=Definition 1=

  A _point_ is that which has no part.

        =Definition 2=

  A _line_ is breadthless length.

        =Definition 3=

  The extremities of a line are points.

        =Definition 4=

  A _straight line_ is a line which lies evenly with the
points on itself.

        =Definition 5=

  A _surface_ is that which has length and breadth only.

        =Definition 6=

  The extremities of a surface are lines.

        =Definition 7=

  A _plane surface_ is a surface which lies evenly with
the straight lines on itself.

        =Definition 8=

  A _plane angle_ is the inclination to one another of
two lines in a plane which meet one another and do not lie in
a straight line.

        =Definition 9=

  And when the lines containing the angle are straight,
the angle is called _rectilineal_.

        =Definition 10=

  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.

        =Definition 11=

  An _obtuse angle_ is an angle greater than a right
angle.

        =Definition 12=

  An _acute angle_ is an angle less than a right angle.

        =Definition 13=

  A _boundary_ is that which is an extremity of any-
thing.

        =Definition 14=

  A _figure_ is that which is contained by any boundary
or boundaries.

        =Definition 15=

  A _circle_ is a plane figure contained by one line such
that all the straight lines falling upon it from one point among
those lying within the figure are equal to one another;

        =Definition 16=

  And the point is called the _centre_ of the circle.

        =Definition 17=

  A _diameter_ of the circle is any straight line drawn
through the centre and terminated in both directions by the
circumference of the circle, and such a straight line also
bisects the circle.

        =Definition 18=

  A _semicircle_ is the figure contained by the diameter
and the circumference cut off by it.  And the centre of the
semicircle is the same as that of the circle.

        =Definition 19=

  _Rectilineal figures_ are those which are contained
by straight lines, _trilateral_ figures being those contained by
three, _quadrilateral_ those contained by four, and _multi-
lateral_ those contained by more than four straight lines.

        =Definition 20=

  Of trilateral figures, an _equilateral triangle_ is that
which has its three sides equal, an _isosceles triangle_ that
which has two of its sides alone equal, and a _scalene
triangle_ that which has its three sides unequal.

        =Definition 21=

  Further, of trilateral figures, a _right-angled tri-
angle_ is that which has a right angle, an _obtuse-angled
triangle_ that which has an obtuse angle, and an _acute- 
angled triangle_ that which has its three angles acute.

        =Definition 22=

  Of quadrilateral figures, a _square_ is that which is
both equilateral and right-angled; an _oblong_ that which is
right-angled but not equilateral; a _rhombus_ that which is
equilateral but not right-angled; and a _rhomboid_ that which
has its opposite sides and angles equal to one another but is
neither equilateral nor right-angled.  And let quadrilaterals
other than these be called _trapezia_.

        =Definition 23=

  _Parallel_ straight lines are straight lines which,
being in the same plane and being produced indefinitely in
both directions, do not meet one another in either direction.


  Let the following be postulated:

        =Postulate 1=

  To draw a straight line from any point to any point.

        =Postulate 2=

  To produce a finite straight line continuously in a
straight line.

        =Postulate 3=

  To describe a circle with any centre and distance.

        =Postulate 4=

  That all right angles are equal to one another.

        =Postulate 5=

  That, if a straight line falling on two straight lines
make the interior angles on the same side less than two right
angles, the two straight lines, if produced indefinitely, meet
on that side on which are the angles less than the two right
angles.


        =Common Notion 1=

  Things which are equal to the same thing are also
equal to one another.

        =Common Notion 2=

  If equals be added to equals, the wholes are equal.

        =Common Notion 3=

  If equals be subtracted from equals, the remainders
are equal.

        =Common Notion 4 [7]=

  Things which coincide with one another are equal to
one another.

        =Common Notion 5 [8]=

  The whole is greater than the part.


