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 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; [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 also proved equal to it.
Therefore etc.
Q. E. D.
1
2
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;