<s>[INST] 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?
 [/INST] $BE$ 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 $ABC=cd$
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