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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 :)

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what are $a,b,c?$ Is this a triangle and $a,b,c$ the sides with $A$ the angle opposite $a$? –  Ross Millikan Nov 12 '12 at 15:58
    
yes, it is the normally notation :) –  Iuli Nov 12 '12 at 16:00
    
It is, but much easier to remember when you are in the middle of a geometry class. We see many subjects here, so it would be helpful to include that in the question. –  Ross Millikan Nov 12 '12 at 16:05
    
Have you tried writing the law of cosines for each of the two small triangles, then eliminating $\cos \frac A2$ between the two equations? It looks like it should work. –  Ross Millikan Nov 12 '12 at 16:09

1 Answer 1

up vote 1 down vote accepted

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.

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