MHB Find: c/(a + b) + b/(a + c) = ?

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In triangle ABC with angle A equal to 60 degrees, the expression c/(a+b) + b/(a+c) simplifies to 1. This is derived using the cosine formula, which leads to the relationship b² + c² = a² + bc. The expression is rewritten and simplified to show that both the numerator and denominator are equal, confirming the result. A request for a geometric proof of the same result was also made, indicating interest in alternative methods of verification. The discussion highlights the interplay between algebraic manipulation and geometric reasoning in solving triangle-related problems.
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$\triangle ABC$ has side lengths $a,b,c$

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

find :

$\dfrac {c}{a+b}+\dfrac {b}{a+c}=?$
 
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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 $b^2+c^2 = a^2+bc$
 
Re: find :c/(a+b) + b/(a+c) =?

Can someone find its value using geometry only ?
 

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