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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 13:06:01 IST
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Answer is A ..
since they are in GP we have
[tanC/2tanA/2]/[tanB/2tanC/2] = [tanA/2tanB/2]/[tanC/2tanA/2]
tanA/2tanC/2 = tan2B/2
[(s-b)(s-c)(s-b)(s-a)/s(s-a)s(s-c)]1/2 = (s-a)(s-c)/s(s-b)
(s-b)/(s-c) = (s-a)/(s-b)
Apply componendo dividendo
(s-b)-(s-c)/(s-b)+(s-c) = (s-a)-(s-b)/(s-a)+(s-b)
(c-b)/(2s-b-c) = (b-a)/(2s-a-b)
(c-b)/a = (b-a)/c
a2+c2 = b(a+c)
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