If point $D$ divides the base $B^{\prime}$; of a $\triangle A B C$ in the ratio $n: m$, then the value of $m B D^2+n C D^2+(m+n) A D^2$ is

• A

  $m A C^2+n A B^2$
• B

  $(m+n)\left(A C^2+A B^2\right)$
• Correct answer
  C

  $n A C^2+m A B^2$
• D

  None of the above

Solution

$m B D^2+n C D^2+(m+n) A D^2$ $=m B D^2+n C D^2+m A D^2+n A D^2$
$\begin{aligned} & =m\left(B D^2+A D^2\right)+n\left(C D^2+A D^2\right) \\ & =m\left(A B^2+2 D M \cdot D B\right)+n\left(A C^2-2 D M \cdot D C\right) \\ & =m A B^2+m \cdot 2 D M \cdot D B+n A C^2-n \cdot 2 D M \cdot D C \\ & =m A B^2+n A C^2+2 D M(m D B-n D C) \\ & =m A B^2+n A C^2+2 D M(m n-n m) \\ & =m A B^2+n A C^2\end{aligned}$

Hence, the correct answer is option: 'C'.