If we are given the equation $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}$ and $l+m+n\ne 0$, prove that each ratio $=\dfrac{a+b}{2}$.

Hint: Assume the given expression of ratios $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}$ equal to k. Now, equate each expression equal to k one by one and form three equations. Add all the three equations and in the L.H.S take the terms together in which ‘a’ is common, similarly take the terms together in which ‘b’ is common. Now, cancel the expression (l + m + n) from both the sides of the inequality and find the value of k to get the desired proof.

Complete step-by-step solution:
Here we have been provided with the equality of ratios $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}$ with the condition that the value $l+m+n\ne 0$. We are asked to prove that each ratio is equal to $\dfrac{a+b}{2}$ or mathematically $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}=\dfrac{a+b}{2}$.
Now, let us assume the ratios $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}$ is equal to k, so we have,
$\Rightarrow \dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}=k$
(1) Equating $\dfrac{al+bm}{l+m}=k$ and cross multiplying the terms we get,
$\begin{align}
  & \Rightarrow al+bm=k\left( l+m \right) \\
 & \Rightarrow al+bm=kl+km............\left( i \right) \\
\end{align}$
(2) Equating $\dfrac{bl+an}{l+n}=k$ and cross multiplying the terms we get,
$\begin{align}
  & \Rightarrow bl+an=k\left( l+n \right) \\
 & \Rightarrow bl+an=kl+kn............\left( ii \right) \\
\end{align}$
(3) Equating $\dfrac{am+bn}{m+n}=k$ and cross multiplying the terms we get,
$\begin{align}
  & \Rightarrow am+bn=k\left( m+n \right) \\
 & \Rightarrow am+bn=km+kn............\left( iii \right) \\
\end{align}$
Taking the sum of equations (i), (ii) and (iii) we get,
$\begin{align}
  & \Rightarrow \left( al+bm \right)+\left( bl+an \right)+\left( am+bn \right)=\left( kl+km \right)+\left( kl+kn \right)+\left( km+kn \right) \\
 & \Rightarrow \left( al+bm \right)+\left( bl+an \right)+\left( am+bn \right)=2kl+2kn+2km \\
 & \Rightarrow \left( al+bm \right)+\left( bl+an \right)+\left( am+bn \right)=2k\left( l+m+n \right) \\
\end{align}$
Taking the terms containing ‘a’ together and the terms containing ‘b’ together in the L.H.S we get,
$\begin{align}
  & \Rightarrow a\left( l+m+n \right)+b\left( l+m+n \right)=2k\left( l+m+n \right) \\
 & \Rightarrow \left( l+m+n \right)\left( a+b \right)=2k\left( l+m+n \right) \\
\end{align}$
Now, we can cancel the expression (l + m + n) from both the sides of the equation because we have been provided with the condition $l+m+n\ne 0$ in the question, so we get,
$\begin{align}
  & \Rightarrow \left( a+b \right)=2k \\
 & \Rightarrow k=\dfrac{a+b}{2} \\
\end{align}$
So, substituting the value of k is the equation we assumed earlier $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}=k$ in the above solution, we get,
$\therefore \dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}=\dfrac{am+bn}{m+n}=\dfrac{a+b}{2}$
Hence, the above expression proves that each ratio is equal to $\dfrac{a+b}{2}$.

Note: There can be different approach also to solve the above question. What you need to do is equate the first two ratios $\dfrac{al+bm}{l+m}=\dfrac{bl+an}{l+n}$ and cross multiply the terms. Cancel the common terms from both the sides and factorize the remaining terms. You will get two equalities, the first will be $a=b$ and the second one will be ${{l}^{2}}=mn$. From here you have to consider the equality $a=b$ and substitute in all the ratios. While doing so you will get each ratio equal to ‘a’. Now, when you will substitute $a=b$ in the expression $\dfrac{a+b}{2}$ you will again get ‘a’. Hence, we will get the required proof.