C}{2}}{\sin \frac{B}{2}}=\sqrt{\frac{ac(s-b)(s-c)(s-b)(s-a)}{(s-a)(s-c)bc\times ab}}=\frac{s-b}{b}\] But a, b and c are in A.
P
. Þ \[2b=a+c\]
HENCE
\[\frac{s-b}{b}=\frac{\frac{3b}{2}-b}{b}=\frac{
1
}{2}\].