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$\frac{a+b-c}{a+b}=\frac{b+c-a}{b+c}=\frac{c+a-b}{c+a}$ এবং $a+b+c \neq 0$ হলে, প্রমাণ কর যে, $a=b=c$

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দেওয়া আছে,
$\frac{a+b-c}{a+b}=$$\frac{b+c-a}{b+c}=$$\frac{c+a-b}{c+a}$

বা, $\frac{a+b-c-\left(a+b\right)}{a+b}=$$\frac{b+c-a-\left(b+c\right)}{b+c}=$$\frac{c+a-b-\left(c+a\right)}{c+a}$
[ বিয়োজন করে ]

বা, $\frac{a+b-c-a-b}{a+b}=$$\frac{b+c-a-b-c}{b+c}=$$\frac{c+a-b-c-a}{c+a}$

বা, $\frac{-c}{a+b}=$$\frac{-a}{b+c}=$$\frac{-b}{c+a}$

বা, $\frac{c}{a+b}=\frac{a}{b+c}=\frac{b}{c+a}$
[ প্রত্যেক পক্ষকে $(-1)$ দ্বারা গুণ করে ]

বা, $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}$
[ ব্যস্তকরণ করে ]

বা, $\frac{a+b+c}{c}=\frac{b+c+a}{a}=\frac{c+a+b}{b}$
[ যোজন করে ]

বা, $\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}$

বা, $\frac{a+b+c}{c\left(a+b+c\right)}=$$\frac{a+b+c}{a\left(a+b+c\right)}=$$\frac{a+b+c}{b\left(a+b+c\right)}$
[ প্রত্যেক পক্ষকে $(a+b+c)$ দ্বারা ভাগ করে ]

বা, $\frac{1}{c}=\frac{1}{a}=\frac{1}{b}$

বা, $c=a=b$

[ব্যস্তকরণ করে]

$\therefore a=b=c$ [প্রমাণিত]

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