Circle in a trapezoid problem

If x = y, then the quadrilateral is a square, and r = sqrt(xy) = x.  So it's true!

Drop a vertical from A meeting DC at E.

DE = (2y - 2x)/2 = y-x.

AD = x + y.

AE = 2R.

Now Pythagoras, AD^2 = AE^2 + DE^2, etc..

Drop  perpendicular  BE   from B to  DC

Then EC  =  (2y  -2x)  /2   =  y  - x

And  by a property of  tangents drawn to a circle,  BC  = x +  y 

So

BE^2    =   BC^2  -  EC^2

BE^2  =  ( x^2 + 2xy + y^2)  - ( x^2  - 2xy + y^2)

BE^2   =  4xy

BE   = 2sqrt (xy)

But the radius is  (1/2)  of BE  =  sqrt (xy)