal to the remaining angles,
namely those which equal sides subtend;
[I. 4]
therefore the angle BHA is equal to
the angle EFD.
But the angle EFD is equal to
the angle BCA;
therefore, in the triangle AHC,
the exterior angle BHA
is equal to the interior and opposite
angle BCA:
which is impossible.
[I. 16]
Therefore BC is not
unequal to EF,
and is therefore equal to it.
But AB is also
equal to DE;
therefore the two sides
AB, BC
are equal to the two sides
DE, EF
respectively, and they contain equal angles;
therefore the base AC
is equal to the base DF,
the triangle ABC equal to
the triangle DEF,
and the remaining angle BAC
equal to the remaining angle
EDF.
[I. 4]
Therefore etc. Q.E.D.
Proposition 6
If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.
Let ABC be a triangle
having the angle ABC equal to
the angle ACB;
I say that the side AB is
also equal to the side AC.
For, if AB is unequal to AC, one of them is greater.
Let AB be greater; and
from AB the greater
let DB
be cut off equal to AC
the less;
let DC be joined.
Then, since DB is equal
to AC, and
BC is common,
the two sides DB,
BC are equal to the
two sides AC,
CB respectively;
and the angle DBC is equal
to the angle ACB;
therefore the base DC is equal
to the base AB,
and the triangle DBC will be equal
to the triangle ACB,
the less to the greater;
which is absurd.
Therefore AB is not unequal
to AC;
it is therefore equal to it.
Therefore etc. Q.E.D.
Proposition 7
Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.
Then, since AC is
equal to AD,
the angle ACD is also equal
to the angle ADC;
[I. 5]
therefore the angle ADC
is greater than the angle
DCB;
therefore the angle CDB
is much greater than the angle DCB.
Again, since CD is
equal to DB,
the angle CDB is also equal
to the angle DCB.
But it was also proved much greater than it:
which is impossible.
Therefore etc. Q.E.D.
Proposition 8
If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.
Let ABC,
DEF be two triangles having the two sides
AB, AC
equal to the two sides DE,
DF respectively, namely
AB to DE
and AC to DF;
and let them have the base BC equal
to the base EF;
I say that the angle BAC is also
equal to the angle EDF.
For if the triangle ABC be applied
to the triangle DEF, and if the
point B be placed on the
point E and the straight line
BC on
EF,
the point C will also
coincide with F,
because BC is equal to
EF.
Then, BC coinciding with
EF,
BA, AC
will also coincide with ED,
DF;
for, if the base BC coincides
with the base EF, and the
sides BA, AC
do not coincide with
ED, DF
but fall beside them as
EG,
GF,
then, given two straight lines constructed on a straight line
(from its extremities) and meeting in a point, there will
have been constructed on the same straight line (from its
extremities), and on the same side of it, two other straight
lines meeting in another point and equal to the former
two respectively, namely each to that which has the same
extremity with it.
But they cannot be so constructed. [I. 7]
Therefore it is not possible that, if the
base BC be applied
to the base EF, the
sides BA,
AC should not coincide
with ED,
DF;
they will therefore coincide,
so that the angle BAC
will also coincide with the angle EDF,
and will be equal to it.
If therefore etc. Q.E.D.
Proposition 9
To bisect a given rectilineal angle.Let the angle BAC be the given rectilineal angle.
Thus it is required to bisect it.
Let a point D be taken
at random on AB;
let AE be cut off from
AC equal to
AD;
[I. 3]
let DE be joined, and on
DE let the equilateral triangle
DEF be constructed;
let AF be joined.
I say that the angle BAC has been bisected by the straight line AF.
For, since AD is equal to
AE,
and AF is common,
the two sides DA,
AF are equal to the
two sides EA,
AF
respectively.
And the base DF is equal to
the base EF;
therefore the angle DAF is equal
to the angle.
EAF
[I. 8]
Therefore the given rectilineal angle BAC has been bisected by the straight line AF. Q.E.F.
Proposition 10
To bisect a given finite straight line.Let AB be the given finite straight line.
Thus it is required to bisect the given straight line AB.
Let the equilateral triangle ABC
be constructed on it,
[I. 1]
and let the angle ACB be bisected
by the straight
line CD;
[I. 9]
I say that the straight line AB
has been bisected at the point D.
For, since AC is equal to
CB, and CD
is common,
the two sides AC,
CD are equal to the
two sides BC,
CD
respectively;
and the angle ACD is equal
to the angle BCD;
therefore the base AD is equal
to the base BD;
[I. 4]
Therefore the given finite straight line AB has been bisected at D. Q.E.F.
Proposition 11
To draw a straight line at right angles to a given straight line from a given point on it.Let AB be the given straight line, and C the given point on it.
Thus it is required to draw from the point C a straight line at right angles to the straight line AB.
Let a point D be taken at random
on AC;
let CE be made equal
to CD;
[I. 3]
on DE let the equilateral
triangle FDE be
constructed,
and let FC be joined;
I say that the straight line FC
has been drawn at right angles to the given straight
line AB from
C the given point on it.
For, since DC is equal to
CE,
and CF is common,
the two sides DC,
CF are equal to the two sides
EC, CF
respectively;
and the base DF is equal
to the base FE;
therefore the angle DCF
is equal to the angle
ECF;
[I. 8]
and they are adjacent angles.
But, when a straight line set up on a straight line makes the adjacent
angles equal to one another, each of the equal angles
is right;
[Def. 10]
therefore each of the angles DCF,
FCE is right.
Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it. Q.E.F.
Proposition 12
To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.
