In $\triangle ABC$, if $C = 60^{\circ}$, then prove that $\dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{3}{a + b + c}$
To prove that (T.P.T) $\;$ $\dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{3}{a + b + c}$
i.e. $\;$ T.P.T $\;$ $\dfrac{b + c + a + c}{\left(a + c\right) \left(b + c\right)} = \dfrac{3}{a + b + c}$
i.e. $\;$ T.P.T $\;$ $\left(a + b + 2c\right) \left(a + b + c\right) = 3 \left(a + c\right) \left(b + c\right)$
i.e. $\;$ T.P.T $\;$ $a^2 + ab + ac + ab + b^2 + bc + 2ac + 2bc + 2c^2$
$\hspace{6cm}$ $= 3 ab + 3 ac + 3 bc + 3 c^2$
i.e. $\;$ T.P.T $\;$ $a^2 + b^2 + 2 c^2 + 2 ab + 3 ac + 3 bc = 3 ab + 3 ac + 3 bc + 3 c^2$
i.e. $\;$ T.P.T $\;$ $a^2 + b^2 - c^2 = ab$ $\;\;\; \cdots \; (1)$
By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$ $\;\;\; \cdots \; (2)$
Given: $\;$ $C = 60^{\circ}$
Substituting the value of $C$ in equation $(2)$ we get,
$\cos 60^{\circ} = \dfrac{a^2 + b^2 - c^2}{2 a b}$
i.e. $\;$ $\dfrac{1}{2} = \dfrac{a^2 + b^2 - c^2}{2 a b}$
i.e. $\;$ $a^2 + b^2 - c^2 = a b$ $\;\;\; \cdots \; (3)$
$\therefore \;$ We have from equations $(1)$ and $(3)$,
$\dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{3}{a + b + c}$
Hence proved.