Cutting proportionally Given any convex quadrilateral $ABCD$ and an inner point $z$ is it possible to draw/construct a line $EF$ passing through $z$ s.t. $$\frac{AE}{EB}=\frac{DF}{FC}$$

 A: The following might work:
Prove this for a square (trivial) Then use the result that any convex non-degenerate quadrilateral can be obtained by a perspectival (projective) transform from the square and that the ratios will be preserved.
A: I don’t think it is always possible.

Let the red line (through Z) be that required line. It cuts AB and DC at E and F respectively. 
Through E draw EG // AD cutting DC at G. It also cuts BD at E’
According to the requirement,
$\frac {DF}{FC} = \frac {AE}{EB} = \frac {DE’}{E’B}$ … [intercept theorem]
Through B, draw BX // E’F cutting DC produced at X.
BY similar triangles, $\frac {DE’}{E’B} = \frac {DF}{FC + CX}$
In other words, FC is too short for the equality to hold unless C is actually X.
A: 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  $AB ,CD$
A: Start with $E=A$ and $F=D$, so that $AE/EB=FD/FC$ because both are equal to $0$, and then move $E$ and $F$ toward $B$ and $C$ respectively, preserving the equality of the ratios (which go to infinity as $E\to B$ and $F\to C$).  By continuity, the line $EF$ sweeps out the entire convex quadrilateral, hence at some point passes through the point $z$.
