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In $\triangle A B C$, if $a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2}$, then
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Verified Answer
The correct answer is:
$2 b=a+c$
In a $\triangle A B C$, if
$$
\begin{aligned}
& a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2} \Rightarrow a \frac{s(s-c)}{a b}+c \frac{s(s-a)}{b c}=\frac{3 b}{2} \\
& \Rightarrow \frac{s}{b}(2 s-a-c)=\frac{3 b}{2} \Rightarrow \frac{s}{b}(b)=\frac{3 b}{2} \\
& \Rightarrow \frac{a+b+c}{b}=3 \quad \Rightarrow 2 b=a+c .
\end{aligned}
$$
$$
\begin{aligned}
& a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2} \Rightarrow a \frac{s(s-c)}{a b}+c \frac{s(s-a)}{b c}=\frac{3 b}{2} \\
& \Rightarrow \frac{s}{b}(2 s-a-c)=\frac{3 b}{2} \Rightarrow \frac{s}{b}(b)=\frac{3 b}{2} \\
& \Rightarrow \frac{a+b+c}{b}=3 \quad \Rightarrow 2 b=a+c .
\end{aligned}
$$
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