Opposite Sides and Angles of Parallelogram are Equal

Theorem
The opposite sides and angles of a parallelogram are equal to one another, and either of its diameters bisects its area.

Proof

 * Euclid-I-34.png

Let $$ACDB$$ be a parallelogram, and let $$BC$$ be a diameter.

By definition, $$AB \| CD$$, and $$BC$$ intersects both.

So by Parallel Implies Equal Alternate Interior Angles, $$\angle ABC = \angle BCD$$.

Similarly, by definition, $$AC \| BD$$, and $$BC$$ intersects both.

So by Parallel Implies Equal Alternate Interior Angles, $$\angle ACB = \angle CBD$$.

So $$\triangle ABC$$ and $$\triangle DCB$$ have two angles equal, and the side $$BC$$ in common.

So by Triangle Angle-Side-Angle Equality, $$\triangle ABC = \triangle DCB$$.

So $$AC = BD$$ and $$AB = CD$$.

Also, we have that $$\angle BAC = \angle BDC$$.

Also, because $$\angle ACB = \angle CBD$$ and $$\angle ABC = \angle BCD$$, we have by Common Notion 2 that $$\angle ACB + \angle BCD = \angle ABC + \angle CBD$$.

So $$\angle ACD = \angle ABD$$.

So we have shown that opposite sides and angles are equal to each other.

Now note that as $$AB = CD$$, and $$BC$$ is common, and $$AC = BD$$, we have that $$\triangle ABC = \triangle BCD$$ by Triangle Side-Side-Side Equality.

So $$BC$$ bisects the parallelogram.