Length of the angle bisector

Dec 2009
54
0
Prove that in every trangle for the length of the angle bisector $l_a,\ l_b,\ l_c$ holds
$$ {l_a}^2=bc\left[1-\left(\frac{a}{b+c}\right)^2\right],\ {l_b}^2=ac\left[1-\left(\frac{b}{a+c}\right)^2\right],\ {l_c}^2=ab\left[1-\left(\frac{c}{a+b}\right)^2\right]$$
Proof 1
First, repeat one well known fact...
Let the angle bisector of angle $\alpha$ intersect side BC at a point D between B and C , and $\overline {BD}= m $ and $\overline{CD}= n $. If $\angle{BDA}=\theta $, aplaing sinus theorem on triangles $ABD $ and $ACD $ we have

$$ \frac{m}{n}=\frac{c}{b}\ \ and\ \ \frac{n}{m}=\frac{b}{c}$$

and adding 1 on both sides it follows

$$m=\frac{ac}{b+c}\ \ and\ \ n=\frac{ab}{b+c}$$

Using cosinus theorem on $\bigtriangleup ABD$ we have

$$l_a^2=m^2+c^2-2amc\cos\beta=$$
$$\frac{a^2c^2}{(b+c)^2} +c^2-\frac{2ac^2}{b+c}\frac{a^2+c^2-b^2}{2ac}=$$
$$\frac{a^2c^2+[c(b+c)]^2-c(b+c)(a^2+c^2-b^2)}{(b+c)^2}=$$
$$\frac{a^2c^2+c(b+c)[c(b+c)-(a^2+c^2-b^2)]}{(b+c)^2}=$$
$$\frac{a^2c^2+c(b+c)(bc+c^2-a^2-c^2+b^2)]}{(b+c)^2}=$$
$$\frac{a^2c^2+(bc+c^2)(bc+b^2-a^2)]}{(b+c)^2}=$$
$$\frac{a^2c^2+b^3c+b^2c^2-a^bc-a^2c^2+b^2c^2+bc^3}{(b+c)^2}=$$
$$\frac{bc(b^2+2bc+c^2-a^2)}{(b+c)^2}=$$
$$\frac{bc[(b+c)^2-a^2)]}{(b+c)^2}=$$
$$bc\left[1-\left(\frac{a}{b+c}\right)^2\right].$$

Similarly for $l_b$ and $l_c$.

Proof 2 (much simpliest, using area)

$$Area(\bigtriangleup ABC)= Area(\bigtriangleup ABD)+ Area(\bigtriangleup ACD)$$

So,

$$bc\sin\alpha=bl_a\sin\frac{\alpha}{2}+cl_a\sin \frac{\alpha}{2}$$

and now easy get

$$l_a=\frac{2bc}{b+c}\cos\frac{\alpha}{2}$$

and rest is following from $2\cos^2\frac{\alpha}{2}=1+\cos\alpha$ and cosinus theorem $\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$.
 
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