<s>[INST] To prove in a triangle: $AD^2=AB\cdot AC- BD\cdot CD$ If $AD$ is an angle bisector of $\triangle ABC$ (with $D\in BC$), then we have to prove that:
$$AD^2=AB\cdot AC- BD\cdot CD$$
I have no idea how to do this, can this be proved with simple geometry? :)
 [/INST] By the bisector theorem,
$$ \frac{BD}{DC}=\frac{BA}{AC}. $$
By Stewart's theorem,
$$ (BD+CD)(AD^2+BD\cdot CD)= CD\cdot AB^2 + BD\cdot AC^2, $$
hence:
$$ AD^2+BD\cdot CD = \frac{CD}{BD+CD}AB^2+\frac{BD}{BD+CD}AC^2$$
$$ AD^2+BD\cdot CD = \frac{AC}{AB+AC}AB^2+\frac{AB}{AB+AC}AC^2$$
$$ AD^2+BD\cdot CD = AB\cdot AC.$$

As an alternative, from the cosine theorem we have:
$$ \cos\widehat{BAD}=\frac{AB^2+AD^2-BD^2}{2\,AB\,AD}=\frac{AC^2+AD^2-CD^2}{2\,AC\,AD}=\cos\widehat{DAC},$$
hence:
$$ AC(AB^2+AD^2-BD^2)=AB(AC^2+AD^2-CD^2) $$
or:
$$ (AC-AB)AD^2 = (AC-AB)AB\,AC+AC\,BD^2-AB\,CD^2.$$
Again, the claim follows from the angle bisector theorem.
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