Laober功力不凡！

2+2^0.5

PD^2 = 4*2^1/2 + 6

let us assume AB is the left side of the square, and P is on the left side of AB, and assume the the with and hight of the triangle of the addition are x and y, then we:

1) u^2= x^2 + y^2

2) v^2 = x^2 + (1-y)^2

3) w^2 = (1-y)^2 + (1+x)^2

4) PD^2 = (1+x)^2 +y^2

since, w^2 = u^2+v^2, from (1), (2) and (3), we can get:

5) (x-1)^2 + y^2 = 2

from (5), the maximum of x is 2^1/2 +1, when y= 0.

Also, substiute (5) into (4), we get PD^2 = 4x +2

so the maximum of PD^2 = 4*2^1/2 + 6