Let AB be the given
infinite straight line, and C
the given point which is not on it;
thus it is required to draw to the given infinite
straight line AB, from
the given point C which is
not on it, a perpendicular straight line.
For let a point D be taken
at random on the other side of the straight
line AB, and with
centre C
and distance CD let the
circle EFG be described;
[Post. 3]
let the straight line EG be bisected
at H,
and let the straight lines CG,
CH, CE
be joined.
[Post. 1]
I say that CH has been drawn
perpendicular to the given infinite
straight line AB from the
given point C which is not on it.
For, since GH is equal
to HE,
and HC is common,
the two sides GH,
HC are equal to the two
sides EH, HC
respectively;
and the base CG is equal to
the base CE;
therefore the angle CHG
is equal to the angle EHC.
[I. 8]
And they are adjacent angles.
But when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. [Def. 10]
Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. Q.E.F.
Proposition 13
If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.
For let any straight line AB set up
on the straight line CD
make the angles CBA,
ABD;
I say that the angles CBA,
ABD are either two right angles or
equal to two right angles.
Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]
But, if not, let BE be drawn
from the point B at right angles to
CD;
[I. 11]
therefore the angles CBE,
EBD are two right angles.
Then, since the angle CBE is equal
to the two angles CBA,
ABE,
let the angle EBD be added
to each;
therefore the angles CBE,
EBD are equal to the three
angles CBA,
ABE, EBD.
[C.N. 2]
Again, since the angle DBA
is equal to the two angles
DBE,
EBA,
let the angle ABC be added
to each;
therefore the angles DBA,
ABC are equal to the three
angles DBE,
EBA, ABC.
[C.N. 2]
But the angles CBE,
EBD were also proved equal to the
same three angles;
and things which are equal to the same thing are also equal
to one another;
[C.N. 1]
therefore the angles CBE,
EBD are also equal to the
angles DBA,
ABC.
But the angles CBE,
EBD are two
right angles;
therefore the angles DBA,
ABC are also equal to two right angles.
Therefore etc. Q.E.D.
Proposition 14
If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.
For with any straight line AB,
and at the point B
on it, let the two straight lines
BC, BD
not lying on the same side make the adjacent angles
ABC, ABD
equal to two right angles;
I say that BD is in a
straight line with CB.
For, if BD is not in a straight line with BC let BE be in a straight line with CB.
Then, since the straight line AB
stands on the straight line
CBE,
the angles ABC,
ABE are equal to two right angles;
[I. 13]
But the angles ABC,
ABD are also equal to two
right angles;
therefore the angles CBA,
ABE are equal to the angles
CBA,
ABD.
[Post. 4 and C.N. 1]
Let the angle CBA be subtracted
from each;
therefore the remaining angle ABE
is equal to the remaining angle
ABD,
the less to the greater: which is impossible.
Therefore BE is not in a straight
line with CB.
Similarly we can prove that neither is any other straight line
except BD.
Therefore CB is in a straight line
with BD.
Therefore etc. Q.E.D.
Proposition 15
If two straight lines cut one another, they make the vertical angles equal to one another.
For let the straight lines AB,
CD cut one another
at the point E;
I say that the angle AEC is equal
to the angle DEB,
and the angle CEB
to the angle AED.
For, since the straight line AE stands
on the straight line CD,
making the angles CEA,
AED,
the angles CEA,
AED are equal to
two right angles.
[I. 13]
Again, since the straight line DE
stands on the straight line
AB, making the
angles AED,
DEB,
the angles AED,
DEB are equal to
two right angles
[I. 13]
But the angles CEA,
AED were also proved equal
to two right angles;
therefore the angles CEA,
AED are equal to the
angles AED,
DEB.
[Post. 4 and C.N. 1]
Let the angle AED be subtracted
from each;
therefore the remaining angle CEA
is equal to the remaining
angle BED.
[C.N. 3]
Similarly it can be proved that the angles CEB, DEA are also equal.
Therefore, etc. Q.E.D.
[PORISM. From this it is manifest that, if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.]
Proposition 16
In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.
Let ABC be a triangle, and let
one side of it BC be
produced to D;
I say that the exterior angle ACD
is greater than either of the interior and
opposite angles CBA,
BAC.
Let AC be bisected
at E
[I. 10]
,
and let BE be joined
and produced in a straight line
to F;
let EF be made equal
to BE
[I. 3]
,
let FC be joined
[Post. 1]
, and let
AC be drawn through
to G.
[Post. 2]
Then, since AE is equal
to EC, and
BE to
EF,
the two sides AE,
EB are equal to the two sides
CE, EF
respectively;
and the angle AEB is equal to
the angle FEC,
for they are vertical angles.
[I. 15]
Therefore the base AB is equal
to the base FC,
and the triangle ABE is equal
to the triangle CFE,
and the remaining angles are equal to the remaining angles
respectively, namely those which the equal sides subtend;
[I. 4]
therefore the angle BAE is equal
to the angle ECF.
But the angle ECD is greater
than the angle ECF;
[C.N. 5]
therefore the angle ACD is greater than
the angle BAE.
Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15] , can be proved greater than the angle ABC as well.
Therefore etc. Q.E.D.
Proposition 17
In any triangle two angles taken together in any manner are less than two right angles.
Let ABC be a triangle;
I say that two angles of the
triangle ABC taken
together in any manner are less than two right angles.
For let BC be produced to D.
Then, since the angle ACD is an
exterior angle of the
triangle ABC,
it is greater than the interior and opposite angle
ABC.
[I. 16]
Let the angle ACB be added
to each;
therefore the angles ACD,
ACB are greater than the angles
ABC, BCA.
