\(\cfrac{x}{b+c}=\cfrac{y}{c+a}=\cfrac{x}{a+b}\)হলে প্রমাণ করো, \((b–c)x+(c–a)y+(a–b)z=0\)
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ধরি, \(\cfrac{x}{b+c}=\cfrac{y}{c+a}=\cfrac{x}{a+b}=k\)
বা, \(x=(b+c)k, y=(c+a)k, z=(a+b)k\)

\((b–c)x+(c–a)y+(a–b)z\)
\(=(b-c)(b+c)k+(c-a)(c+a)k+(a-b)(a+b)k\)
\(=k(\cancel{b^2}-\cancel{c^2}+\cancel{c^2}-\cancel{a^2}+\cancel{a^2}-\cancel{b^2})\)
\(=k.0\)
\(=0\) [প্রমানিত]

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