Greatest distance one point can have from a vertice of a square given following conditions

A point P lies in the same plane as a given square of side 1.Let the vertices of the square,taken counterclockwise,be A,B,C, and D.Also,let the distances from P to A,B, and C, respectively, be u,v and w.

What is the greatest distance that P can be from D if ${u}^{2}+{v}^{2}={w}^{2}$?

Some thoughts I had:

1) Given a pair of vertices I could construct an ellipse with P as a point on the ellipse.

2) From the equality ${u}^{2}+{v}^{2}={w}^{2}$. I think that I have to consider the case where the angle between u and v is ${90}^{\circ}$. In this case I would have $w=1$ and $PD<2$.

That being said,I still fail to come at a concrete solution of the problem,it might be that none of my thoughts are right...

A point P lies in the same plane as a given square of side 1.Let the vertices of the square,taken counterclockwise,be A,B,C, and D.Also,let the distances from P to A,B, and C, respectively, be u,v and w.

What is the greatest distance that P can be from D if ${u}^{2}+{v}^{2}={w}^{2}$?

Some thoughts I had:

1) Given a pair of vertices I could construct an ellipse with P as a point on the ellipse.

2) From the equality ${u}^{2}+{v}^{2}={w}^{2}$. I think that I have to consider the case where the angle between u and v is ${90}^{\circ}$. In this case I would have $w=1$ and $PD<2$.

That being said,I still fail to come at a concrete solution of the problem,it might be that none of my thoughts are right...