Using vectors, prove that if the diagonals of a…

Using vectors, prove that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Step 1: Setup

Let $\vec a,\vec b,\vec c,\vec d$ be the position vectors of $A,B,C,D$. The midpoints of the diagonals are

\[ M_{AC}=\frac{\vec a+\vec c}{2},\qquad M_{BD}=\frac{\vec b+\vec d}{2}. \]

Step 2: The hypothesis

"The diagonals bisect each other" means $M_{AC}=M_{BD}$: $\vec a+\vec c=\vec b+\vec d \qquad (*)$.

Step 3: Rearranging $(*)$

\[ \vec c-\vec b=\vec d-\vec a\ \Longrightarrow\ \overrightarrow{BC}=\overrightarrow{AD}, \]

\[ \vec b-\vec a=\vec c-\vec d\ \Longrightarrow\ \overrightarrow{AB}=\overrightarrow{DC}. \]

Equal vectors are parallel and have equal lengths, so both pairs of opposite sides are parallel and equal: $ABCD$ is a parallelogram. $\blacksquare$

Step 4: Converse

Every step is reversible: \[\overrightarrow{AB}=\overrightarrow{DC}\iff \vec b-\vec a=\vec c-\vec d\iff \vec a+\vec c=\vec b+\vec d\iff M_{AC}=M_{BD}.\]

\[ABCD\text{ is a parallelogram}\iff\text{its diagonals bisect each other}\]

Step 5: Same argument from the midpoint

Taking the common midpoint $M$ as origin gives $\overrightarrow{MC}=-\overrightarrow{MA}$ and $\overrightarrow{MD}=-\overrightarrow{MB}$. Hence

\[ \overrightarrow{DC}=\overrightarrow{MC}-\overrightarrow{MD}=-\overrightarrow{MA}+\overrightarrow{MB}=\overrightarrow{AB}. \]

This is the vector form of the congruence $\triangle AMB\cong\triangle CMD$ (central symmetry about $M$).