If the bisector of $\angle A$ in $\triangle ABC$ meets $BC$ at $U$, prove that $$AU^2 = bc\left(1-\frac{a^2}{(b+c)^2}\right)$$
I see no other way to bring in $AU$ into the equation besides constructing additional triangles. So I tried extending $AU$ till $A'$ such that $CA' || AB$. Then $\triangle ABU$ and $\triangle A'UC$ can be proven similar. I tried to use this in tandem with angle bisector theorem, and got $$\frac{A'U}{AU} = \frac{CU}{BU} = \frac{AC}{AB}$$
Now all I need to do from here is somehow write $A'U$ in terms of a,b, and c. I am unable to proceed further. Any help is appreciated.
Best Answer
One method of proof is to use the circumcircle. First off, knowing that $U$ divides $BC$ in ratio $c:b$, we have $$BU = \frac{c}{b+c}\cdot a \quad , \quad UC = \frac{b}{b+c}\cdot a$$
Let $AU$ extended meets the circumcircle at $M$. We have $\triangle ABM \sim \triangle AUC $, $$\frac{AB}{AM}=\frac{AU}{AC}\Rightarrow AB \cdot AC = AM \cdot AU$$ $$AB \cdot AC = (AU+UM)\cdot AU$$ By intersecting chord theorem, $AU \cdot UM = BU \cdot UC$. Therefore $$AU^2 = AB\cdot AC - BU \cdot UC = bc - \frac{bc}{(b+c)^2}\cdot a^2 $$
Alternatively, area($\triangle ABU$) $+$ area($\triangle AUC$) $=$ area($\triangle ABC$) $$\frac{1}{2}AB \cdot AU \sin \frac{A}{2} + \frac{1}{2}AU \cdot AC \sin \frac{A}{2} = \frac{1}{2}AB \cdot AC \sin A$$ simplifies to $$AU = \frac{2bc}{b+c}\cos\frac{A}{2}$$ Squaring both sides and using $$2\cos^2\frac{A}{2}=1+\cos A = 1 + \frac{b^2+c^2-a^2}{2bc}$$ gives the sought result again.