Quadrilateral with Bisecting Diagonals is Parallelogram

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

• Diameters of Parallelogram Bisect each other

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: $66$