Diameters of Parallelogram Bisect each other/Proof 2

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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-Complex.png


Let $\Box ABCD$ be embedded in the complex plane such that $B$ is identified with the origin $0 + 0 i$.

Let $A$ be identified with the complex number $z_1$.

Let $C$ be identified with the complex number $z_2$.

By Geometrical Interpretation of Complex Subtraction:

$z_1 - z_2 = AC$

Then:

$AE = m \paren {z_1 - z_2}$

for some $m$ where $0 \le m \le 1$.

Similarly, by Geometrical Interpretation of Complex Addition:

$z_1 + z_2 = BD$

Then:

$BE = n \paren {z_1 + z_2}$

for some $n$ where $0 \le n \le 1$.

Then:

\(\ds BA + AE\) \(=\) \(\ds BE\)
\(\ds \leadsto \ \ \) \(\ds z_1 + m \paren {z_1 - z_2}\) \(=\) \(\ds n \paren {z_1 + z_2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {1 - m - n} z_1 + \paren {m - n} z_2\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds 1 - m - n\) \(=\) \(\ds 0\) Linear Combination of Non-Parallel Complex Numbers is Zero if Factors are Both Zero
\(\ds m - n\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds \frac 1 2\)
\(\ds n\) \(=\) \(\ds \frac 1 2\)

and so $E$ is at the midpoint of the diameters of $\Box ABCD$.

$\blacksquare$


Sources

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