Diameters of Parallelogram Bisect each other/Proof 1

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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 Congruence:

$\triangle ABC = \triangle ADC$
$\triangle BAD = \triangle BCD$

Thus:

$\angle ADE = \angle CBE$
$\angle DAE = \angle BCE$

We have $AD = BC$.

So from Triangle Angle-Side-Angle Congruence:

$\triangle ADE = \triangle CBE$

and so:

$DE = BE$
$AE = CE$

Hence the result.

$\blacksquare$


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