In triangle ABC,with usual notations,if cos f | vedclass

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(C) Given $\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}$.Squaring both sides,we get $\cos^2 \frac{B}{2} = \frac{c+a}{2a}$.Using the half-angle formula $\c

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$b^2 + c^2$

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(C) Given $\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}$.Squaring both sides,we get $\cos^2 \frac{B}{2} = \frac{c+a}{2a}$.Using the half-angle formula $\cos^2 \frac{B}{2} = \frac{s(s-b)}{ac}$,where $s = \frac{a+b+c}{2}$ is the semi-perimeter.So,$\frac{s(s-b)}{ac} = \frac{c+a}{2a}$.Multiplying both sides by $2ac$,we get $2s(s-b) = c(c+a)$.Substitute $2s = a+b+c$ and $s-b = \frac{a+c-b}{2}$:$(a+b+c) 	imes \frac{a+c-b}{2} = c(c+a)$.$(a+c+b)(a+c-b) = 2c(c+a)$.$(a+c)^2 - b^2 = 2c^2 + 2ac$.$a^2 + c^2 + 2ac - b^2 = 2c^2 + 2ac$.$a^2 - b^2 = c^2$.Therefore,$a^2 = b^2 + c^2$.

In $\triangle ABC$,with usual notations,if $\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}$,then $a^2 =$

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