But the angles ACD,
ACB are equal to two
right angles.
[I. 13]
Therefore the angles ABC,
span class="geofig">BCA are less than two right angles.
Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well.
Therefore etc. Q.E.D.
Proposition 18
In any triangle the greater side subtends the greater angle.
For let ABC be a triangle having
the side AC greater than
AB;
I say that the angle ABC
is also greater than the angle BCA.
For, since AC is greater than AB, let AD be made equal to AB [I. 3] , and let BD be joined.
Then, since the angle ADB
is an exterior angle of the triangle
BCD,
it is greater than the interior and opposite angle
DCB.
[I. 16]
But the angle ADB is equal to the
angle ABD,
since the side AB
is equal to AD;
therefore the angle ABD
is also greater than the
angle ACB;
therefore the angle ABC
is much greater than the angle ACB.
Therefore etc. Q.E.D.
Proposition 19
In any triangle the greater angle is subtended by the greater side.
Let ABC be a triangle having
the angle ABC greater
than the angle BCA;
I say that the side AC
is also greater than the side
AB.
For, if not, AC is either equal to AB or less.
Now AC is not equal
to AB;
for then the angle ABC
would also have been equal
to the angle ACB;
[I. 5]
but it is not;
therefore AC
is not equal to AB.
Neither is AC less than
AB,
for then the angle ABC
would also have been less
than the angle ACB;
[I. 18]
but it is not;
therefore AC
is not less than AB.
And it was proved that it is not equal either.
Therefore AC is
greater than AB.
Therefore etc. Q.E.D.
Proposition 20
In any triangle two sides taken together in any manner are greater than the remaining one.
For let ABC be
a triangle;
I say that in the triangle ABC two sides
taken together in any manner are greater than the remaining
one, namely
BA, AC greater than BC,
AB, BC greater than AC,
BC, CA greater than AB.
For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined.
Then, since DA is
equal to AC,
the angle ADC is also equal to the
angle ACD;
[I. 5]
therefore the angle BCD is greater than the
angle ADC.
[C.N. 5]
And, since DCB is a triangle
having the angle BCD greater
than the angle BDC,
and the greater angle is subtended by the greater side,
[I. 19]
therefore DB is greater
than BC.
But DA is equal
to AC;
therefore BA,
AC are greater
than BC.
Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.
Therefore etc. Q.E.D.
Proposition 21
If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.
On BC, one of the sides of
the triangle ABC, from
its extremities, B,
C, let the two straight lines
BD, DC
be constructed meeting within the triangle;
I say that BD,
DC are less than the remaining two
sides of the triangle BA,
AC, but contain an
angle BDC greater than the
angle BAC.
For let BD be drawn through to E.
Then, since in any triangle two sides are greater than
the remaining one,
[I. 20]
therefore, in the triangle ABE,
the two sides
AB, AE
are greater than BE.
Let EC be added to each,
therefore BA,
AC are greater than
BE, EC.
Again, since, in the
triangle CED,
the two sides CE,
ED are greater
than CD,
let DB be added to each;
therefore CE,
EB are greater
than CD, DB.
But BA, AC,
were proved greater than
BE,
EC;
therefore BA,
AC are
much greater than BD,
DC.
Again, since in any triangle the exterior angle is
greater than the interior and opposite angle,
[I. 16]
therefore in the
triangle CDE,
the exterior angle BDC is
greater than the angle CED.
For the same reason, moreover, in the
triangle ABE also,
the exterior angle CEB is greater
than the angle BAC.
But the angle BDC was proved greater than
the angle CEB;
therefore the angle BDC is
much greater than the angle BAC.
Therefore, etc. Q.E.D.
Proposition 22
Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.
Let the three given straight lines
be A, B,
C,
and of these let two taken together in any manner be
greater than the remaining one,
namely
A, B greater than C,thus it is required to construct a triangle out of straight lines equal to A, B, C.
A, C greater than B,
and
B, C greater than A;
Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C.
With centre F and
distance FD let the
circle DKL be described;
again, with centre G and
distance GH let the
circle KLH be
described;
and let KF,
FG be joined;
I say that the triangle KFG has
been constructed out of three straight lines equal to
A, B,
C.
For, since the point F
is the centre of the circle
DKL,
FD is equal to
FK.
But FD is equal to
A;
therefore KF is also
equal to A.
Again, since the point G
is the centre fo the circle LKH,
GH is equal
to GK.
But GH is
equal to C;
therefore GK is
equal to C.
And FG is also
equal to B;
therefore the three straight lines
KF, FG,
GK are equal to
the three straight lines A,
B, C.
Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constucted. Q.E.F.
Proposition 23
On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.
Let AB be the given straight
line, A the point on
it, and the angle DCE the given
rectilineal angle;
thus it is required to construct on the given straight
line AB, and at the
point A on it, a rectilineal
angle equal to the given rectilineal
angle DCE.
On the straight lines CD,
CE respectively let the points
D, E
be taken at random;
let DE be joined,
and out of three straight lines which are equal to the three
straight lines CD,
DE, CE
let the triangle AFG be
constructed in such a way that CD
is equal to AF,
CE to AG,
and further DE
to FG.
Then, since the two sides DC,
CE are equal to the two
sides FA,
AG respectively,
and the base DE is equal to the
base FG,
the angle DCE is equal to the
angle FAG.
[I. 8]
Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE. Q.E.F.
Proposition 24
If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.
Let ABC, DEF
be two triangles having the two sides
AB, AC
equal to the two sides DE,
DF
respectively, namely AB
to DE, and
AC to DF,
and let the angle at A be greater than
the angle at D;
I say that the base BC is also
greater than the base EF.
