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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 May 2007 19:35:21 IST
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If in a triangle, angleC = 60 degraees, then prove that 1/(a+c) + 1/(b+c) = 3/(a+b+c)
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Given,
angle C =pi/3
We have to prove,
1/(a+c) + 1/(b+c) = 3/(a+b+c) (ie)
1 + b/(a+c) + 1 + a/(b+c) =3 (ie)
b2 + bc = (a+c)(b+c-a) (ie)
a2 + b2 -c2 = ab (ie)
(a2 + b2 -c2 )/2ab = 1/2 = cos(pi/3) which is cosine formula for angle C.
therefore proved.
That soles the problem!
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