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 1
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.
$\blacksquare$
Proof 2
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 $\Box ABCD$.
$\blacksquare$