For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC [I. 23] ; let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined.
Then, since AB is equal to
DE, and AC
to DG, the two
sides BA, AC
are equal to the two sides
ED, DG
respectively;
and the angle BAC
is equal to the
angle EDG;
therefore the base BC is equal
to the
base EG.
[I. 4]
Again, since DF is
equal to DG,
the angle DGF is also equal to
the angle DFG;
[I. 5]
therefore the angle DFG
is greater than the angle EGF.
Therefore the angle EFG is much greater than the angle EGF.
And, since EFG is a triangle
having the angle EFG greater
than the angle EGF,
and the greater angle is subtended by the greater side,
[I. 19]
the side EG is also greater
than EF.
But EG is equal to
BC.
Therefore BC is also
greater than EF.
Therefore etc. Q.E.D.
Proposition 25
If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.
Let ABC, DEF
be two triangles having the two sides
AB, AC
equal to the two sides DE,
DF respectively, namely
AB to DE
and AC to DF;
and let the base BC be
greater than the
base EF;
I say that the angle BAC is also
greater than the angle EDF.
For, if not, it is either equal to it or less.
Now the angle BAC is not equal to
the angle EDF;
for then the base BC would also have been
equal to the base EF,
[I. 4]
but it is not;
therefore the angle BAC is not equal to
the angle DEF.
Neither again is the angle BAC
less than the angle EDF;
for then the base BC would also have been
less than the base EF,
[I. 24]
but it is not;
therefore the angle BAC is not
less than the angle EDF.
But it was proved that it is not equal either;
therefore the angle BAC
is greater than the angle EDF.
Therefore, etc. Q.E.D.
Proposition 26
If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.
Let ABC,
DEF be two triangles
having the two angles ABC,
BCA equal to the two
angles DEF,
EFD respectively, namely
the angle ABC to the
angle DEF, and the angle
BCA to the angle
EFD; and let them also have
one side equal to one side, first that adjoining the
equal angles, namely BC to
EF;
I say that they will also have the remaining
sides equal to the remaining sides respectively,
namely AB to
DE and
AC to
DF, and the remaining
angle to the remaining angle, namely the angle
BAC to the angle
EDF.
For if AB is unequal to DE, one of them is greater.
Let AB be greater, and let
BG be made equal to
DE;
and let GC be joined.
Then, since BG
is equal to DE,
and BC to
EF,
the two sides GB,
BC are equal to the two sides
DE, EF
respectively;
and the angle GBC is equal to
the angle DEF;
therefore the base GC is equal
to the base DF,
and the triangle GBC is equal
to the triangle DEF,
and the remaining angles will be equal to the
remaining angles, namely those which the equal sides
subtend;
[I. 4]
therefore the angle GCB is
equal to the angle DFE.
But the angle DFE
is by hypothesis equal to the angle
BCA;
therefore the angle BCG is
equal to the angle BCA,
the less to the greater: which is impossible.
Therefore AB is not unequal
to DE,
and is therefore equal to it.
But BC is also equal to
EF;
therefore the two sides AB,
BC are equal to the
two sides DE,
EF respectively,
and the angle ABC is equal to
the angle DEF;
therefore the base AC is equal
to the base DF,
and the remaining angle BAC
is equal to the remaining angle
EDF.
[I. 4]
Again, let sides subtending equal angles be
equal, as AB to
DE;
I say again that the remaining sides will
be equal to the remaining sides, namely
AC to DF
and BC to
EF, and further the remaining angle
BAC is equal to the remaining
angle EDF.
For if BC is unequal to EF, one of them is greater.
Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.
Then, since BH
is equal to EF,
and AB to
DE,
the two sides AB,
BH are equal to the two sides
DE, EF
respectively, and they contain equal angles;
therefore the base AH is equal
to the base DF,
and the triangle ABH is equal
to the triangle DEF,
and the remaining angles will be equal to the
remaining angles, namely those which equal sides
subtend;
[I. 4]
therefore the angle BHA is
equal to the angle EFD.
But the angle EFD is equal to
the angle BCA;
therefore, in the triangle AHC,
the exterior angle BHA
is equal to the interior and opposite angle
BCA:
which is impossible.
[I. 16]
Therefore BC is not unequal
to EF,
and is therefore equal to it.
But AB is also equal to
DE;
therefore the two sides AB,
BC are equal to the
two sides DE,
EF respectively, and they
contain equal angles;
therefore the base AC is equal
to the base DF,
the triangle ABC equal to the
triangle DEF,
and the remaining angle BAC
equal to the remaining angle
EDF.
[I. 4]
Therefore etc. Q.E.D.
Proposition 27
If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.
For let the straight line EF
falling on the two straight lines AB,
CD make the alternate angles
AEF, EFD
equal to one another;
I say that AB is parallel to
CD.
For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.
Let them be produced and meet, in the direction of B, D, at G.
Then, in the triangle GEF,
the exterior angle AEF is equal
to the interior and opposite
angle EFG:
which is impossible.
[I. 16]
Therefore
AB, CD
when produced will not meet in the direction
of B, D.
Similarly it can be proved that neither will they meet towards A, C.
But straight lines which do not meet in either direction are
parallel;
[Def. 23]
therefore AB is
parallel to CD.
Therefore etc. Q.E.D.
Proposition 28
If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.
For let the straight line EF
falling on the two straight lines
AB, CD
make the exterior angle EGB
equal to the interior and opposite
angle GHD, or the
interior angles on the same side, namely
BGH, GHD,
equal to two right angles;
I say that AB is parallel
to CD.
