If the bisector of $\angle A$ in $\triangle ABC$ meets $BC$ at $U$, prove that $AU^2 =bc\left(1-\frac{a^2}{(b+c)^2}\right)$

Question

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.

Answer 1

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 $$AU^2 = AB\cdot AC - BU \cdot UC = bc - \frac{bc}{(b+c)^2}\cdot a^2 $$

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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.

Answer 2

Let us use the displayed idea to solve the problem, i always try to use a given start, this encourages in future tries, and should give trust in own thoughts. The picture is:

We need no more, and will use only the generalized Pythagoras to finish. In the given picture, $A'\in AU$ is as in OP with $CA'\|AB$. The $x$-marked angles are equal, their common value is $\frac 12\hat A$. So $\Delta ACA'$ is isosceles, Pythagoras gives: $$ \begin{aligned} {AA'}^2&=b^2+b^2-2b^2\cos\widehat{ACA'} =2b^2(1-\underbrace{\cos(180^\circ-2x)}_{=\cos(180^\circ-\hat A)=-\cos \hat A})=2b^2(1+\cos\hat A) \\ &=2b^2\left(1+\frac{b^2+c^2-a^2}{2bc}\right)=\frac bc\Big(\ (b+c)^2-a^2\ \Big)\ . \\[3mm] &\qquad\text{ Recall the similarity $\Delta AUB\sim \Delta A'UC$:} \\[3mm] \frac{AU}{UA'} &=\frac{AB}{CA'}=\frac cb\ ,\qquad\text{ so }\qquad \frac{AU}{AA'} =\frac{AU}{AU+UA'}=\frac c{b+c}\ ,\qquad\text{ which gives:} \\[3mm] AU^2 &= \frac{c^2}{(b+c)^2}\cdot {AA'}^2 = \frac{c^2}{(b+c)^2}\cdot \frac bc\Big(\ (b+c)^2-a^2\ \Big) = \bbox[lightyellow]{\ bc\left(1-\frac{a^2}{(b+c)^2}\right)\ }\ . \end{aligned} $$ $\square$