Prove that here $BE×DF$ is fixed

Question

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.

Answer 1

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 $$CF=AB\left({\small{\frac{CE}{BE}}}\right)=x\left(\frac{1-t}{t}\right)$$ so $$DF=x+CF=x+x\left(\frac{1-t}{t}\right)=\frac{x}{t}$$ hence $$(BE)(DF)=(ty)\left(\frac{x}{t}\right)=xy$$ which is constant.

Answer 2

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