For, since the angle EGB
is equal to the
angle GHD,
while the angle EGB
is equal to the angle AGH,
[I. 15]
the angle AGH is also equal to
the angle GHD;
and they are alternate;
therefore AB is parallel to
CD.
[I. 27]
Again, since the angles
BGH, GHD
are equal to two right angles, and the angles
AGH, BGH
are also equal to two right angles,
[I. 13]
the angles AGH,
BGH are equal to the angles
BGH, GHD.
Let the angle BGH be subtracted from
each;
therefore the remaining angle AGH
is equal to the remaining
angle GHD;
and they are alternate;
therefore AB is parallel to
CD.
[I. 27]
Therefore etc. Q.E.D.
Proposition 29
A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.
For let the straight line EF
fall on the parallel straight lines
AB,
CD;
I say that it makes the alternate angles
AGH, GHD
equal, the exterior angle EGB
equal to the interior and opposite
angle GHD, and the interior angles on
the same side, namely BGH,
GHD, equal to two right angles.
For, if the angle AGH is unequal to the angle GHD, one of them is greater.
Let the angle AGH be greater.
Let the angle BGH be added
to each;
therefore the angles AGH,
BGH are greater than angles
BGH, GHD.
But the angles AGH,
BGH are equal to two right angles;
[I. 13]
therefore the angles
BGH, GHD
are less than two right angles.
But straight lines produced indefinitely from angles less than
two right angles meet
[Post. 5]
;
therefore
AB, CD,
if produced indefinitely, will meet;
but they do not meet, because they are by hypothesis parallel.
Therefore the angle AGH is not
unequal to the
angle GHD,
and is therefore equal to it.
Again, the angle AGH is equal to
the angle EGB;
[I. 15]
therefore the angle EGB is also
equal to the
angle GHD.
[C.N. 1]
Let the angle BGH be added
to each;
therefore the angles
EGB, BGH
are equal to the angles
BGH, GHD.
[C.N. 1]
But the angles
EGB, BGH
are equal to two right angles
[I. 13]
therefore the angles BGH,
GHD are also equal to two right angles.
Therefore etc. Q.E.D.
Proposition 30
Straight lines parallel to the same straight line are also parallel to one another.
Let each of the straight lines AB,
CD be parallel
to EF;
I say that AB is also
parallel to CD.
For let the straight line GK fall upon them.
Then, since the straight line GK
has fallen on the parallel straight lines
AB,
EF,
the angle AGK is equal to the
angle GHF.
[I. 29]
Again, since the straight line GK
has fallen on the parallel straight lines
EF,
CD,
the angle GHF is equal to the
angle GKD.
[I. 29]
But the angle AGK was also proved
equal to the angle
GHF;
therefore the angle AGK is also
equal to the angle
GKD;
[C.N. 1]
and they are alternate.
Therefore AB is parallel to CD. Q.E.D.
Proposition 31
Through a given point to draw a straight line parallel to a given straight line.
Let A be the given point, and
BC the given straight
line;
thus it is required to draw through the
point A a
straight line parallel to the straight line
BC.
Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23] ; and let the straight line AF be produced in a straight line with EA.
Then, since the straight line AD
falling on the two straight lines
BC, EF
has made the alternate angles
EAD, ADC
equal to one another,
therefore EAF is parallel
to BC.
[I. 27]
Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC. Q.E.F.
Proposition 32
In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.
Let ABC be a triangle,
and let one side of it BC be
produced to D;
I say that the exterior angle ACD
is equal to the two interior and opposite angles
CAB, ABC,
and the three interior angles of the triangle
ABC, BCA,
CAB are equal to two right angles.
For let CE be drawn through the point C parallel to the straight line AB. [I. 31]
Then since AB is parallel
to CE,
and AC has fallen
upon them,
the alternate angles BAC,
ACE are
equal to one another.
[I. 29]
Again, since AB is
parallel to CE,
and the straight line BD has fallen
upon them,
the exterior angle ECD is equal to
the interior and opposite
angle ABC.
[I. 29]
But the angle ACE was also proved
equal to the angle BAC;
therefore the whole angle ACD
is equal to the two interior and opposite angles
BAC, ABC.
Let the angle ACB be added
to each;
therefore the angles ACD,
ACB are equal to the three
angles ABC,
BCA, CAB.
But the angles ACD, ACB are equal to two right angles; [I. 13] therefore the angles ABC, BCA, CAB are also equal to two right angles.
Therefore etc. Q.E.D.
Proposition 33
The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.
Let AB, CD
be equal and parallel, and let the straight
lines AC, BD
join them (at the extremities which are) in the same directions
(respectively);
I say that
AC, BD
are also equal and parallel.
Let BC be joined.
Then, since AB is parallel to
CD, and
BC has fallen
upon them,
the alternate angles
ABC, BCD
are equal to one another.
[I. 29]
And, since AB is equal to
CD,
and BC is common,
the two sides
AB, BC
are equal to the two sides
DC,
CB;
and the angle ABC is equal to
the angle BCD;
therefore the base AC is equal to
the base BD,
and the triangle ABC is equal to
the triangle DCB,
and the remaining angles will be equal to the remaining angles
respectively, namely those which the equal sides subtend;
[I. 4]
therefore the angle ACB
is equal to the angle CBD.
And, since the straight line BC
falling on the two straight lines
AC, BD
has made the alternate angles equal to one another,
AC is parallel to
BD.
[I. 27]
And it was proved equal to it.
Therefore etc. Q.E.D.
Proposition 34
In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.
Let ACDB be a parallelogrammic area,
and BC its diameter;
I say that the opposite sides and angles of the
parallelogram ACDB are equal to
one another, and the
diameter BC bisects it.
