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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;
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.
Being (the rest of) what it was required to prove.
