In a triangle A B C , the median from B to C A is perpendicular to...

Let G be centroid, $\mathrm{D}, \mathrm{E}, \mathrm{F}$ be mid-points of $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ \& M be mid-point of FE . Let $\mathrm{BE}=3 \mathrm{x}, \mathrm{CF}=3 \mathrm{y}$, given $\mathrm{AD}=30$ Hence $\mathrm{AM}=5, \mathrm{GM}=5, \mathrm{GD}=10, \mathrm{BG}=2 \mathrm{x},$\mathrm{GE}=\mathrm{x}$, \mathrm{CG}=2 \mathrm{y}, \mathrm{GF}=\mathrm{y}$ Now $D$ is mid-point of hypotenuse of right angle triangle $B G C$. So $D$ is circum centre of the triangle. $\mathrm{SO}$\mathrm{BD}=\mathrm{GD}$=10 \Rightarrow \mathrm{BC}=20$ Hence $4\left(x^2+y^2\right)=400 \Rightarrow x^2+y^2=100$ Now $9\left(x^2+y^2\right)=\frac{1}{4}\left\{2 B C^2+2 A B^2-A C^2+2 B C^2+2 A C^2-A B^2\right\}$ $$\begin{aligned} & \Rightarrow 900 \times 4=4 B C^2+A B^2+A C^2 \\ & \Rightarrow A B^2+A C^2=3600-1600=2000 \end{aligned}$$ Hence $\frac{A B^2+B C^2+C A^2}{100}=\frac{2400}{100}=24$