For, since AB is
parallel to CD,
and the straight line BC
has fallen upon them,
the alternate angles
ABC, BCD
are equal to one another.
[I. 29]
Again, since AC is parallel to
BD,
and BC has fallen
upon them,
the alternate angles
ACB, CBD
are equal to one another.
[I. 29]
Therefore
ABC, DCB
are two triangles having the two angles
ABC, BCA
equal to the two angles DCB,
CBD respectively,
and one side equal to one side, namely that adjoining the equal
angles and common to both of them,
BC;
therefore they will also have the remaining sides equal to the
remaining sides respectively, and the remaining angle to the
remaining angle;
[I. 26]
therefore the side AB
is equal to CD,
and AC
to BD,
and further the angle BAC
is equal to the angle CDB.
And, since the angle ABC
is equal to the
angle BCD,
and the angle CBD to the
angle ACB,
the whole angle ABD is equal to the whole
angle ACD.
[C.N. 2]
And the angle BAC
was also proved equal to the angle
CDB.
Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.
I say, next, that the diameter also bisects the areas.
For since AB is equal to
CD,
and BC is common,
the two sides AB,
BC are equal to the two sides
DC, CB
respectively;
and the angle ABC is equal to
the angle BCD;
therefore the base AC is also
equal to DB,
and the triangle ABC is equal to the
triangle DCB.
[I. 4]
Therefore the diameter BC bisects the parallelogram ACDB. Q.E.D.
Proposition 35
Parallelograms which are one the same base and in the same parallels are equal to one another.
Let ABCD, EBCF
be parallelograms on the same base
BC and in the same parallels
AF,
BC;
I say that ABCD is equal to the
parallelogram EBCF.
For, since ABCD is a
parallelogram,
AD is equal to
BC.
[I. 34]
For the same reason also
EF is
equal to BC,
so that AD is also equal to
EF;
[C.N. 1]
and DE is common;
therefore the whole AE is equal to
the whole
DF.
[C.N. 2]
But AB is also equal to
DC;
[I. 34]
therefore the two sides EA,
AB are equal to the two sides
FD, DC
respectively,
and the angle FDC is equal to
the angle EAB, the exterior
to the interior;
[I. 29]
therefore the base EB is equal to
the base FC,
and the triangle EAB will be equal to the
triangle FDC.
[I. 4]
Let DGE be subtracted
from each;
therefore the trapazium ABGD
which remains is equal to the trapezium
EGCF
which remains.
[C.N. 3]
Let the triangle GBC
be added to each;
therefore the thole parallelogram ABCD
is equal to the whole
parallelogram EBCF.
[C.N. 2]
Therefore, etc. Q.E.D.
Proposition 36
Parallelograms which are on equal bases and in the same parallels are equal to one another.
Let ABCD, EFGH
be parallelograms which are on equal bases
BC, FG
and in the same parallels AH,
BG;
I say that the parallelogram ABCD
is equal to EFGH.
For let BE, CH be joined.
Then, since BC is equal to
FG
while FG is equal to
EH,
BC is also equal to
EH.
[C.N. 1]
But they are also parallel.
And EB, HC
join them;
but straight lines joining equal and parallel straight lines
(at the extremities which are) in the same directions
(respectively) are equal and parallel.
[I. 33]
Therefore EBCH is a parallelogram.
[I. 34]
And it is equal to ABCD;
for it has the same base BC with it, and
is in the same parallels
BC, AH
with it.
[I. 35]
For the same reason also EFGH
is equal to the same;
EBCH
[I. 35]
so that the parallelogram ABCD
is also equal to EFGH.
[C.N. 1]
Therefore etc. Q.E.D.
Proposition 37
Triangles which are on the same base and in the same parallels are equal to one another.
Let ABC, DBC
be triangles on the same base BC
and in the same parallels AD,
BC;
I say that the triangle ABC
is equal to the triangle DBC.
Let AD be produced in both directions
to E,
F;
through B let
BE be drawn parallel to
CA,
[I. 31]
and through C
let CF be drawn parallel to
BD.
[I. 31]
Then each of the figures EBCA,
DBCF is a parallelogram;
and they are equal,
for they are on the same base BC
and in the same parallels BC,
EF.
[I. 35]
Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34]
And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34]
[But halves of equal things are equal to one another.]
Therefore the triangle ABC is equal to the triangle DBC.
Therefore, etc. Q.E.D.
Proposition 38
Triangles which are on equal bases and in the same parallels are equal to one another.
Let ABC, DEF
be triangles on equal bases
BC, EF
and in the same parallels
BF,
AD;
I say that the triangle ABC
is equal to the triangle DEF.
For let AD be produced in
both directions to
G,
H;
through B let
BG be drawn parallel to
CA,
[I. 31]
and through
F let FH
be drawn parallel to DE.
Then each of the figures
GBCA, DEFH
is a parallelogram;
and GBCA is equal to
DEFH;
for they are on equal bases
BC, EF
and in the same parallels
BF, GH.
[I. 36]
Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]
And the triangle ABC is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]
[But the halves of equal things are equal to one another.]
Therefore the triangle ABC is equal to the triangle DEF.
Therefore etc. Q.E.D.
Proposition 39
Equal triangles which are on the same base and on the same side are also in the same parallels.
Let ABC, DBC
be equal triangles which are on the same base
BC and on the same
side of it;
[I say that they are also in the same parallels.] And [For]
let AD be joined;
I say that AD is parallel to
BC.
For, if not, let AE be drawn through
the point A
parallel to the straight line BC,
[I. 31]
and let EC be joined.
