Question

$\frac{a^2+b^2}{b^2+c^2}=\frac{\left(a+b\right)^2}{\left(b+c\right)^2}$ হলে, দেখাও যে, $a$, $b$, $c$ ক্রমিক সমানুপাতী।

Answer 1

দেওয়া আছে,
$\frac{a^2+b^2}{b^2+c^2}=$$\frac{\left(a+b\right)^2}{\left(b+c\right)^2}$

বা, $\frac{\left(b+c\right)^2}{b^2+c^2}=$$\frac{\left(a+b\right)^2}{a^2+b^2}$

বা, $\frac{b^2+c^2+2bc}{b^2+c^2}=$$\frac{a^2+b^2+2ab}{a^2+b^2}$

বা, $\frac{b^2+c^2+2bc-\left(b^2+c^2\right)}{b^2+c^2}=$$\frac{a^2+b^2+2ab-\left(a^2+b^2\right)}{a^2+b^2}$

বা, $\frac{b^2+c^2+2bc-b^2-c^2}{b^2+c^2}=$$\frac{a^2+b^2+2ab-a^2-b^2}{a^2+b^2}$

বা, $\frac{2bc}{b^2+c^2}=$$\frac{2ab}{a^2+b^2}$

বা, $\frac{2bc}{b^2+c^2}\times\frac{1}{2b}=$$\frac{2ab}{a^2+b^2}\times\frac{1}{2b}$

বা, $\frac{c}{b^2+c^2}=$$\frac{a}{a^2+b^2}$

বা, $c\left(a^2+b^2\right)=$$a\left(b^2+c^2\right)$

বা, $a^2c+b^2c=$$ab^2+ac^2$

বা, $a^2c-ac^2=ab^2-b^2c$

বা, $ac\left(a-c\right)=b^2\left(a-c\right)$

বা, $ac=\frac{b^2\left(a-c\right)}{\left(a-c\right)}$

বা, $ac=b^2$

বা, $a\cdot c=b\cdot b$

বা, $\frac{a}{b}=\frac{b}{c}$

$\therefore a:b=b:c$