        =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 con-
struct 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 dis-
tance *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*.
          =q.~ e.~ f.=


        =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. [1.~ 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; [Post.~ 3] 
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*.
          =q.~ e.~ f.=


        =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 un-
equal 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*; [1.~ 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.
          =q.~ e.~ f.=


        =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 sides 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.
          =q.~ e.~ d.=


        =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; [1.~ 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*. [1.~ 4]
  And, since the whole *AF* is equal to the whole *AG*,
and in these *AB* is equal to *AC*,
the remainder *BF* is equal to the remainder *CG*.
  But *FC* was also proved equal to *GB*;
therefore the two sides *BF*, *FC* are equal to the two sides
*CG*, *GB* respectively;
and the angle *BFC* is equal to the angle *CGB*,
while the base *BC* is common to them;
therefore the triangle *BFC* is also equal to the triangle *CGB*,
and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides
subtend;
therefore the angle *FBC* is equal to the angle *GCB*,
and the angle *BCF* to the angle *CBG*.
  Accordingly, since the whole angle *ABG* was proved
equal to the angle *ACF*,
and in these the angle *CBG* is equal to the angle *BCF*,
the remaining angle *ABC* is equal to the remaining angle
*ACB*;
and they are at the base of the triangle *ABC*.
  But the angle *FBC* was also proved equal to the angle *GCB*;
and they are under the base.
  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* con-
structed 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*; [1.~ 5] 
therefore the angle *ADC* is greater than the angle *DCB*;
therefore the angle *CDB* is much greater than the angle
*DCB*.
  Again, since *CB* 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. [1.~ 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*; [1.~ 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*.
[1.~ 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 finite straight line *AB*.
  Let the equilateral triangle *ABC* be
constructed on it, [1.~ 1] 
and let the angle *ACB* be bisected by the
straight line *CD*; [1.~ 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*. [1.~ 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 ran-
dom on *AC*;
let *CE* be made equal to *CD*; [1.~ 3] 
on *DE* let the equilateral triangle
*FDE* be constructed, [1.~ 1] 
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*;
[1.~ 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 per-
pendicular 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*, [1.~ 10] 
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*.
[1.~ 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*; [1.~ 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.
[1.~ 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*, [C.N.~ 3] 
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.
[1.~ 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.
[1.~ 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*, [1.~ 10] 
and let *BE* be joined and produced
in a straight line to *F*;
let *EF* be made equal to *BE*, [1.~ 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. [1.~ 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; [1.~ 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*, [1.~ 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*. [Post.~ 2]
  Then, since the angle *ACD* is an exterior angle of the
triangle *ABC*,
it is greater than the interior and opposite angle *ABC*.
[1.~ 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.
[1.~ 13]
  Therefore the angles *ABC*, *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*, [1.~ 3]  and let *BD* bejoined.
  Then, since the angle *ADB*
is an exterior angle of the triangle
*BCD*,
it is greater than the interior
and opposite angle *DCB*. [1.~ 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*; [1.~ 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*; [1.~ 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*; [1.~ 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,
[1.~ 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 con-
structed 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, [1.~ 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, [1.~ 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._ [1.~ 20]

  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*,
*A*, *C* greater than *B*,
and *B*, *C* greater than *A*;
thus it is required to construct a triangle out of straight lines
equal to *A*, *B*, *C*.
  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*. [1.~ 3]
  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*, *KG* 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 of the circle *LKH*,
*GH* is equal to *GK*.
  But *GH* is equal to *C*;
therefore *KG* is also 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 constructed.
          =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 con-
structed in such a way that *CD* is equal to *AF*, *CE* to *AG*,
and further *DE* to *FG*. [1.~ 22]
  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*. [1.~ 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*; [1.~ 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*. [1.~ 4]
  Again, since *DF* is equal to *DG*,
the angle *DGF* is also equal to the angle *DFG*; [1.~ 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,
[1.~ 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*, [1.~ 4] 
but it is not;
therefore the angle *BAC* is not equal to the angle *EDF*.
  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*, [1.~ 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; [1.~ 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*. [1.~ 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 the equal sides subtend; [1.~ 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. [1.~ 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*. [1.~ 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 pro-
duced 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. [1.~ 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*, [1.~ 15] 
the angle *AGH* is also equal to the
angle *GHD*;
and they are alternate;
therefore *AB* is parallel to *CD*. [1.~ 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, [1.~ 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*. [1.~ 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 the angles *BGH*,
*GHD*.
  But the angles *AGH*, *BGH* are equal to two right angles;
[1.~ 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*; [1.~ 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.~ 2]
  But the angles *EGB*, *BGH* are equal to two right angles;
[1.~ 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*. [1.~ 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*. [1.~ 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*; [1.~ 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*. [1.~ 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*. [1.~ 31]
  Then, since *AB* is parallel to *CE*,
and *AC* has fallen upon them,
the alternate angles *BAC*, *ACE* are
equal to one another. [1.~ 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*. [1.~ 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;
[1.~ 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. [1.~ 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 griangle *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; [1.~ 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*. [1.~ 27]
  And it was also 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. [1.~ 29]
  Again, since *AC* is parallel to *BD*,
and *BC*has fallen upon them,
the alternate angles *ACB*, *CBD* are equal to one
another. [1.~ 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; [1.~ 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*. [1.~ 4]
  Therefore the diameter *BC* bisects the parallelogram
*ACDB*.
          =q.~ e.~ d.=


