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-

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

\frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c} .

Answered by som(math1967) last updated on 15/Sep/24

\frac{a^{2} + b^{2} - c^{2}}{2 a b} = c o s 60

\Rightarrow a^{2} + b^{2} - c^{2} = \frac{1}{2} \times 2 a b

\Rightarrow a^{2} + b^{2} = a b + c^{2}

\Rightarrow a^{2} + b^{2} + b c + c a = a b + b c + c a + c^{2}

\Rightarrow b (b + c) + a (a + c) = (a + c) (b + c)

\Rightarrow \frac{b}{a + c} + \frac{a}{b + c} = 1

\Rightarrow \frac{b}{a + c} + 1 + \frac{a}{b + c} + 1 = 1 + 2

\Rightarrow \frac{a + b + c}{a + c} + \frac{a + b + c}{b + c} = 3

\Rightarrow \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}