Let P be the point of intersection of AC and BD.

Let MPN be a perpendicular line to AB and DC.

Let PN = h1 and MP = h2.

Triangle ABP and DEP are similar. So

AB/EC = h2/h1. AB = x+y, and EC = y

So (x+y)/y = h2/h1........(1)

Area of ABCD isA = DC*(h1 + h2) = (x+y)(h1+h2).....(2)

Area PEC ia a = 1/2*EC*h1 = 1/2*y*h1.....(3)

Rewrite eq.(1) as

1 + (x+y)/y = 1 + h2/h1

(2y+x)/y = (h1+h2)/h1

h1 = (h1+h2)y/(x +2y)

Substitute this value of h1 in eq. 3.

a= 1/2*y*y(h1+h2)/(x+2y) .......(4)

Now find the ratio of a/A