        =Proposition 35=

  _Parallelograms which are on 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*. [1.~ 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*; [1.~ 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; [1.~ 29] 
therefore the base *EB* is equal
to the base *FC*,
and the triangle *EAB* will be equal to the triangle *FDC*.
[1.~ 4]
  Let *DGE* be subtracted from each;
therefore the trapezium *ABGD* which remains is equal to
the trapezium *EGCF* which remains. [C.N.~ 3]
  Let the triangle *GBC* be added to each;
therefore the whole 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. [1.~ 33]
  Therefore *EBCH* is a parallelogram. [1.~ 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. [1.~ 35]
  For the same reason also *EFGH* is equal to the same
*EBCH*; [1.~ 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*, [1.~ 31] 
and through *C* let *CF* be drawn
parallel to *BD*. [1.~ 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*. [1.~ 35]
  Moreover the triangle *ABC* is half of the parallelogram
*EBCA*; for the diameter *AB* bisects it. [1.~ 34]
  And the triangle *DBC* is half of the parallelogram *DBCF*;
for the diameter *DC* bisects it. [1.~ 34]
  [But the 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*, [1.~ 31] 
and through *F* let *FH* be drawn parallel to *DE*.
  Then each of the figures *GBCA*, *DEFH* is a parallelo-
gram;
and *GBCA* is equal to *DEFH*;
for they are on equal bases *BC*, *EF* and in the same
parallels *BF*, *GH*. [1.~ 36]
  Moreover the triangle *ABC* is half of the parallelogram
*GBCA*; for the diameter *AB* bisects it. [1.~ 34]
  And the triangle *FED* is half of the parallelogram *DEFH*;
for the diameter *DF* bisects it. [1.~ 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*, [1.~ 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. [1.~ 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*, [1.~ 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*. [1.~ 38]
  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*.
[1.~ 37]
  But the parallelogram *ABCD* is double of the triangle
*ABC*;
for the diameter *AC* bisects it; [1.~ 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 recti-
lineal angle;
thus it is required to construct in the rectilineal 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*; [1.~ 23] 
through *A* let *AG* be drawn parallel to *EC*, and [1.~ 31]
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*; [1.~ 38] 
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; [1.~ 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*. [1.~ 34]
  Again, since *EH* is a parallelo-
gram, 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*; [1.~ 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*. [1.~ 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.
[1.~ 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*, [1.~ 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*. [1.~ 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*,
[1.~ 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*; [1.~ 42] 
let the parallelogram *GM* equal to the triangle *DBC* be
applied to the straight line *GH*, in the angle *GHM* which is
equal to *E*. [1.~ 44]
  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 *FKH*, *KHG* are equal to the angles
*KHG*, *GHM*.
  But the angles *FKH*, *KHG* are equal to two right angles;
[1.~ 29] 
therefore 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 *KH* is in a straight line with *HM*. [1.~ 14]
  And, since the straight line *HG* falls upon the parallels
*KM*, *FG*, the alternate angles *MHG*, *HGF* are equal to one
another. [1.~ 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;
[1.~ 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*. [1.~ 14]
  And, since *FK* is equal and parallel to *HG*, [1.~ 34] 
and *HG* to *ML* also,
*KF* is also equal and parallel to *ML*; [C.N.~ 1 and 1.~ 30] 
and the straight lines *KM*, *FL* join them (at their extremities);
therefore *KM*, *FL* are also equal and parallel. [1.~ 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, [1.~ 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*.
[1.~ 31]
  Therefore *ADEB* is a parallelogram;
therefore *AB* is equal to *DE*, and *AD* to *BE*. [1.~ 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.
[1.~ 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; [1.~ 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*; [1.~ 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,
it 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*. [1.~ 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*. [1.~ 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*. [1.~ 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*. [1.~ 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 *AC* 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; [1.~ 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*. [1.~ 8]
  But the angle *DAC* is right;
therefore the angle *BAC* is also right.
  Therefore etc.
          =q.~ e.~ d.=