In a triangle `A B C ,/_C=60^0,`then prove that: `1/(a+c)+1/(b+c)=3/(a+b+c)dot`

In a triangle `A B C ,/_C=60^0,`then prove that: `1/(a+c)+1/(b+c)=3/(a

1.	In a triangle `A B C ,/_C=60^0,`then prove that: `1/(a+c)+1/(b+c)=3/(a+b+c)dot`
Answer» Let `1/(a+b)+1/(b+c)=3/(a+b+c)` `rArr (b+c+a+b)/((a+b)(b+c))=3/(a+b+c)` `rArr (a+2b+c)(a+b+c)=3(a+b)(b+c)` `rArr a^(2)+2b^(2)+c^(2)+3ab+3bc+2ac = 3ab + 3b^(2)+3ac+3bc` `rArr a^(2)+c^(2)-b^(2)=ac` `rArr (a^(2)+c^(2)-b^(2))/(2ac)=1/2` `rArr cosB=cos60^(@)` `rArr B=60^(2)` `therefore` In `DeltaABC, angleB=60^(@)` `rArr 1/(a+b)+1/(b+c)=3/(a+b+c)`. Hence Proved.