---

\(\cfrac{b}{a+b}=\cfrac{a+c-b}{b+c-a}=\cfrac{a+b+c}{2a+b+2c}\)
বা, \(\cfrac{b}{a+b}\times \cfrac{2}{2}=\cfrac{a+c-b}{b+c-a}=\cfrac{a+b+c}{2a+b+2c}\times \cfrac{2}{2}\)
বা, \(\cfrac{2b}{2a+2b}=\cfrac{a+c-b}{b+c-a}=\cfrac{2a+2b+2c}{4a+2b+4c}\)
\(=\cfrac{2b+a+c-b+2a+2b+2c}{2a+2b+b+c-a+4a+2b+4c}\) [যোগভাগ পক্রিয়া করে পাই]
\(=\cfrac{3a+3b+3c}{5a+5b+5c}\)
\(=\cfrac{3\cancel{(a+b+c)}}{5\cancel{(a+b+c)}}\)
\(=\cfrac{3}{5}\)

\(\therefore \cfrac{b}{a+b}=\cfrac{3}{5}\)
বা, \(3a+3b=5b\)
বা, \(3a=5b-3b\)
বা, \(3a=2b-----(i)\)
বা, \(\cfrac{a}{2}=\cfrac{b}{3} -----(ii)\)

এবং \(\cfrac{a+b+c}{2a+b+2c}=\cfrac{3}{5}\)
বা, \(5a+5b+5c=6a+3b+6c\)
বা, \(5a-6a+5b-3b=6c-5c\)
বা, \(-a+2b=c\)
বা, \(-a+3a=c\) [\((ii)\) নং থেকে পাই, \(3a=2b\)]
বা, \(2a=c\)
বা, \(\cfrac{2a}{4}=\cfrac{c}{4}\)
বা, \(\cfrac{a}{2}=\cfrac{c}{4}-----(iii)\)

\(\therefore (ii)\) এবং \((iii)\) নং থেকে পাই, \(\cfrac{a}{2}=\cfrac{b}{3}=\cfrac{c}{4}\) [প্রমানিত]