To prove in a triangle: $AD^2=AB\cdot AC- BD\cdot CD$

Question

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? :)

Answer 1

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: $$ AD^2+BD\cdot CD = \frac{CD}{BD+CD}AB^2+\frac{BD}{BD+CD}AC^2$$ $$ AD^2+BD\cdot CD = \frac{AC}{AB+AC}AB^2+\frac{AB}{AB+AC}AC^2$$ $$ AD^2+BD\cdot CD = AB\cdot AC.$$

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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.

Answer 2

$$\triangle {ABD} \sim \triangle {D_1BC}$$ Then $$\frac{AB}{BD_1}=\frac{BD}{BC}$$ $$AB \cdot BC= BD \cdot BD_1$$ Because $BD_1=BD+DD_1$, then $AB \cdot BC= BD^2+BD \cdot DD_1$

$BD \cdot DD_1= AD \cdot DC$.

Then $AB \cdot BC= BD^2+AD \cdot DC \Rightarrow$ $$BD^2=AB\cdot BC-AD \cdot DC$$