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

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Discussion Overview

The discussion revolves around the expression $\dfrac {c}{a+b}+\dfrac {b}{a+c}$ in the context of triangle $\triangle ABC$ with side lengths $a$, $b$, and $c$, and with angle $A$ set to $60^\circ$. Participants explore various methods to find the value of this expression, including algebraic and geometric approaches.

Discussion Character

  • Mathematical reasoning, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant applies the cosine formula to derive a relationship between the sides of the triangle, leading to the conclusion that the expression equals 1 under the given conditions.
  • Another participant requests a geometric method to find the value of the expression, indicating a desire for alternative approaches beyond algebraic manipulation.

Areas of Agreement / Disagreement

There is no consensus on the value of the expression, as one participant provides an algebraic solution while another seeks a geometric interpretation, suggesting multiple competing views remain.

Contextual Notes

The discussion does not resolve the assumptions underlying the geometric approach or the implications of the algebraic solution, leaving these aspects open for further exploration.

Albert1
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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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