PROPOSITION 34.¶
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,
Again, since AC is parallel to BD, and BC has fallen upon them,
Therefore ABC, DCB[2] 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;
And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB,
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;
See Commentary on Proposition I.34.