Book 1 – 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 parallelogram, and BC be its diameter.

We will first prove that AB is equal to CD, AC is equal to BD, angle BAC is equal to CDB, and angle ABD is equal to ACD.

Since AC is parallel to BD and the line BC falls upon them, the alternate angles ABC and BCD are equal to one another by Proposition 29. (If a straight line falls upon two parallel lines, then the alternate angles are equal to one another.)
By the same reasoning, the alternate angles ACB and CBD are equal to one another.
So triangles ABC and DCB have angle ABC equal to angle BCD, angle ACB equal to angle CBD, and share the common side BC between those angles. By Proposition 26 (angle-side-angle) the remaining sides are equal to the remaining sides and the remaining angle is equal to the remaining angle.
Side AB is equal to side CD, side AC is equal to side CD, and angle BAC is equal to angle CDB.
All that remains is to prove that angle ABD is equal to angle ACD.
Angle ABC is equal to angle BCD and angle CBD is equal to is equal to angle ACB.
Thus angle ABD (ABC & CBD) is equal to angle ACD (BCD & ACB).

Therefore the opposite sides and angles are equal to one another.

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Now we will prove that the diameter bisects the areas.

We will look to triangles ABC and BCD.
Side AB is equal to side CD, and side BC is common to the two triangles. The angles contained between the equal sides, ABC and BCD, are equal.
By Proposition 4 (side-angle-side), the triangle ABC is equal to the triangle DCB.

Therefore the diameter BC bisects the parallelogram ACDB.

Q.E.D.

The next proposition also involves parallelograms, comparing two parallelograms on the same base and in the same parallels.