Let ABCD and PQRS be squares of the same side length such that P is the center of ABCD (i.e., the intersection of its diagonals). Suppose that BC and PQ intersect each other at X. Suppose also that PS and DC intersect each other at Y (so that C is inside PQRS). Prove that the area of the quadrilateral PXCY is 1/4.
Can I let ABCD be a unit square such that the area would be = 1. Then since ABCD and PQRS are congruent, the area of PQRS is 1 also. Then PY is congruent in PX since they are radii of the same circle. So there exists a point, W, on AD such that W is between A and D, PW is congruent to PY and PX; and PW is perpendicular to AD. Then there is another point, Z, on DC such that PZ is congruent to PX, PY, and PW and is also perpendicular to DC? Then would the area of PXBY be 1/4? Or is my proof not adequate/true?
Your proof is incorrect as it currently stands, as such $W$ and $Z$ as you described does not exist.
However you are on the right tract by trying to consider perpendicular lines. Just draw two perpendicular lines, $PM$ perpendicular to $BC$ at $M$ and $PN$ perpendicular to $DC$ at $N$, you can easily show triangles $PMX$ and $PNY$ are congruent by $ASA$.