Draw a quadrilateral given its four side perpendicular bisectors. I can't think of anything. I guess that for every two of them, we draw a circle with center the intersection point and same for all the pairs but I can't think of anything more specific.
 A: Given a quadrilateral with vertices A, B, C, D at generic and unknown positions.
The only information provided is the four perpendicular bisectors together with their association with the sides.

Let 


*

*$A', B', C', D'$ be the midpoints of sides $AB$, $BC$, $CD$ and $DA$.

*$E$ be the intersection of bisectors passing through $A'$ and $B'$.

*$F$ be the intersection of bisectors passing through $C'$ and $D'$.

*$G$ be the intersection of bisectors passing through $B'$ and $C'$.

*$H$ be the image of $G$ under reflection with respect to angular bisector of $\angle A'EF$.

*$I$ be the image of $G$ under reflection with respect to angular bisector of $\angle EFD'$.


Recall for any triangle, the three perpendicular bisectors intersect at the 
circumcenter. Apply this to triangle $ABC$ and $ACD$, we find line $EF$
is the perpendicular bisectors for $AC$. 
Playing around with the angles in triangle $ABC$ and $ACD$, one will discover $A$ is collinear with the pair of points $E, H$ and the pair of points $F, I$. This means we can construct $A$ as the intersection of lines $EH$ and $FI$. The vertex $C$ can be constructed in a similar manner. For the remaining two vertices $B$ and $D$, one can obtain them from $A$ by reflection with respect to the perpendicular bisectors passing through $A'$ and $D'$.
Please note that above procedure doesn't always work. If any pairs of bisectors are parallel, some members of $E, F, G$ will move to infinity. This will
explain the observation pointed out in comment: when $ABCD$ is a rectangle, there are infinitely many solutions.
