(Edited by Sir Thomas L. Heath, 1908)

[Euclid, ed. Heath, 1908, on

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.

Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)

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