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\(\cfrac{y+z-x}{b+c-a}=\cfrac{z+x-y}{c+a-b}=\cfrac{x+y-z}{a+b-c}\)
\(=\cfrac{y+z-x+z+x-y+x+y-z}{b+c-a+c+a-b+a+b-c}\)

[সংযোগ প্রক্রিয়ার সাহায্যে পাই]

\(\therefore\) প্রতিটি অনুপাত \(=\cfrac{x+y+z}{a+b+c}\)
সুতরাং, \(\cfrac{y+z-x}{b+c-a}=\cfrac{x+y+z}{a+b+c}\)
\(=\cfrac{(y+z-x)-(x+y+z)}{(b+c-a)-(a+b+c)}=\cfrac{-2x}{-2a}=\cfrac{x}{a}\)

অনুরূপে, \(\cfrac{z+x-y}{c+a-b}=\cfrac{x+y+z}{a+b+c}\)
\(=\cfrac{(z+x-y)-(x+y+z)}{(c+a-b)-(a+b+c)}=\cfrac{-2y}{-2b}=\cfrac{y}{b}\)

এবং, \(\cfrac{x+y-z}{a+b-c}=\cfrac{x+y+z}{a+b+c}\)
\(=\cfrac{(x+y-z)-(x+y+z)}{(a+b-c)-(a+b+c)}=\cfrac{-2z}{-2c}=\cfrac{z}{c}\)

\(\therefore \cfrac{x}{a}=\cfrac{y}{b}=\cfrac{z}{c}\) [প্রমাণিত]