Albert1
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$\triangle ABC (with \,\,side \,\,length \,\, a,b,c), \,\, given :\angle A+\angle C =2 \angle B $
prove :
$\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c}$
prove :
$\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c}$