If in a triangle $ABC,c\left( a+b \right)\cos \dfrac{B}{2}=b\left( c+a \right)\cos \dfrac{C}{2}$, then the triangle is,
A. Isosceles
B. Right-angled
C. Isosceles or right-angled
D. None of the above

Hint: To solve this question, we should know a few half angle triangle formulae of any triangle, that is, $\cos \dfrac{B}{2}=\sqrt{\dfrac{s\left( s-b \right)}{ac}}s$ and $\cos \dfrac{C}{2}=\sqrt{\dfrac{s\left( s-c \right)}{ab}}$, where $s=\dfrac{a+b+c}{2}$ and $a,b,c$ are the lengths of the sides of the triangle $ABC$.

Complete step-by-step answer:

In this question, we have been asked to find the type of triangle that satisfies $c\left( a+b \right)\cos \dfrac{B}{2}=b\left( c+a \right)\cos \dfrac{C}{2}$. Now, we know that for any triangle $ABC$, having sides $a,b,c$, follow the relation of,
$\cos \dfrac{B}{2}=\sqrt{\dfrac{s\left( s-b \right)}{ac}}\ldots \ldots \ldots \left( i \right)$ and,
$\cos \dfrac{C}{2}=\sqrt{\dfrac{s\left( s-c \right)}{ab}}\ldots \ldots \ldots \left( ii \right)$.
We have been given, $c\left( a+b \right)\cos \dfrac{B}{2}=b\left( c+a \right)\cos \dfrac{C}{2}$, which can also be written as,
$\dfrac{\cos \dfrac{B}{2}}{\cos \dfrac{C}{2}}=\dfrac{b\left( c+a \right)}{c\left( a+b \right)}\ldots \ldots \ldots \left( iii \right)$
Now, we will put the values of $\cos \dfrac{B}{2}$ and $\cos \dfrac{C}{2}$ from equations (i) and (ii) in equation (iii). So, by substituting the values in equation (iii), we get,
$\begin{align}
  & \dfrac{\sqrt{\dfrac{s\left( s-b \right)}{ac}}}{\dfrac{s\left( s-c \right)}{ab}}=\dfrac{b\left( c+a \right)}{c\left( a+b \right)} \\
 & \Rightarrow \sqrt{\dfrac{\dfrac{s\left( s-b \right)}{ac}}{\dfrac{s\left( s-c \right)}{ab}}}=\dfrac{b\left( c+a \right)}{c\left( a+b \right)} \\
 & \Rightarrow \sqrt{\dfrac{s\left( s-b \right)ab}{s\left( s-c \right)ac}}=\dfrac{b\left( c+a \right)}{c\left( a+b \right)} \\
 & \Rightarrow \sqrt{\dfrac{\left( s-b \right)b}{\left( s-c \right)c}}=\dfrac{b\left( c+a \right)}{c\left( a+b \right)} \\
\end{align}$
Now, we square both the sides of the equation. By doing so, we get,
$\begin{align}
  & \dfrac{\left( s-b \right)b}{\left( s-c \right)c}={{\left[ \dfrac{b\left( c+a \right)}{c\left( a+b \right)} \right]}^{2}} \\
 & \Rightarrow \dfrac{\left( s-b \right)b}{\left( s-c \right)c}=\dfrac{{{b}^{2}}{{\left( c+a \right)}^{2}}}{{{c}^{2}}{{\left( a+b \right)}^{2}}} \\
\end{align}$
Now, we can see that $\dfrac{b}{c}$ is common in both the sides of the equation, so by cancelling them we get,
$\dfrac{s-b}{s-c}=\dfrac{b{{\left( c+a \right)}^{2}}}{c{{\left( a+b \right)}^{2}}}\ldots \ldots \ldots \left( iv \right)$
Now, we know that, $s=\dfrac{a+b+c}{2}$. So, we can write $s-b=\dfrac{a+b+c}{2}-b=\dfrac{a+c-b}{2}$. Also, we can write $s-c=\dfrac{a+b+c}{2}-c=\dfrac{a+b-c}{2}$. By substituting these values in equation (iv), we get,
$\begin{align}
  & \dfrac{\dfrac{a+c-b}{2}}{\dfrac{a+b-c}{2}}=\dfrac{b{{\left( c+a \right)}^{2}}}{c{{\left( a+b \right)}^{2}}} \\
 & \Rightarrow \dfrac{a+c-b}{a+b-c}=\dfrac{b{{\left( c+a \right)}^{2}}}{c{{\left( a+b \right)}^{2}}} \\
 & \Rightarrow \dfrac{\left( a+c \right)-b}{\left( a+b \right)-c}=\dfrac{b{{\left( c+a \right)}^{2}}}{c{{\left( a+b \right)}^{2}}} \\
\end{align}$
By cross multiplying the above equation, we get,
$c\left( a+c \right){{\left( a+b \right)}^{2}}-bc{{\left( a+b \right)}^{2}}=b{{\left( a+c \right)}^{2}}\left( a+b \right)-bc{{\left( a+c \right)}^{2}}$
Simplifying it further, we get,
$c\left( a+c \right){{\left( a+b \right)}^{2}}-b{{\left( a+c \right)}^{2}}\left( a+b \right)=bc{{\left( a+b \right)}^{2}}-bc{{\left( a+c \right)}^{2}}$
Now, we will take $\left( a+b \right)\left( a+c \right)$ common from LHS and $bc$ common form RHS. So, we get,
$\begin{align}
  & \left( a+b \right)\left( a+c \right)\left[ c\left( a+b \right)-b\left( a+c \right) \right]=bc\left[ {{\left( a+b \right)}^{2}}-{{\left( a+c \right)}^{2}} \right] \\
 & \Rightarrow \left( a+b \right)\left( a+c \right)\left[ ac+bc-ab-bc \right]=bc\left[ {{a}^{2}}+{{b}^{2}}+2ab-{{a}^{2}}-{{c}^{2}}-2ac \right] \\
\end{align}$
By cancelling the similar terms, we get,
$\left( a+b \right)\left( a+c \right)\left[ ac-ab \right]=bc\left[ {{b}^{2}}-{{c}^{2}}+2a\left( b-c \right) \right]$
We know that ${{b}^{2}}-{{c}^{2}}=\left( b+c \right)\left( b-c \right)$. So, by applying that in the above equation, we get,
$\begin{align}
  & \left( a+b \right)\left( a+c \right)a\left( c-b \right)=bc\left[ \left( b+c \right)\left( b-c \right)+2a\left( b-c \right) \right] \\
 & \Rightarrow a\left( a+b \right)\left( a+c \right)\left( c-b \right)=bc\left( b-c \right)\left[ b+c+2a \right] \\
\end{align}$
Let us consider, $b=c=\lambda $, and so, by substituting it in the LHS of the above equation, we get LHS as,
$\begin{align}
  & =a\left( a+\lambda \right)\left( a+\lambda \right)\left( \lambda -\lambda \right) \\
 & =0 \\
\end{align}$
Similarly, we will get the RHS as,
$\begin{align}
  & {{\lambda }^{2}}\left( \lambda -\lambda \right)\left( \lambda +\lambda +2a \right) \\
 & =0 \\
\end{align}$
Hence, we observe that LHS = RHS for $b=c$.
Therefore triangle $ABC$ is an isosceles triangle, so the correct answer is option A.

Note: One might think that the triangle can also be a right angled triangle, but for a right angled triangle to satisfy the given condition, angle B should be equal to angle C, where angle A is a right angle. So, this would become a special of the isosceles triangle. So, a right angled triangle would be the wrong answer.