ABCD is a cyclic trapezium where AB||CD. if AB=4, AD=5, BD=7, find CD

The answer is CD=6. 

Solution: draw BE parallel to AD such that BE=5 and DE=4. Let CE=x. 

then, let angle BEC=Z

so, angle ADE and angle BCD are also Z(corresponding angles and cyclic trapezium).

From triangles BED and BEC, 

cos 180-z=-cos z=(4^2+5^2-7^2)/2.4.5=-(5^2+x^2-5^2)/2.5.x

                             triangle BED                   Triangle BEC

on solving we get x=2 . so, CD=4+X=6.

thankyou so much!!!

but i didnt get why you took 

cos z=(4^2+5^2-7^2)/2.4.5=-(5^2+x^2-5^2)/2.5.x

                             triangle BED                   Triangle BEC

i think basically i donno cos funcn in obtuse triangle