Diameters of Parallelogram Bisect each other

Theorem
Let $\Box ABCD$ be a parallelogram with diameters $AC$ and $BD$.

Let $AC$ and $BD$ intersect at $E$.

Then $E$ is the midpoint of both $AC$ and $BD$.

Proof

 * DiametersOfParallelogramBisect.png

By definition of parallelogram:

By Opposite Sides and Angles of Parallelogram are Equal:
 * $AB = CD$
 * $AD = BC$
 * $\angle ABC = \angle ADC$
 * $\angle BAD = \angle BCD$

Therefore by Triangle Side-Angle-Side Equality:
 * $\triangle ABC = \triangle ADC$
 * $\triangle BAD = \triangle BCD$

Thus:
 * $\angle ADE = \angle CBE$
 * $\angle DAE = \angle BCE$

We have $AD = BC$.

So from Triangle Angle-Side-Angle Equality:
 * $\triangle ADE = \triangle CBE$

and so:
 * $DE = BE$
 * $AE = CE$

Hence the result.