Question

If $\left(11a^2+13b^2\right)\left(11c^2-13d^2\right)=\left(11a^2-13b^2\right)\left(11c^2+13d^2\right)$, prove that the value of $a:b::c:d$

Answer 1

Given: $\left(11a^2+13b^2\right)\left(11c^2-13d^2\right)=\left(11a^2-13b^2\right)\left(11c^2+13d^2\right)$

Rewriting the above equation,

$\Rightarrow \frac{11a^2+13b^2}{11c^2+13d^2}=\frac{11a^2-13b^2}{11c^2-13d^2}$

$\Rightarrow \frac{11a^2+13b^2}{11a^2-13b^2}=\frac{11c^2+13d^2}{11c^2-13d^2}$

Applying componendo and dividendo, we get;

$\Rightarrow \frac{11a^2+13b^2+11a^2-13b^2}{11a^2+13b^2-11a^2+13b^2}=\frac{11c^2+13d^2+11c^2-13d^2}{11c^2+13d^2-11c^2+13d^2}$

$\Rightarrow \frac{2\times 11a^2}{2\times 13b^2}=\frac{2\times 11c^2}{2\times 13d^2}$

$\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}$

Taking Square root on both the sides, we have,

$\frac{a}{b}=\frac{c}{d}$

Thus, $a,b,c,d$ are in proportion