Prove relation about area of triangle 
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 ${\cal A}=AB\times AD$.
