# Math Help - Bisecting Diagonals of Quadrilateral means Parallelogram Proof

1. ## Bisecting Diagonals of Quadrilateral means Parallelogram Proof

I am trying to prove: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram, by using vectors.

This is what I have:
Let ABCD be a quadrilateral with diagonals AC, BD bisecting each other.
Then DC =1/2 (AC - BD),
and AB = 1/2 (DB - CA).

How do I equate these? Am I allowed to say -AB = BA = 1/2 (AC - BD)?

2. Hello,

How do I equate these? Am I allowed to say -AB = BA = 1/2 (AC - BD)?
Yes you can ^^
The line I'm going to write is not very formal, but it helps you memorize..
Inverting the two points of a vector will make its direction different, so it will be the opposit of the initial vector

$\vec{MN}=-\vec{NM}$

This goes for any points M and N

3. Hello, Ultros88!

If the diagonals of a quadrilateral bisect each other,
then the quadrilateral is a parallelogram. (vector proof)
Code:
            A                 B
* - - - - - - - - *
/  *           *  /
/     *     *     /
/        *        /
/     *   E *     /
/  *           *  /
* - - - - - - - - *
D                 C

We are given: . $\begin{array}{cccc}\overrightarrow{AE} &=& \overrightarrow{EC} & {\color{blue}[1]}\\ \overrightarrow{EB} &=& \overrightarrow{DE} & {\color{blue}[2]}\end{array}$

We see that: . $\begin{array}{cccc}\overrightarrow{AB} &=& \overrightarrow{AE} + \overrightarrow{EB} & {\color{blue}[3]}\\ \overrightarrow{DC} &=& \overrightarrow{DE} + \overrightarrow{EC} & {\color{blue}[4]} \end{array}$

Substitute [1] and [2] into [3]: . $\overrightarrow{AB} \:=\:\overrightarrow{EC} + \overrightarrow{DE}$

. . Hence: . $\overrightarrow{AB} \:=\:\overrightarrow{DC}$

Since $\overrightarrow{AB} \parallel \overrightarrow{DC}\text{ and }|\overrightarrow{AB}| = |\overrightarrow{DC}|$, then $ABCD$ is a parallelogram.

Theorem: If two sides of a quadrilateral are parallel and equal,
. . . . . . . the quadrilateral is a parallelogram.
.