Let polygon ACDB be a parallelogrammic area, and line BC its diameter.

I say that the opposite sides and angles of the parallelogram ACDB equal one another, and the diameter BC bisects it.

Since line AB is parallel to line CD, and the straight line BC falls upon them, therefore the alternate angles ABC and BCD equal one another.

Again, since line AC is parallel to line BD, and line BC falls upon them, therefore the alternate angles ACB and CBD equal one another.

Therefore triangle ABC and triangle DCB are two triangles having the two angles ABC and BCA equal to the two angles DCB and CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, line BC. Therefore they also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle. Therefore the side line AB equals line CD, and line AC equals line BD, and further the angle BAC equals the angle CDB.

Since the angle ABC equals the angle BCD, and the angle CBD equals the angle ACB, therefore the whole angle ABD equals the whole angle ACD.

And the angle BAC was also proved equal to the angle CDB.

Therefore in parallelogrammic areas the opposite sides and angles equal one another.

I say, next, that the diameter also bisects the areas.

Since line AB equals line CD, and line BC is common, the two sides line AB and line BC equal the two sides line CD and line BC respectively, and the angle ABC equals the angle BCD. Therefore the base AC also equals line BD, and the triangle ABC equals the triangle DCB.

Therefore the diameter BC bisects the parallelogram ACDB.

Therefore in parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas.

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