How can I prove that :
$b_{A}=\frac{1}{b+c} \cdot \sqrt{bc\left[(b+c)^2-a^2\right]} ?$ where $b_{A}$ is the length of the bisector from $A$ .
Thanks :)
How can I prove that :
$b_{A}=\frac{1}{b+c} \cdot \sqrt{bc\left[(b+c)^2-a^2\right]} ?$ where $b_{A}$ is the length of the bisector from $A$ .
Thanks :)
Let $P$ be the point where the bisector of $\angle A$ meets $BC$. Let $x=BP$ and $y=PC$. Then $x+y=a$. Moreover, by a standard theorem on angle bisectors, we have $\dfrac{x}{c}=\dfrac{y}{b}$, or equivalently $bx=cy$.
We have two linear equations in two unknowns. Solve for $x$ and $y$. We obtain $x=\frac{ac}{b+c}\qquad\text{and}\qquad y=\frac{ab}{b+c}.$
Now let $\theta=\angle APB$ and $\phi=\angle APC$. Then $\cos\phi=-\cos\theta$. By the Cosine Law on $\triangle BPA$, we have $c^2=b_A^2+x^2 -2b_A \,x\cos\theta.\tag{$1$}$ Similarly $b^2=b_A^2 +y^2+2b_A \,y\cos\theta.\tag{$2$}$ Multiply both sides of $(1)$ by $y$, and both sides of $(2)$ by $x$, and add. We get $yc^2+xb^2=(x+y)b_A^2+xy(x+y).$ Substituting our values for $x$ and $y$, and noting that $x+y=a$, we obtain $\frac{abc^2}{b+c}+\frac{acb^2}{b+c}=ab_A^2+\frac{a^3bc}{(b+c)^2}.$ This simplifies to $bc=b_A^2+\frac{a^2bc}{(b+c)^2},$ or equivalently $b_A^2=\frac{1}{(b+c)^2}\left(bc\left[(b+c)^2-a^2\right]\right),$ which is what we wanted to show.