If $a,b,c,d$ are in proportion then prove that $\left( m{{a}^{2}}+n{{b}^{2}} \right):\left( m{{c}^{2}}+n{{d}^{2}} \right)=\left( m{{a}^{2}}-n{{b}^{2}} \right):\left( m{{c}^{2}}-n{{d}^{2}} \right)$\[\].

Hint: We take $\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk$ . We put the value of $a,c$ in the left hand side of the statement of proof , $\left( m{{a}^{2}}+n{{b}^{2}} \right):\left( m{{c}^{2}}+n{{d}^{2}} \right)$ to obtain a simplified value. We do the same in right hand side $\left( m{{a}^{2}}-n{{b}^{2}} \right):\left( m{{c}^{2}}-n{{d}^{2}} \right)$ too. We compare both the obtained values to prove the statement.

Complete step by step answer:
We know that a proportion is an equality relation among more than one ratio. It means when we say that four numbers $a,b,c,d$are in proportion which we symbolize as $a:b::c:d$ the relation is given by
\[\dfrac{a}{b}=\dfrac{c}{d}\]
We call $b,c$ means of mean terms and $c,d$ are called extreme or extreme terms. The property of componendo is
\[\dfrac{a+b}{b}=\dfrac{c+d}{d}\]
The property of componendo and dividendo is
\[\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\]
We are given that $a,b,c,d$are in proportion. So let us assume
\[\begin{align}
  & \dfrac{a}{b}=\dfrac{c}{d}=k \\
 & \Rightarrow a=bk,c=dk \\
\end{align}\]
Where $k$ is a real number. We are given to prove $\left( m{{a}^{2}}+n{{b}^{2}} \right):\left( m{{c}^{2}}+n{{d}^{2}} \right)=\left( m{{a}^{2}}-n{{b}^{2}} \right):\left( m{{c}^{2}}-n{{d}^{2}} \right)$ which we can write as
\[\dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}}\]
Let us put $a=bk,c=dk$ in the left hand side of the given equation
\[\dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{m{{\left( bk \right)}^{2}}+n{{b}^{2}}}{m{{\left( dk \right)}^{2}}+n{{d}^{2}}}=\dfrac{m{{b}^{2}}{{k}^{2}}+n{{b}^{2}}}{m{{d}^{2}}{{k}^{2}}+n{{d}^{2}}}\]
Let us take ${{b}^{2}}$ common in the numerator and ${{d}^{2}}$ common in the denominator and get
\[\dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{m{{b}^{2}}{{k}^{2}}+n{{b}^{2}}}{m{{d}^{2}}{{k}^{2}}+n{{d}^{2}}}=\dfrac{{{b}^{2}}\left( m{{k}^{2}}+n \right)}{{{d}^{2}}\left( m{{k}^{2}}+n \right)}\]
Let us divide both numerator and denominator by $m{{k}^{2}}+n$and get
\[\dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{{{b}^{2}}\left( m{{k}^{2}}+n \right)}{{{d}^{2}}\left( m{{k}^{2}}+n \right)}=\dfrac{{{b}^{2}}}{{{d}^{2}}}.....(1)\]
Now Let us put $a=bk,c=dk$ in the right hand side of the given equation
\[\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}}=\dfrac{m{{\left( bk \right)}^{2}}-n{{b}^{2}}}{m{{\left( dk \right)}^{2}}-n{{d}^{2}}}=\dfrac{m{{b}^{2}}{{k}^{2}}-n{{b}^{2}}}{m{{d}^{2}}{{k}^{2}}-n{{d}^{2}}}\]
Let us take ${{b}^{2}}$ common in the numerator and ${{d}^{2}}$ common in the denominator and get
\[\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}}=\dfrac{m{{b}^{2}}{{k}^{2}}-n{{b}^{2}}}{m{{d}^{2}}{{k}^{2}}-n{{d}^{2}}}=\dfrac{{{b}^{2}}\left( m{{k}^{2}}-n \right)}{{{d}^{2}}\left( m{{k}^{2}}-n \right)}\]
Let us divide both numerator and denominator by $m{{k}^{2}}-n$and get
\[\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}}=\dfrac{{{b}^{2}}\left( m{{k}^{2}}-n \right)}{{{d}^{2}}\left( m{{k}^{2}}-n \right)}=\dfrac{{{b}^{2}}}{{{d}^{2}}}.....(2)\]
 So we observe from equation (1) and (2) that we have values at left hand side and right hand side are equal. So we have
\[\dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}}\Rightarrow \left( m{{a}^{2}}+n{{b}^{2}} \right):\left( m{{c}^{2}}+n{{d}^{2}} \right)=\left( m{{a}^{2}}-n{{b}^{2}} \right):\left( m{{c}^{2}}-n{{d}^{2}} \right)\]
Hence it is proved.
The converse of the given statement may not be true. Let us assume $\left( m{{a}^{2}}+n{{b}^{2}} \right):\left( m{{c}^{2}}+n{{d}^{2}} \right)=\left( m{{a}^{2}}-n{{b}^{2}} \right):\left( m{{c}^{2}}-n{{d}^{2}} \right)$ and try to prove $a,b,c,d$ are in proportion. So we have
\[\dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}}\]
 We also know that from the law of alternendo and get
\[\begin{align}
  & \dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{c}^{2}}+n{{d}^{2}}}=\dfrac{m{{a}^{2}}-n{{b}^{2}}}{m{{c}^{2}}-n{{d}^{2}}} \\
 & \Rightarrow \dfrac{m{{a}^{2}}+n{{b}^{2}}}{m{{a}^{2}}-n{{b}^{2}}}=\dfrac{m{{c}^{2}}+n{{d}^{2}}}{m{{c}^{2}}-n{{d}^{2}}} \\
\end{align}\]
We use componendo and dividendo and get ,
\[\begin{align}
  & \Rightarrow \dfrac{m{{a}^{2}}+n{{b}^{2}}+\left( m{{a}^{2}}-n{{b}^{2}} \right)}{m{{a}^{2}}+n{{b}^{2}}-\left( m{{a}^{2}}-n{{b}^{2}} \right)}=\dfrac{m{{c}^{2}}+n{{d}^{2}}+\left( m{{c}^{2}}-n{{d}^{2}} \right)}{m{{c}^{2}}+n{{d}^{2}}-\left( m{{c}^{2}}-n{{d}^{2}} \right)} \\
 & \Rightarrow \dfrac{m{{a}^{2}}}{n{{b}^{2}}}=\dfrac{m{{c}^{2}}}{n{{d}^{2}}} \\
 & \Rightarrow \dfrac{{{a}^{2}}}{{{b}^{2}}}=\dfrac{{{c}^{2}}}{{{d}^{2}}} \\
\end{align}\]
It is not necessary that $\dfrac{{{a}^{2}}}{{{b}^{2}}}=\dfrac{a}{b}$. So we cannot assert $a:b::c:d$
\[\]

Note:
We could divide $m{{k}^{2}}+n$ or $m{{k}^{2}}-n$ as they cannot be zero simultaneously as they are present in ratios in the proportion. The difference between ratio and proportion is that ratio is a fraction between two relative primes and proportion is a relation of equality between ratios.