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Question Number 211651 by MATHEMATICSAM last updated on 15/Sep/24
In triangle ABC, ∠C = 60°. If the length  of opposite sides of ∠A, ∠B and ∠C are  a, b and c respectively then prove that  (1/(a + c)) + (1/(b + c)) = (3/(a + b + c)) .
$$\mathrm{In}\:\mathrm{triangle}\:\mathrm{ABC},\:\angle\mathrm{C}\:=\:\mathrm{60}°.\:\mathrm{If}\:\mathrm{the}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{of}\:\angle\mathrm{A},\:\angle\mathrm{B}\:\mathrm{and}\:\angle\mathrm{C}\:\mathrm{are} \\ $$$${a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{respectively}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}\:+\:{c}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{3}}{{a}\:+\:{b}\:+\:{c}}\:. \\ $$
Answered by som(math1967) last updated on 15/Sep/24
 ((a^2 +b^2 −c^2 )/(2ab))=cos60  ⇒a^2 +b^2 −c^2 =(1/2)×2ab  ⇒a^2 +b^2 =ab+c^2   ⇒a^2 +b^2 +bc+ca=ab+bc+ca+c^2   ⇒b(b+c)+a(a+c)= (a+c)(b+c)  ⇒(b/(a+c)) +(a/(b+c))=1  ⇒(b/(a+c))+1+(a/(b+c))+1=1+2  ⇒((a+b+c)/(a+c))+((a+b+c)/(b+c))=3  ⇒ (1/(a+c)) +(1/(b+c))=(3/(a+b+c))
$$\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}={cos}\mathrm{60} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{ab} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={ab}+{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{bc}+{ca}={ab}+{bc}+{ca}+{c}^{\mathrm{2}} \\ $$$$\Rightarrow{b}\left({b}+{c}\right)+{a}\left({a}+{c}\right)=\:\left({a}+{c}\right)\left({b}+{c}\right) \\ $$$$\Rightarrow\frac{{b}}{{a}+{c}}\:+\frac{{a}}{{b}+{c}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{b}}{{a}+{c}}+\mathrm{1}+\frac{{a}}{{b}+{c}}+\mathrm{1}=\mathrm{1}+\mathrm{2} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{a}+{c}}+\frac{{a}+{b}+{c}}{{b}+{c}}=\mathrm{3} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{a}+{c}}\:+\frac{\mathrm{1}}{{b}+{c}}=\frac{\mathrm{3}}{{a}+{b}+{c}} \\ $$

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