Find: c/(a + b) + b/(a + c) = ?

[Albert]

$\triangle ABC$ has side lengths $a,b,c$

$\text{if}\,\, \angle A=60^o$

find :

$\dfrac {c}{a+b}+\dfrac {b}{a+c}=?$

 

[jacks]

Re: find :c/(a+b) + b/(a+c) =?

Using Cosine formula $\displaystyle \cos (A) = \frac{b^2+c^2-a^2}{2bc}$

Now Given $A = 60^0$ . So $\displaystyle \frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$

So $b^2+c^2-a^2 = bc \Rightarrow b^2+c^2 = a^2+bc$

Now Given $\displaystyle \frac{c}{a+b}+\frac{b}{a+c} = \frac{c\cdot(a+c)+b\cdot (a+b)}{(a+b)\cdot (a+c)}$

So $\displaystyle = \frac{c^2+ac+b^2+ab}{a^2+ab+bc+ca} = \frac{a^2+ab+bc+ca}{a^2+ab+bc+ca} = 1$

Using $b^2+c^2 = a^2+bc$

 

[Albert]

Re: find :c/(a+b) + b/(a+c) =?

Can someone find its value using geometry only ?

 

[Albert]

Re: find :c/(a+b) + b/(a+c) =?