In $\triangle A B C$, if $a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2}$, then

In $\triangle A B C$, if $a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2}$, then
  1. $2 b=a+c$
  2. $b^2=a c$
  3. $\frac{1}{b}=\frac{1}{a}+\frac{1}{c}$
  4. $a=c$

Solution

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} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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