EDIT1: If we only want to prove if $ EF$ exist,that $z$ is on $EF$, it is easy.
let $A(0,0),D(x_1,0),C(x_2,y_2),B(x_3,y_3),Z(a,b),H(p,q),H$ is on $AC$with $FH//AD \to EH//BC$, if we can find $H(p,q)$, then we can draw $EF$ and the problem is solved.
with the conditions, we have:
$((x_2-x_1)y_3-x_3y_2)p^2+((x_1-a)x_2y_3+ax_2y_2+bx_2x_3-bx_2^2+bx_1x_2)p-bx_1x_2^2=0 \\\ q=\dfrac{y_2p}{x_2}$
which means $H$ is constructible even by compass and straight $\implies EF$ is constructable also.
The $p$ has two solutions because we didn't limit $E,F$ 's position and another solution is for $E,F$ out of segments$AB ,CD$ but on lines $AB ,CD$