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In $\triangle \mathrm{ABC}$, if $\mathrm{a}^2 \sin ^2 \frac{\mathrm{C}}{2}+\mathrm{c} \sin ^2 \frac{\mathrm{A}}{2}=\frac{\mathrm{b}}{2}$, then $\mathrm{a}+\mathrm{c}: \mathrm{b}=$
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$2: 1$
$\because a \sin ^2 \frac{C}{2}+C \sin ^2 \frac{A}{2}=\frac{b}{2}$
$\begin{aligned} & \frac{a(s-b)(s-a)}{a b}+\frac{c(s-b)(s-c)}{b c}=\frac{b}{2} \\ & \Rightarrow 2(s-b)[s-a+s-c]=b^2 \\ & \Rightarrow 2(s-b)[a+b+c-a-c]=b^2 \\ & \Rightarrow(2 s-2 b) b=b^2 \\ & \Rightarrow(a+b+c-2 b)=b\end{aligned}$
$\begin{aligned} & \Rightarrow \quad a+c-b=b \Rightarrow a+c=2 b \\ & \Rightarrow \quad \frac{a}{b} \frac{2}{1} \\ & \therefore a+c: b=2: 1\end{aligned}$
$\begin{aligned} & \frac{a(s-b)(s-a)}{a b}+\frac{c(s-b)(s-c)}{b c}=\frac{b}{2} \\ & \Rightarrow 2(s-b)[s-a+s-c]=b^2 \\ & \Rightarrow 2(s-b)[a+b+c-a-c]=b^2 \\ & \Rightarrow(2 s-2 b) b=b^2 \\ & \Rightarrow(a+b+c-2 b)=b\end{aligned}$
$\begin{aligned} & \Rightarrow \quad a+c-b=b \Rightarrow a+c=2 b \\ & \Rightarrow \quad \frac{a}{b} \frac{2}{1} \\ & \therefore a+c: b=2: 1\end{aligned}$
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