Find vertex coordinates of a square given their distances $p,\>s,\>q,\>r$ to an inner point 
The figure represents a square, with P a point inside the square. The four segments are drawn from P to the four vertices of the square and they are named p, q, r, s. 
If the bottom-left vertex is at origin and the sides of the square are parallel to the axes (square is drawn in the first quadrant), find the co-ordinates of other three vertices if the measures of p, q, r, s are given.
I'm seeking your help.

Well, as it,s a square $C=(0,0); A=(0,w); E=(w,w); D=(w,0)$ where $w$ is the side of the square.
Law of cosines applied to $\triangle APE$ says that $q^2 = r^2 + w^2 - 2rw \cos \angle PAE$.
Applied to to $\triangle APC$ we have $p^2 = r^2+ w^2 - 2rw \cos (90 -\angle PAE)$.
By my reckoning   that should be enough to solve.  If not we can keep going
$r^2 = p^2 + w^2 -2pw \cos(\angle ACP)$.
etc.
If you are into that whole symmetry thing but not into a $w$ coefficient in quadratic equation jazz you can do the interior angles.
$w^2 = r^2 + q^2 - 2rq\cos \theta$
$= r^2 + p^2 - 2rp \cos \eta$
$= p^2 +s^2 - 2ps\cos \zeta$
$= s^2 +q^2 - 2sq\cos \beta$
$\theta + \eta + \zeta + \beta = 360^\circ$.
