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
BChas 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.
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