Quadrilateral with Bisecting Diagonals is Parallelogram
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Theorem
Let $ABCD$ be a quadrilateral.
Let the diagonals of $ABCD$ bisect each other.
Then $ABCD$ is a parallelogram.
Proof
The diagonals of $ABCD$ bisect each other if the position vectors of the midpoints of the diagonals are the same point.
Let $z_1, z_2, z_3, z_4$ be the position vectors of the vertices of $ABCD$.
Thus:
\(\ds z_1 + \frac {z_3 - z_1} 2\) | \(=\) | \(\ds z_2 + \frac {z_4 - z_2} 2\) | condition for bisection | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {z_1 + z_3} 2\) | \(=\) | \(\ds \frac {z_2 + z_4} 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z_1 + z_3\) | \(=\) | \(\ds z_2 + z_4\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z_1 - z_2 + z_3 - z_4\) | \(=\) | \(\ds 0\) |
The result follows from Condition for Points in Complex Plane to form Parallelogram.
$\blacksquare$
Also see
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $66$