Prove relation about area of triangle

As you have noted, the area $\cal A$ can be written as $AC\times BF/2$. Proving ${\cal A}=AB\times AD$ is equivalent to proving $$AB^2\times 4AD^2=AC^2\times BF^2\tag1$$ on squaring both sides.

The median $AD$ has the property that $4AD^2=2AC^2+2AB^2-BC^2$ as shown here. Using Pythagoras on $\triangle CFB$ yields $$BC^2=(AF+AC)^2+BF^2=AF^2+AC^2+2AF\times AC+BF^2.$$ Since $AF^2+BF^2=AB^2=AF\times AC$ we obtain $BC^2=3AB^2+AC^2$ so that $$4AD^2=AC^2-AB^2\tag2.$$ Finally, using the same identity, we have $BF^2=AF\times(AC-AF)$ so that $$AC^2\times BF^2=AF\times AC\times(AC^2-AF\times AC)=AB^2\times(AC^2-AB^2).$$ Substituting in $(2)$ gives $(1)$ so it follows that ${\cal A}=AB\times AD$.

$BE$ parallel to $DA$, $E$ on the line $CA$

Let $AB=c$, $AC=b$, $BC=2a$ , $AD=d$

we have $BE=2d$

Given $AF = \frac{c^2}{b}$

We get $EF=b-\frac{c^2}{b}$, $CF=b+\frac{c^2}{b}$

using Pythagoras, express $BF^2$ in two different ways

$4d^2-\left(b-\frac{c^2}{b}\right)^2=4a^2-\left(b+\frac{c^2}{b}\right)^2$

it follows from this that $d^2+c^2=a^2$

and so $BAD$ is a right angle, area $ABD = \frac{1}{2} cd$ and therefore area $ABC=cd$