Solveeit Logo
Login

Question

Question: If in a triangle ABC \(\angle C=60{}^\circ \), then prove that \(\dfrac{1}{a+c}+\dfrac{1}{b+c}=\dfra...

If in a triangle ABC ∠C=60∘\angle C=60{}^\circ , then prove that 1a+c+1b+c=3a+b+c\dfrac{1}{a+c}+\dfrac{1}{b+c}=\dfrac{3}{a+b+c}

Explanation

Solution

Hint: Apply cosine rule on angle C, i.e. use cos⁡C=a2+b2−c22ab\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}. Use the fact that cos⁡60∘=12\cos 60{}^\circ =\dfrac{1}{2}. Simplify the resulting expression to get the result.

Complete step-by-step answer:

Here as per usual notation, we have a = BC, b = AC and c = AB.
We know that in any triangle ABC, we have cos⁡C=a2+b2−c22ab\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} (cosine rule)
Applying cosine rule on triangle ABC, we get
cos⁡C=a2+b2−c22ab\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}
We know that cos⁡60∘=12\cos 60{}^\circ =\dfrac{1}{2}
Since C=60∘C=60{}^\circ , we have cos⁡C=12\cos C=\dfrac{1}{2}
Hence, we have
12=a2+b2−c22ab\dfrac{1}{2}=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}
Multiplying by 2 on both sides, we get
a2+b2−c2ab=1\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{ab}=1
Adding and subtracting 2ab in numerator, we get
a2+b2−c2+2ab−2abab=1 ⇒a2+2ab+b2−c2−2abab=1 \begin{aligned} & \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}+2ab-2ab}{ab}=1 \\\ & \Rightarrow \dfrac{{{a}^{2}}+2ab+{{b}^{2}}-{{c}^{2}}-2ab}{ab}=1 \\\ \end{aligned}
We know that a2+2ab+b2=(a+b)2{{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}
Using the above identity, we get
(a+b)2−c2−2abab=1\dfrac{{{\left( a+b \right)}^{2}}-{{c}^{2}}-2ab}{ab}=1
We know that a+bc=ac+bc\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}
Using the above identity, we get
(a+b)2−c2ab−2abab=1 ⇒(a+b)2−c2ab−2=1 ⇒(a+b)2−c2ab=3 \begin{aligned} & \dfrac{{{\left( a+b \right)}^{2}}-{{c}^{2}}}{ab}-\dfrac{2ab}{ab}=1 \\\ & \Rightarrow \dfrac{{{\left( a+b \right)}^{2}}-{{c}^{2}}}{ab}-2=1 \\\ & \Rightarrow \dfrac{{{\left( a+b \right)}^{2}}-{{c}^{2}}}{ab}=3 \\\ \end{aligned}
We know that a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
Using the above identity, we get
(a+b+c)(a+b−c)ab=3\dfrac{\left( a+b+c \right)\left( a+b-c \right)}{ab}=3
Dividing both sides by a+b+c, we get
a+b−cab=3a+b+c\dfrac{a+b-c}{ab}=\dfrac{3}{a+b+c}
Also, we have
a2+b2−c22ab=2 ⇒ \begin{aligned} & \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=2 \\\ & \Rightarrow \\\ \end{aligned}
Now, we have
a+b−cab=3a+b+c\dfrac{a+b-c}{ab}=\dfrac{3}{a+b+c}
We know that a+bc=ac+bc\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}
Using the above identity, we have
aab+bab−cab=3a+b+c ⇒1b+1a−cab=3a+b+c \begin{aligned} & \dfrac{a}{ab}+\dfrac{b}{ab}-\dfrac{c}{ab}=\dfrac{3}{a+b+c} \\\ & \Rightarrow \dfrac{1}{b}+\dfrac{1}{a}-\dfrac{c}{ab}=\dfrac{3}{a+b+c} \\\ \end{aligned}
In the first fraction on LHS, multiply the numerator and denominator by b+c and in the second fraction on LHS, multiply the numerator and denominator by a+c, we get
b+cb(b+c)+a+ca(a+c)−cab=3a+b+c ⇒1b+c[b+cb]+1a+c[a+ca]−cab=3a+b+c \begin{aligned} & \dfrac{b+c}{b\left( b+c \right)}+\dfrac{a+c}{a\left( a+c \right)}-\dfrac{c}{ab}=\dfrac{3}{a+b+c} \\\ & \Rightarrow \dfrac{1}{b+c}\left[ \dfrac{b+c}{b} \right]+\dfrac{1}{a+c}\left[ \dfrac{a+c}{a} \right]-\dfrac{c}{ab}=\dfrac{3}{a+b+c} \\\ \end{aligned}
Now we know that a+bc=ac+bc\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}
Applying the above identity, we get
1b+c[1+cb]+1a+c[1+ca]−cab=3a+b+c ⇒1b+c+1a+c+ca(a+c)+cb(b+c)−cab=3a+b+c ⇒1b+c+1a+c+cab(a+c)(b+c)[b(b+c)+a(a+c)−(a+c)(b+c)]=3a+b+c \begin{aligned} & \dfrac{1}{b+c}\left[ 1+\dfrac{c}{b} \right]+\dfrac{1}{a+c}\left[ 1+\dfrac{c}{a} \right]-\dfrac{c}{ab}=\dfrac{3}{a+b+c} \\\ & \Rightarrow \dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{c}{a\left( a+c \right)}+\dfrac{c}{b\left( b+c \right)}-\dfrac{c}{ab}=\dfrac{3}{a+b+c} \\\ & \Rightarrow \dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{c}{ab\left( a+c \right)\left( b+c \right)}\left[ b\left( b+c \right)+a\left( a+c \right)-\left( a+c \right)\left( b+c \right) \right]=\dfrac{3}{a+b+c} \\\ \end{aligned}
Expanding the terms, we get
1a+c+1b+c+cab(a+c)(b+c)(b2+bc+a2+ac−ab−ac−bc−c2)=3a+b+c ⇒1a+c+1b+c+cab(b+c)(a+c)(a2+b2−c2−ab)=3a+b+c \begin{aligned} & \dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{c}{ab\left( a+c \right)\left( b+c \right)}\left( {{b}^{2}}+bc+{{a}^{2}}+ac-ab-ac-bc-{{c}^{2}} \right)=\dfrac{3}{a+b+c} \\\ & \Rightarrow \dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{c}{ab\left( b+c \right)\left( a+c \right)}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}}-ab \right)=\dfrac{3}{a+b+c} \\\ \end{aligned}
We know that
a2+b2−c2−ab=0{{a}^{2}}+{{b}^{2}}-{{c}^{2}}-ab=0 (proved above)
Hence we have
1a+c+1b+c−0=3a+b+c ⇒1a+c+1b+c=3a+b+c \begin{aligned} & \dfrac{1}{a+c}+\dfrac{1}{b+c}-0=\dfrac{3}{a+b+c} \\\ & \Rightarrow \dfrac{1}{a+c}+\dfrac{1}{b+c}=\dfrac{3}{a+b+c} \\\ \end{aligned}

Note: In the questions of the above type, we need to understand which formula to apply and which not. Cosine rule is generally applied when on the basis of a known angle we want to find the relations between the sides of the triangle as is the case above.
[2] Alternatively, we can solve the above question by showing that 1a+c+1b+c=3a+b+c⇔a2+b2−c2=ab\dfrac{1}{a+c}+\dfrac{1}{b+c}=\dfrac{3}{a+b+c}\Leftrightarrow {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=ab
Multiply both sides of the equation by (a+b)(b+c)(a+b+c) and simplify to prove the above result.