Compare the sum of the squares of the median of a triangle to the sum of the squares of sides You have to compare the sum of the squares of the median of a triangle to the sum of the squares of sides?
 A: 
$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$

AB , BC , CD are lengths of sides of triangle and AD , BE , CF are lengths of medians of triangle
HINT : 
start with appollonius theorem and add all 
Apollonius Theorem
so you will get relation as 
3 (sum of square of sides) = 4 ( sum of squares of medians )
Proof : 

$$ AB^2 + AC^2 = 2 ( AD^2 + \frac{BC^2}{4} )$$
$$ AB^2 + BC^2 = 2 ( BE^2 + \frac{AC^2}{4} )$$
$$ AC^2 + BC^2 = 2 ( CF^2 + \frac{AB^2}{4} )$$
Add all above and bring sides to LHS and medians to RHS to get
$$ 3( AB^2 + BC^2 + AC^2 ) = 4 ( AD^2 + BE^2 + CF^2)$$
A: Let $ABC$ be a triangle and let $I,J,K$ be the midpoints of $[BC],[AC],[AB]$ respectively.
$\overrightarrow{AI}^2=\frac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)^2=\frac{1}{4}\left(AB^2+AC^2+2\overrightarrow{AB}\cdot\overrightarrow{AC}\right)$
The same for $BJ^2$ and $CK^2$. Adding the three equations, we obtain
$$AI^2+BJ^2+CK^2=\frac{1}{4}\left(2AB^2+2AC^2+2BC^2+(\overrightarrow{AB}\cdot\overrightarrow{AC}+\overrightarrow{BA}\cdot\overrightarrow{BC})+(\overrightarrow{AB}\cdot\overrightarrow{AC}+\overrightarrow{CA}\cdot\overrightarrow{CB})+(\overrightarrow{BA}\cdot\overrightarrow{BC}+\overrightarrow{CA}\cdot\overrightarrow{CB})\right)=\frac{3}{4}\left(AB^2+AC^2+BC^2\right)$$
A: The solution below is probably basically the same as the one by metacompactness. Let our triangle be $ABC$, with the usual conventions about the naming of sidelengths.
Draw the median from vertex $A$ to side $BC$, meeting $BC$ at $M$. By the Cosine Law, we have
$$c^2=m^2+\frac{a^2}{4}-2m\cdot\frac{a}{2}\cos(\angle AMB).$$
In the same way, we get
$$b^2=m^2+\frac{a^2}{4}-2m\cdot\frac{a}{2}\cos(\angle AMC).$$
Add. The two angles are supplementary, so their cosines have sum $0$. We obtain
$$m^2=\frac{2b^2+2c^2-a^2}{4}.$$
We can write down analogous expressions for the squares of the other medians. Add  up. The sum of the squares of the medians is $\frac{3}{4}$  times the sum of the squares of the sides.
A: In vector terms consider three vectors $0,a,b$. Let $c=a-b.$ The median-vectors are $\{m_1,m_2,m_3\}=\{b-a/2,a-b/2,(a+b)/2\}$. We have $$\sum_{i=1}^{i=3}\|m_i\|^2=$$ $$=\frac {3}{2}(\|a\|^2+\|b|^2-a\cdot b)=$$ $$=\frac {3}{4}(2\|a\|^2+2\|b\|^2)-2 a\cdot b)=$$ $$=\frac {3}{4}(\|a\|^2+\|b\|^2+\|a\|^2+\|b\|^2-2a\cdot b)=$$ $$=\frac {3}{4}(\|a\|^2+\|b\|^2+\|c\|^2).$$
