In the below picture, angle A is obtuse, AD is a median. We are also given the relation $AB^2 = AF*AC$. We want to prove that area of triangle $(ABC) = AB*AD$.
What I have tried:
Area of triangle is $A = \frac {1}{2}*AC*BF$.
Replacing AC from the given relation $AB^2 = AF*AC$, we get $A = \frac {1}{2}*\frac{AB^2*BF}{AF}$.
Also from Pythagoras, we replace $AB^2$ with $BF^2+AF^2$.
But I can't see how to involve AD and prove the required relation!
Any help please?
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 $ABC=cd$
