By alternate angles ADC is also equal to the two equal corners at point A. By isosceles triangle, ABD is also the same angle.

Therefore triangles ADB and ACD are similar. AD:AB = CD:AD.

Let CD = x^2 and AB = y^2. It follows that AD =xy.

We are told y^2 – x^2 = 13, therefore (y+x)(y-x) = 13.

The only integer solutions to this is y=7, x=6, therefore AD = 42.

(Pedantic note: x and y don’t necessarily need to be integers for x^2, xy, y^2 and (y^2)-(x^2) all to be integers, eg sqrt(2) and sqrt(8) would satisfy. Basically either x and y are integers, or y/x is an integer, (or both). However since they represent the sides of a triangle, y/x must be between 1 and 2, so y=7, x=6 is the only solution).