Therefore the triangle ABC is equal
to the triangle EBC;
for it is on the same base BC with it
and in the same parallels.
[I. 37]
But ABC is equal to
DBC;
therefore DBC is also equal to
EBC,
[C.N. 1]
the greater to the less: which is impossible.
Therefore AE is not
parallel to BC.
Similarly we can prove that neither is any other straight line except
AD;
therefore AD is
parallel to BC.
Therefore etc. Q.E.D.
Proposition 40
Equal triangles which are on equal bases and on the same side are also in the same parallels.Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.
I say that they are also in the same parallels.
For let AD be joined;
I say that AD is
parallel to BE.
For, if not, let AF be drawn through A parallel to BE [I. 31] , and let FE be joined.
Therefore the triangle ABC is equal to
the triangle FCE;
for they are on equal bases
BC, CE
and in the same parallels
BE, AF.
But the triangle ABC is equal to
the triangle DCE;
therefore the triangle DCE
is also equal to the
triangle FCE,
[C.N. 1]
the greater to the less: which is impossible.
Therefore AF is not parallel to BE.
Similarly we can prove that neither is any other straight line
except AD;
therefore AD is
parallel to BE.
Therefore etc. Q.E.D.
Proposition 41
If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.
For let the parallelogram ABCD
have the same base
BC with the triangle
EBC, and let it be in the same parallels
BC,
AE;
I say that the parallelogram ABCD
is double of the triangle BEC.
For let AC be joined.
Then the triangle ABC is equal to
the triangle EBC;
for it is on the same base BC with it
and in the same parallels
BC, AE.
[I. 37]
But the parallelogram ABCD is double of
the triangle ABC;
for the diameter AC bisects it;
[I. 34]
so that the parallelogram ABCD is also
double of the triangle EBC.
Therefore, etc. Q.E.D.
Proposition 42
To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.
Let ABC be the given triangle,
and D the given
rectilineal angle;
thus it is required to construct in the rectilinear
angle D a parallelogram
equal to the triangle ABC.
Let BC be bisected
at E, and let
AE be joined;
on the straight line EC, and
at the point E on it, let
the angle CEF be constructed equal to the
angle D;
[I. 23]
through A let
AG be drawn
parallel to EC,
[I. 31]
and through C let
CG be drawn parallel to
EF.
Then FECG is a parallelogram.
And, since BE is equal to
EC,
the triangle ABE is also equal
to the triangle
AEC,
for they are on equal bases BE,
EC and in the same
parallels BC,
AG;
therefore the triangle ABC
is double of the triangle AEC.
But the parallelogram FECG
is also double of the triangle AEC,
for it has the same base with it and is in the same
parallels with it;
[I. 41]
therefore the parallelogram FECG
is equal to the triangle
ABC.
And it has the angle CEF
equal to the given angle D.
Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q.E.F.
Proposition 43
In any parallelogram the complements of the parallelograms about the diameter are equal to one another.
Let ABCD be a parallelogram,
and AC its diameter;
and about AC let
EH, FG
be parallelograms, and
BK, KD
the so-called complements;
I say that the complement BK
is equal to the complement KD.
For since ABCD is a parallelogram,
and AC its diameter,
the triangle ABC is equal to
the triangle ACD.
[I. 34]
Again, since EH is a parallelogram,
and AK is its diameter,
the triangle AEK is equal to
the triangle AHK.
For the same reason
the triangle KFC is also
equal to KGC.
Now, since the triangle AEK
is equal to the
triangle AHK,
and KFC to
KGC,
the triangle AEK together with
KGC is equal to the
triangle AHK together with
KFC.
[C.N. 2]
And the whole triangle ABC is also
equal to the whole
ADC;
therefore the complement BK
which remains is equal to the
complement KD which remains.
[C.N. 3]
Therefore etc. Q.E.D.
Proposition 44
To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.
Let AB be the given straight
line, C the given triangle and
D the given rectilineal
angle;
thus it is required to apply to the given straight line
AB, in an angle equal to the
angle D, a parallelogram equal
to the given triangle C.
Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42] ; let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF [I. 31]
Let HB be joined.
Then, since the straight line HF
falls upon the parallels AH,
EF,
the angles AHF,
HFE are equal to two right angles.
[I. 29]
Therefore the angles BHG,
GFE are less than two
right angles;
and straight lines produced indefinitely from angles less
than two right angles meet;
[Post. 5]
therefore
HB, FE,
when produced, will meet.
Let them be produced and meet at K;
through the point K let
KL be drawn parallel to either
EA or FH,
[I. 31]
and let HA, GB
be produced to the points
L, M.
Then HLKF is a
parallelogram,
HK is its diameter,
and AG, ME
are parallelograms, and
LB, BF
the so-called complements, about
HK;
therefore LB is equal to
BF.
[I. 43]
But BF is equal to the
triangle C;
therefore LB is also
equal to C.
[C.N. 1]
And, since the angle GBE
is equal to the angle ABM,
[I. 15]
while the angle GBE is equal to
D,
the angle ABM is also equal to
the angle D.
Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D. Q.E.F.
Proposition 45
To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.
Let ABCD be the given rectilineal
figure and E the given
rectilineal angle;
thus it is required to construct, in the given
angle E, a parallelogram
equal to the rectilineal figure ABCD.
Let DB be joined, and let the
parallelogram FH be
constructed equal to the triangle
ABD, in the angle
HKF which is equal
to E;
let the parallelogram GM
equal to the trangle DBC be
applied to the straight line GH,
in the angle GHM which
is equal to E.
