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In a triangle ABC ∠A=60°.. How to prove that (1/(a+ b) )+ ( 1/(a+ c) ) =3/(a + b + c)?
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- 4 years ago
Cosine law
a²=b²+c²-2bcCos60°
a²=b²+c²-bc
(1/(a+b))+(1/(a+c))=3/(a+b+c)
((a+c)+(a+b))/(a+b)(a+c)=3/(a+b+c)
(2a+b+c)/(a²+ab+ac+bc)=3/(a+b+c)
2a²+b²+c²+3ab+3ac+2bc=3a²+3ab+3ac+3bc
b²+c²-bc=a²
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