I1.1 $a, b$ and $c$ are the lengths of the opposite sides $\angle A, \angle B$ and $\angle C$ of the $\triangle A B C$ respectively. If $\angle C=60^{\circ}$ and $\frac{a}{b+c}+\frac{b}{a+c}=P$, find the value of $P$.
$\begin{array}{l}c^{2}=a^{2}+b^{2}-2 a b \cos 60^{\circ}=a^{2}+b^{2}-a b \Rightarrow a^{2}+b^{2}=c^{2}+a b \\ P=\frac{a}{b+c}+\frac{b}{a+c}=\frac{a(a+c)+b(b+c)}{(b+c)(a+c)} \\ P=\frac{a^{2}+a c+b^{2}+b c}{a b+a c+b c+c^{2}}=\frac{a b+a c+b c+c^{2}}{a b+a c+b c+c^{2}}=1\end{array}$