Question

If $(2ma + 6mb + 3nc + 9nd)(2ma - 6mb - 3nc + 9nd) = $$(2ma - 6mb + 3nc - 9nd)(2ma + 6mb - 3nc - 9nd)$, prove that $a,b,c,d$ are proportional.

Answer 1

Hint: In this problem, we have to prove that $a,b,c,d$ are proportional. That is, we have to prove that $\dfrac{a}{b} = \dfrac{c}{d}$ or $\dfrac{a}{c} = \dfrac{b}{d}$. For this, we will use componendo and dividendo rules. After using this rule, we will simplify the obtained expression.


Complete step-by-step answer:
In this problem, it is given that $(2ma + 6mb + 3nc + 9nd)(2ma - 6mb - 3nc + 9nd) = $$(2ma - 6mb + 3nc - 9nd)(2ma + 6mb - 3nc - 9nd)$.
First, we will rewrite the above equation. Therefore, we get $\dfrac{{(2ma + 6mb + 3nc + 9nd)}}{{(2ma - 6mb + 3nc - 9nd)}} = \dfrac{{(2ma + 6mb - 3nc - 9nd)}}{{(2ma - 6mb - 3nc + 9nd)}} \cdots \cdots \left( 1 \right)$
If the ratio of $m$to $n$ is equal to the ratio of $p$ to $q$ then the ratio of $m + n$ to $m - n$ is equal to the ratio of $p + q$ to $p - q$. This is called the componendo and dividendo rule.
Now we are going to use the componendo and dividendo rule in the equation $(1)$. Therefore, we get
$\dfrac{{2ma + 6mb + 3nc + 9nd + 2ma - 6mb + 3nc - 9nd}}{{2ma + 6mb + 3nc + 9nd - 2ma + 6mb - 3nc + 9nd}} = \dfrac{{2ma + 6mb - 3nc - 9nd + 2ma - 6mb - 3nc + 9nd}}{{2ma + 6mb - 3nc - 9nd - 2ma + 6mb + 3nc - 9nd}}$After cancellation of equal terms with opposite signs from numerator and denominator, we get
$\dfrac{{2(2ma + 3nc)}}{{2(6mb + 9nd)}} = \dfrac{{2(2ma - 3nc)}}{{2(6mb - 9nd)}}$
After cancellation of number $2$ from numerator and denominator, we get
$\dfrac{{2ma + 3nc}}{{6mb + 9nd}} = \dfrac{{2ma - 3nc}}{{6mb - 9nd}}$
Let us rewrite the above equation. Therefore, we get
$\dfrac{{2ma + 3nc}}{{2ma - 3nc}} = \dfrac{{6mb + 9nd}}{{6mb - 9nd}} \cdots \cdots \left( 2 \right)$
Now one more time we are going to use the componendo and dividendo rule in equation $(2)$. Therefore, we get
$\dfrac{{\left( {2ma + 3nc} \right) + \left( {2ma - 3nc} \right)}}{{\left( {2ma + 3nc} \right) - \left( {2ma - 3nc} \right)}} = \dfrac{{\left( {6mb + 9nd} \right) + \left( {6mb - 9nd} \right)}}{{\left( {6mb + 9nd} \right) - \left( {6mb - 9nd} \right)}}$
Let us simplify the above equation by opening brackets.
$\dfrac{{2ma + 3nc + 2ma - 3nc}}{{2ma + 3nc - 2ma + 3nc}} = \dfrac{{6mb + 9nd + 6mb - 9nd}}{{6mb + 9nd - 6mb + 9nd}}$
After cancellation of equal terms with opposite signs from numerator and denominator, we get $\dfrac{{4ma}}{{6nc}} = \dfrac{{12mb}}{{18nd}}$
After simplifying the above equation, we get $\dfrac{a}{c} = \dfrac{b}{d}$. Also we can write $\dfrac{a}{b} = \dfrac{c}{d}$. Therefore, we can say that $a,b,c,d$ are proportional.


Note: Componendo and dividendo is the rule on proportionality which is very useful to simplify complex expression. Converse of componendo and dividendo rule is also true. That is, if the ratio of $m + n$ to $m - n$ is equal to the ratio of $p + q$ to $p - q$ then the ratio of $m$to $n$ is equal to the ratio of $p$ to $q$. That is, $\dfrac{{m + n}}{{m - n}} = \dfrac{{p + q}}{{p - q}} \Rightarrow \dfrac{m}{n} = \dfrac{p}{q}$.