Prove With Vectors That a Parallelogram's Diagonals Bisect

Question

I designed a proof for a problem set but I'm unsure whether the proof is actually conclusive. The proof is as follows:

If a parallelogram's diagonals bisect each other, then...

$\frac{1}{2}(\lvert\vec{A}\rvert + \lvert\vec{B}\rvert) + \frac{1}{2}(\lvert\vec{A}\rvert - \lvert\vec{B}\rvert) - \lvert\vec{A}\rvert = \frac{1}{2}(\lvert\vec{A}\rvert + \lvert\vec{B}\rvert) + \frac{1}{2}(\lvert\vec{B}\rvert - \lvert\vec{A}\rvert) - \lvert\vec{B}\rvert = 0$

With a little algebra, it can be seen that the above equation is true. I'm wondering if there is any way for the equation to be true and not have the diagonals of quadrilateral ABCD (if proof is conclusive, then parallelogram ABCD) bisect.

Answer 1

HINT: If one vertex of the parallelogram is at the origin and $\vec A$ and $\vec B$ are adjacent edges, one diagonal is $\vec A + \vec B$. What is its midpoint? The other diagonal starts, say, at $\vec A$ and proceeds to $\vec B$. What's the vector from the origin to the midpoint of that diagonal?

Answer 2

Draw a parallegram, label the corners from lower left, counter clockwise:

A(lower left), B(lower right),C(Upper right),D(Upper left).

Let $AB$ be vector $\vec b$, $AD$ be vector $\vec a$.

Diagonal $AC$ : $\vec a + \vec b$.

Diagonal $BD$ : $\vec a - \vec b.$

Label the point of intersection of the diagonals $O$.

Consider $\triangle$ $ABO$:

$\alpha (\vec a + \vec b) = \vec b + \beta (\vec a - \vec b), \alpha, \beta \in \mathbb{R}$.

Collect coefficients of $\vec a$ and $\vec b$:

$(\alpha - \beta) \vec a + (\alpha + \beta - 1) b = 0$

Since $\vec a$ and $\vec b$ are independent (not collinear) it follows that the coefficients of $\vec a$ and $\vec b$ must vanish.

1) $\alpha - \beta = 0$, and

2) $\alpha + \beta - 1 = 0.$

Adding and subtracting:

$\alpha = 1/2$, $\beta = 1/2.$