|Book I Proposition 34|
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 BC, CB respectively; and the angle ABC is equal to the angle BCD;
the base AC is also equal to DB,
the triangle ABC is equal to the triangle DCB. [I.4]
the diameter BC bisects the parallelogram ACDB.
Being (the rest of) what it was required to prove.
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