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
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.
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.