Prove that 3times the sum of …

To prove :
{tex}3 \left( A B ^ { 2 } + B C ^ { 2 } + C A ^ { 2 } \right) = 4 \left( A D ^ { 2 } + B E ^ { 2 } + C F ^ { 2 } \right){/tex}

In triangle sum of squares of any two sides is equal to twice the square of half of the third side, together with twice the square of median bisecting it.
If AD is the median 
{tex}A B ^ { 2 } + A C ^ { 2 } = 2 \left\{ A D ^ { 2 } + \frac { B C ^ { 2 } } { 4 } \right\}{/tex}
or, {tex}2 \left( A B ^ { 2 } + A C ^ { 2 } \right) = 4 A D ^ { 2 } + B C ^ { 2 }{/tex} ...(i)
Similarly by taking BE & CF as medians, 
{tex}2 \left( A B ^ { 2 } + B C ^ { 2 } \right) = 4 B E ^ { 2 } + A C ^ { 2 }{/tex} ...(ii)
& {tex}2 \left( A C ^ { 2 } + B C ^ { 2 } \right) = 4 C F ^ { 2 } + A B ^ { 2 }{/tex} ...(iii)
By adding, (i), (ii) and (iii), we get 
{tex}3 \left( A B ^ { 2 } + B C ^ { 2 } + A C ^ { 2 } \right) = 4 \left( A D ^ { 2 } + B E ^ { 2 } + C F ^ { 2 } \right){/tex}
Hence proved.