Lemma. Let $ABCD$ be any quadrilateral with $E, F, G, H$ the midpoints of sides $AB, BC, CD, DA$, respectively. Then $[ABCD] = 2[EFGH]$.
Proof. Let $P$ be the intersection of diagonals $AC$ and $BD$. $HE$ is the midline of triangle $ABD$ so $HE$ bisects segment $AP$. It follows that $2[HPE] = [AHPE]$. Similarly, we obtain $2[EPF] = [BEPF]$, $2[FPG] = [CFPG]$, and $2[GPH] = [DGPH]$. Summing these up yields the desired result. $\square$

Let $O$ denote the circumcenter of $ABCD$, and let $P, Q, R, S$ denote the midpoints of $HE, EF, FG, GH$ respectively. Applying the lemma to $ABCD$ and then $EFGH$, we get $[ABCD] = 2[EFGH] = 4[PQRS]$.

We have $WH\parallel OE$ since they are both perpendicular to $AB$, and similarly $WE\parallel OH$, so $WHOE$ is a parallelogram. It follows that $W$ is the image of the midpoint of $P$ (i.e. the midpoint of $HE$) under dilation from $O$ by a factor of $2$. Similarly $X, Y, Z$ are the images of $R, S, T$ under the same transformation, so $WXYZ$ is the image of $PQRS$ under a dilation of factor $2$. Thus, it follows that $[WXYZ] = 4[PQRS] = [ABCD]$ as desired.