Cfrgtkky

We have a parallelogram namely $ABCD$. Then we draw a line from the vertex $A$ to:
1- Cross a point (namely $E$)on the side $BC$.
2- Cross a point (namely $F$) along the side DC.
Now we must prove and show that here $BE×DF$ has a fixed and changeless amount if the line $AF$ cuts any other points as $E$ and $F$
I mean that we must prove that $BE×DF$ never changes, even if $E$ and $F$ be different.
I think it's better to take a look here at this picture:

What I obtained was:
1- $$ triangle ABE thicksim triangle CEF $$
2- $$ frac{AB}{CF}= frac{BE}{EC}= frac{AE}{EF}$$
3- Thales's theorem works for $triangle AFD$

4- $$ frac{CE}{AD}= frac{FE}{FA}= frac{FC}{FD}$$

Then we obtain that :
$$BE=frac{AB×EC}{CF}=frac{AE×EC}{EF}$$
$$DF=frac{FD×EC}{AD}=frac{AF×FC}{EF}$$
So any ideas to prove:
$$BE×DF=x$$ and $x$ is changeless?
Note:C AND D MUST BE SWAPPED.

Your original diagram had the points $C,D$ (perhaps accidentally) swapped.

Swapping them back,

we can argue as follows . . .

Let $x=AB,;y=BC,;$and let $t={large{frac{BE}{BC}}}$.

Thus, we have $BE=ty$, and $CE=(1-t)y$.

Since triangles $ABE$ and $FCE$ are similar, we get
$$CF=ABleft({small{frac{CE}{BE}}}right)=xleft(frac{1-t}{t}right)$$
so
$$DF=x+CF=x+xleft(frac{1-t}{t}right)=frac{x}{t}$$
hence
$$(BE)(DF)=(ty)left(frac{x}{t}right)=xy$$
which is constant.

This is not true: $$ triangle ABE thicksim triangle CEF $$
so your work is not correct.