# Bimedians and Diagonals in a Quadrilateral

Denote the midpoints of the sides $AB$, $BC$, $CD$, and $DA$ of quadrilateral $ABCD$ as $X$, $Y$, $Z$, $W$, respectively. Segments $XZ$ and $YW$ are known as the bimedians of the quadrilateral. We have an engaging statement

1. The bimedians of a quadrilateral are equal if and only if its diagonals are perpendicular.

2. The bimedians of a quadrilateral are perpendicular if and only if its diagonals are equal.

Proof

1. The bimedians of a quadrilateral are equal if and only if its diagonals are perpendicular.

2. The bimedians of a quadrilateral are perpendicular if and only if its diagonals are equal.

### Proof

The easiest way to establish the statement is by employing vector techniques. Choose an arbitrary origin $O$ and associate with every point $P$ the vector $p=\overline{OP}$, denoted by the corresponding low case.

Thus we have, $x=\frac{a+b}{2}$, $y=\frac{b+c}{2}$, $z=\frac{c+d}{2}$, $w=\frac{a+d}{2}$. The statement at hand can be reformulated as

1. $|x - z|=|y - w|$ iff $(a - c)\perp (b-d)$.

2. $(x - z)\perp (y - w)$ iff $|a-c|=|b-d|$.

The absolute value of vector $m$ is defined as the square root of the scalar product: $|m|=\sqrt{m\cdot m}$; while the condition of orthogonality of two vectors $m$ and $n$ is also expressible in terms of the scalar product: $m\perp n$ iff $m\cdot n=0$. Using the scalar product we are able to once more reformulate our statement

1. $(x - z)(x - z)=(y - w)(y - w)$ iff $(a - c)(b-d)=0$.

2. $(x - z)(y - w)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$.

Substituting $x$, $y$, $z$, $w$ from the definition reduces the above to

1. $(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$ iff $(a - c)(b-d)=0$.

2. $(a+b - c-d)(b+c-d-a)=0$ iff $(a-c)(a-c)=(b-d)(b-d)$.

All that remains is pure algebra. For example,

$(a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)$

without the squares that are certainly the same on both sides, appears as

$2(ab-ac-ad-bc-bd+cd)=2(bc-bd-ab-cd-ac+ad)$,

which is equivalent to $(a - c)(b-d)=0$.

Similarly, written explicitly, $(a+b - c-d)(b+c-d-a)=0$ gives

\begin{align}ab+ac & -ad-aa +bb+bc-bd-ab \\& -bc-cc+cd+ac -bd-cd+dd+ad \\& =2ac-aa+bb-2bd-cc+dd=0.\end{align}

which is equivalent to $(a-c)(a-c)=(b-d)(b-d)$.