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

Hint:
First, we will divide the whole equation by $(2ma + 6mb - 3nc - 9nd)(2ma - 6mb - 3nc + 9nd)$. After that, we will get fractions on both sides. Then, we will apply componendo and dividendo to the fraction TWO times and then simplify it to get to the final answer.

Complete step by step solution:
According to the question, we are given that
$(2ma + 6mb + 3nc + 9nd)(2ma - 6mb - 3nc + 9nd) = (2ma - 6mb + 3nc - 9nd)(2ma + 6mb - 3nc - 9nd)$
On dividing the equation by $(2ma + 6mb - 3nc - 9nd)(2ma - 6mb - 3nc + 9nd)$ , we get
⇒$\dfrac{{(2ma + 6mb + 3nc + 9nd)(2ma - 6mb - 3nc + 9nd)}}{{(2ma + 6mb - 3nc - 9nd)(2ma - 6mb - 3nc + 9nd)}} = \dfrac{{(2ma - 6mb + 3nc - 9nd)(2ma + 6mb - 3nc - 9nd)}}{{(2ma + 6mb - 3nc - 9nd)(2ma - 6mb - 3nc + 9nd)}}$
On cancelling the common terms in numerator and denominator, we get
⇒$\dfrac{{(2ma + 6mb + 3nc + 9nd)}}{{(2ma + 6mb - 3nc - 9nd)}} = \dfrac{{(2ma - 6mb + 3nc - 9nd)}}{{(2ma - 6mb - 3nc + 9nd)}}$ … (1)
According to componendo dividendo, for a fraction $\dfrac{a}{b} = \dfrac{c}{d}$ ,
$\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}$
Hence, using componendo dividendo in (1), we get
⇒$\dfrac{{(2ma + 6mb + 3nc + 9nd) + (2ma + 6mb - 3nc - 9nd)}}{{(2ma + 6mb + 3nc + 9nd) - (2ma + 6mb - 3nc - 9nd)}} = \dfrac{{(2ma - 6mb + 3nc - 9nd) + (2ma - 6mb - 3nc + 9nd)}}{{(2ma - 6mb + 3nc - 9nd) - (2ma - 6mb - 3nc + 9nd)}}$
On opening the bracket, we get
⇒$\dfrac{{2ma + 6mb + 3nc + 9nd + 2ma + 6mb - 3nc - 9nd}}{{2ma + 6mb + 3nc + 9nd - 2ma - 6mb + 3nc + 9nd}} = \dfrac{{2ma - 6mb + 3nc - 9nd + 2ma - 6mb - 3nc + 9nd}}{{2ma - 6mb + 3nc - 9nd - 2ma + 6mb + 3nc - 9nd}}$
Now on further solving, we get
⇒\[\dfrac{{2ma + 6mb + 2ma + 6mb}}{{3nc + 9nd + 3nc + 9nd}} = \dfrac{{2ma - 6mb + 2ma - 6mb}}{{3nc - 9nd + 3nc - 9nd}}\]
On adding like terms, we get
⇒\[\dfrac{{4ma + 12mb}}{{6nc + 18nd}} = \dfrac{{4ma - 12mb}}{{6nc - 18nd}}\]
On taking ‘4m’ common from the numerator and ‘6n’ common from the denominator, we get
⇒\[\dfrac{{4m(a + 3b)}}{{6n(c + 3d)}} = \dfrac{{4m(a - 3b)}}{{6n(c - 3d)}}\]
Hence, cancelling out common products from both sides, we get
⇒\[\dfrac{{a + 3b}}{{c + 3d}} = \dfrac{{a - 3b}}{{c - 3d}}\]
On rearranging the terms, we get
⇒\[\dfrac{{a + 3b}}{{a - 3b}} = \dfrac{{c + 3d}}{{c - 3d}}\]
Since we can no more simplify this, we will again apply componendo dividendo
⇒\[\dfrac{{(a + 3b) + (a - 3b)}}{{(a + 3b) - (a - 3b)}} = \dfrac{{(c + 3d) + (c - 3d)}}{{(c + 3d) - (c - 3d)}}\]
On opening the bracket, we get
⇒\[\dfrac{{a + 3b + a - 3b}}{{a + 3b - a + 3b}} = \dfrac{{c + 3d + c - 3d}}{{c + 3d - c + 3d}}\]
Now on further solving, we get
⇒\[\dfrac{{a + a}}{{3b + 3b}} = \dfrac{{c + c}}{{3d + 3d}}\]
On adding like terms, we get
⇒\[\dfrac{{2a}}{{6b}} = \dfrac{{2c}}{{6d}}\]
Hence, cancelling out common products from both sides, we get
⇒\[\dfrac{a}{b} = \dfrac{c}{d}\]
Hence, we can say that
⇒\[a:b = c:d\]
Therefore, a, b, c and d are proportional
Hence, proved

Note:
When using the componendo dividendo method, keep in mind that the fraction is in the form of $\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}$
So that the terms get cancelled and we get near to our answer, if ‘componendo and dividendo’ is not carefully used, it can result in a very complex solution which would consume a lot of time and effort.