Then, since the angle E is equal
to each of the angles
HKF,
GHM,
the angle HKF is also equal to
the angle GHM.
[C.N. 1]
Let the angle KHG be added
to each;
therefore the angles KFH,
KHG are equal to the angles
KHG, GHM.
But the angles FKH,
KHG are equal to two right angles;
[I. 29]
thereore the angles KHG,
GHM are also equal to two right angles.
Thus, with a straight line GH,
and at the point H on it,
two straight lines
KH, HM
not lying on the same side make the adjacent angles
equal to two right angles;
therefore HK is in a
straight line with HM
[I. 14]
And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29]
Let the angle HGL be added
to each;
therefore the angles
MHG, HGL
are equal to the angles
HGF, HGL.
[C.N. 2]
But the angles
MHG, HGL
are equal to two right angles;
[I. 29]
therefore the angles
HGF, HGL
are also equal to two right angles.
[C.N. 1]
Therefore FG is in a straight line with
GL
[I. 14]
And, since FK is equal and
parallel to HG,
[I. 34]
and HG to
ML also,
KF is also equal and parallel
to ML;
[C.N. 1; ]
[I. 30]
and the straight lines
KM, FL
join them (at their extremities); therefore
KM, FL
are also equal and parallel.
[I. 33]
Therefore KFLM is a parallelogram.
And, since the triangle ABD is equal
to the parallelogram
FH,
and DBC to
GM,
the whole rectilineal figure ABCD
is equal to the whole
parallelogram KFLM.
Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E. Q.E.F.
Proposition 46
On a given straight line to describe a square.
Let AB be the given
straight line;
thus it is required to describe a square on the
straight line AB.
Let AC be drawn at right angles
to the straight line AB
from the point A on it,
[I. 11]
and let AD be made
equal to AB;
through the point D let
DE be drawn parallel to
AB,
and through the point B let
BE be drawn parallel
to AD.
[I. 31]
Therefore ADEB is a
parallelogram;
therefore AB is equal to
DE, and AD
to BE
[I. 34]
.
But AB is equal to
AD;
therefore the four straight lines
BA, AD,
DE, EB are
equal to one another;
therefore the parallelogram
ADEB is equilateral.
I say next that it is also right-angled.
For, since the straight line AD
falls upon the parallels AB,
DE,
the angles BAD,
ADE are equal to two right angles.
[I. 29]
But the angle BAD
is right;
therefore the angle ADE
is also right.
And in parallelogrammic areas the opposite sides and angles
are equal to one another;
[I. 34]
therefore each of the opposite angles
ABE, BED
is also right.
Therefore ADEB is right-angled.
And it was also proved equilateral.
Therefore it is a square; and it is described on the straight line AB. Q.E.F.
Proposition 47
In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
Let ABC be a right-angled triangle
having the angle
BAC right;
I say that the square on BC
is equal to the squares on
BA, AC.
For let there be described on BC
the square BDEC,
and on BA,
AC the squares
GB, HC;
[I. 46]
through A let
AL be drawn parallel to either
BD or CE,
and let AD, FC
be joined.
Then, since each of the angles
BAC, BAG
is right, if follows that with a straight line
BA, and at the point
A on it, the two straight lines
AC, AG
not lying on the same side make the adjacent angles equal to two right
angles;
therefore CA is in a straight line with
AG.
[I. 14]
For the same reason
BA is also in a straight line
with AH.
And, since the angle DBC
is equal to the angle FBA: for
each is right:
let the angle ABC
be added to each;
therefore the whole angle DBA
is equal to the whole angle
FBC.
[C.N. 2]
And, since DB is equal to
BC, and
FB to
BA,
the two sides AB,
BD are equal to the two sides
FB, BC
respectively;
and the angle ABD is equal to
the angle FBC;
therefore the base AD is equal to
the base FC,
and the triangle ABD is equal to
the triangle FBC.
[I. 4]
Now the parallelogram BL
is double of the triangle ABD,
for they have the same base BD
and are in the same parallels
BD, AL.
[I. 41]
And the square GB is double of the
triangle FBC,
for they again have the same base FB
and are in the same parallels.
FB, GC
[I. 41]
[But the doubles of equals are equal to one another.]
Therefore the parallelogram BL
is also equal to the square
GB.
Similarly, if
AE, BK
be joined,
the parallelogram CL can also be
proved equal to the
square HC;
therefore the whole square BDEC
is equal to the two
squares GB, HC.
[C.N. 2]
And the square BDEC
is described on BC,
and the squares
GB, HC
on BA, AC.
Therefore the square on the side BC is equal to the squares on the sides BA, AC.
Therefore etc. Q.E.D.
Proposition 48
If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.
For in the triangle ABC let the square
on one side BC be equal to the
squares on the sides BA,
AC;
I say that the angle BAC is right.
For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined.
Since DA is equal to
AB,
the square on DA is also equal to
the square on AB.
Let the square on AD be
added to each;
therefore the squares on
DA, AC
are equal to the squares on
BA, AC.
But the square on DC
is equal to the squares on DA,
AC,
for the angle DAC is right;
[I. 47]
and the square on BC is equal to
the squares on BA,
AC, for this is
the hypothesis;
therefore the square on DC
is equal to the square on
BC,
so that the side DC
is also equal to BC.
And since DA is equal to
AB,
and AC is common,
the two sides
DA, AC
are equal to the two sides
BA,
AC;
and the base DC is equal to
the base BC;
therefore the angle DAC
is equal to the angle BAC.
[I. 8]
But the angle DAC
is right;
therefore the angle BAC is also right.
Therefore etc. Q.E.D.