This question has already been asked before, but the answer gives the solution involving trigonometry and Stewart's theorem which I wanted to avoid.
In a triangle $\triangle ABC$, the bisector of angle from the point $A$ intersects $\overline {BC}$ in point $D$. Prove: $|AD|^2=|AB|\cdot |AC|-|DB|\cdot |DC|$.
My approach:
Let $c$ be the circumcircle of $\triangle ABC$ and let $E$ be the intersection of the line $AD$ and circle $c$.
We obtain the following:
$\begin{aligned}\measuredangle ABC=\measuredangle AEC\ \land\ \measuredangle EAB=\measuredangle CAE&\implies\boxed{\triangle ABD\sim\triangle AEC}\\&\implies\frac{|AC|}{|AE|}=\frac{|AD|}{|AB|}\\&\implies|AB|\cdot|AC|=|AD|\cdot(|AD|+|DE|)=|AD|^2+|AD|\cdot|DE|\\&\implies\boxed{ |AD|^2=|AB|\cdot|AC|-|AD|\cdot|DE|}\end{aligned}$
On the other hand:
$\begin{aligned}\measuredangle CBE=\measuredangle CAE\ \land\ \measuredangle EDB=\measuredangle ADC&\implies\boxed{\triangle DBE\sim\triangle ADC}\\&\implies\frac{|BD|}{|AD|}=\frac{|DE|}{|DC|}\\&\implies\boxed{|BD|\cdot|DC|=|AD|\cdot|DE|}\end{aligned}$
Finally,
$|AD|^2=|AB|\cdot|AC|-|AD|\cdot|DE|=|AB|\cdot|AC|-|BD|\cdot|DC|$
Picture:
May I ask if this is valid? If so, is there anything I could do to improve my proof?
Thank you in advance!
Note the second step ($|BD|\cdot|DC|=|AD|\cdot|DE|$ is a well-known theorem (Intersecting Chords Theorem), so you might just as well refer to it instead of proving it yourself. Other than that, this proof is perfectly valid, and being very short, I can't see how it could be improved.